Differential Equations and Engineering Applications

Transcription

1 Differential Equations and Engineering Applications My Students Fall 200 It is mostly based on the textbook, Peter V O Neil, Advanced Engineering Mathematics, 5th Edition, and has been reorganized and retyped by Jae Lee

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3 CONTENTS First Order Differential Equations 5 Preliminary Concepts 5 General and Particular Solutions 5 2 Implicitly Defined Solutions 6 3 Integral Curves 6 4 Initial Value Problem 7 5 Direction Fields 7 2 Separable Equations 8 2 Extended Method: Reduction to Separable Form 0 22 Some Applications of Separable Differential Equations 3 Linear Differential Equations 2 4 Exact Differential Equations 4 4 Appendix: Formula 20 5 Integrating Factors 22 5 Separable Equations and Integrating Factors Linear Equations and Integrating Factors Appendix: Formula 28 6 Homogeneous, Bernoulli, and Riccati Equations 29 6 Homogeneous Differential Equations Bernoulli Differential Equation 3 63 Appendix: Formula 33 7 Applications to Mechanics, Electrical Circuits, and Orthogonal Trajectories 33 8 Existence and Uniqueness for Solutions of Initial Value Problems 33 2 Second Order Differential Equations 35 2 Preliminary Concepts Theory of Solutions of y + p(x)y + q(x)y = f (x) Reduction of Order Constant Coefficient Homogeneous Linear Equation Case A 2 4B > Case 2 A 2 4B = Case 3 A 2 4B < Alternative General Solution in Case of Complex Roots Appendix Euler s Equation 5 25 Appendix Nonhomogeneous Equation y + p(x)y + q(x)y = f (x) Method of Variation of Parameters Method of Undetermined Coefficients 62 3

4 263 Principle of Superposition Higher Order Differential Equations Application of Second Order Differential Equations to a Mechanical System 69 3 Laplace Transform 7 3 Definition and Basic Properties 7 3 Appendix (Table of Laplace Transform) Solution of Initial Value Problem Using the Laplace Transform Shifting Theorems and the Heaviside Function The First Shifting Theorem The Heaviside Function and Pulses The Second Shifting Theorem Analysis of Electrical Circuits Convolution Unit Impulses and the Dirac Delta Function Laplace Transform Solution of Systems Differential Equations with Polynomial Coefficients 93 Page 4 of 93

5 Chapter First Order Differential Equations A differential equation is an equation that contains one or more derivatives For example, y (x) + x 3 y (x) + y 5 (x) = 0cos(4x), d 4 w dt 4 (w(t))3 = e 5t If the differential equation is involved with only total derivatives, then it is called an ordinary differential equation If it contains the partial derivatives, then we call it a partial differential equation In this chapter, we study only the ordinary differential equation The order of a differential equation is the order of its highest derivative For instance, the differential equation xy y 3 = e x has the first order derivative, which is the highest derivative, and so it is the first order differential equation The differential equation y 4 y xy + y 3 = e x has the second order derivative and so it is the second order differential equation In this chapter, we study only the first order differential equation and in this course, we lay our concern on only the first order and second order differential equations The solution of a differential equation is any function that satisfies it For example, y = cos(3x) is a solution of the second order differential equation y + 9y = 0 PROOF Differentiating y = cos(3x) twice, we get y = 3sin(3x), y = 9cos(3x) Putting it into the given equation, we observe y + 9y = 9cos(3x) + 9cos(3x) = 0, ie, y = cos(3x) satisfies the given differential equation y + 9y = 0, hence it is a solution of the given differential equation A solution may be defined on the entire real line or on only part of the real line, often an interval For example, y = xlnx x is a solution of y = y/x + Obviously the solution y = xlnx x is defined for x > 0, because lnx has the domain x > 0 Preliminary Concepts General and Particular Solutions Recall that a first order differential equation is an equation involving a first derivative The general form of a first order differential equation is given by where y is the function of x and y is the derivative of y with respect to x Example The followings are first order differential equations F(x,y,y ) = 0, () y y 2 e y = 0, y 2 = 0, y cosx = 0, x 5 y + y 00 ln(xy 4 ) = 0 A solution of equation () on an interval I is a function φ(x) satisfying the equation for all x in I That is, φ(x) is a solution of equation () on the interval I if and only if F(x,φ(x),φ (x)) = 0 for all x in I Example 2 The function φ(x) = 2 + ke x is a solution of the differential equation y + y = 2 for all real x and for any number k 5

6 PROOF Differentiating φ(x) and putting it into the given equation, we observe φ (x) = ke x, φ (x) + φ(x) = ke x ke x = 2, ie, φ (x) + φ(x) = 2, That is, the function φ(x) = 2 + ke x satisfies the given equation, hence it is a solution of the equation In this example, as we can see, the solution contains an arbitrary constant k Such a solution is called a general solution of the differential equation That is, φ(x) = 2 + ke x is the general solution of y + y = 2 Each choice of the constant in the general solution gives a particular solution For example, φ(x) = 2 + e x and φ(x) = 2 0e x also satisfy the equation y + y = 2 and so they are particular solutions of the equation As one can see, those particular solutions can be obtained by putting k = and k = 0 into the general solution One of the main goals in this course is to find the general solution of a differential equation We will develop the techniques/methods to find the general solutions of the first order and second order differential equations in the later chapters 2 Implicitly Defined Solutions Example 3 The differential equation y = y has the general solution y = ke x Example 4 The differential equation y = 2xy x 2 y 2 + 8e 4y has the general solution y implicitly defined by the equation where k is an arbitrary constant x 2 y 3 + 2x + 2e 4y = k, In the Example 3, the solution is explicitly obtained: y = ke x However, in the Example 4, the solution is implicitly obtained by the equation x 2 y 3 + 2x + 2e 4y = k How do we verify that y implicitly defined by x 2 y 3 + 2x + 2e 4y = k is the solution of the differential equation y = 2xy x 2 y 2 + 8e 4y? We differentiate the whole equation x 2 y 3 + 2x + 2e 4y = k implicitly (CALCULUS I FOR ENGINEERS) and deduce the given differential equation 3 Integral Curves A graph of a solution of a differential equation is called an integral curve of the equation So if we know the general solution, then we obtain infinitely many integral curves, because of the arbitrary constant in the general solution Example 5 The general solution of y + y = 2 is y = 2 + ke x for all x The figure shows the integral curves with k = 6, 3,0,3,6, ie, the graphs of y = 2 6e x, y = 2 3e x, y = 2, y = 2 + 3e x and y = 2 + 6e x Exercise 6 () Show that the differential equation y + y/x = e x has the general solution (2) Show that y + xy = 2 has the general solution y = x (xex e x + e), x 0 y = e x2 /2 ˆ x 0 2e w2 /2 dw + ke x2 /2 Use the Matlab or Mathematica, sketch the integral curves for various values of k Page 6 of 93

7 y x Figure : Integral curves of y + y = 2 for k = 6, 3,0,3,6 4 Initial Value Problem Since a differential equation can have infinitely many integral curves, if we fix a point (x 0,y 0 ) on the plane, then we may find only one integral curve passing through the fixed point (x 0,y 0 ), which gives only one solution From this observation, the problem solving a first order differential equation F(x,y,y ) = 0, y(x 0 ) = y 0, where x 0 and y 0 are given numbers, is called an initial value problem and the condition y(x 0 ) = y 0 is called an initial condition We need to keep in mind that the initial value problem has unique solution Example 7 Solve the initial value problem y = e x and y(0) = 2 ANSWER It is straightforward to see that the general solution of the equation y = e x is y = e x + k, where k is any constant Because of the initial condition, we have 2 = y(0) = e 0 + k = + k 2 = + k k = 3 Hence, the solution of the initial value problem is y = e x Direction Fields Skip Please read the textbook the single one of its kind Page 7 of 93

8 2 Separable Equations Definition 2 A differential equation is called separable if it can be written y = A(x)B(y) How to solve a separable differential equation y = A(x)B(y)? Theorem 22 (STRATEGY) We separate the variables and integrate as follows: ˆ ˆ dy dx = y = A(x)B(y), dy = A(x)B(y)dx, dy = A(x)dx, B(y) B(y) dy = A(x) dx After integrating it, we simplify and get the general solution of the differential equation Be careful! It may be or may not be possible to solve explicitly for y(x) Example 23 Solve the differential equation y = e x y 2 ANSWER Since the equation has the form of a separable equation, so it is a separable differential equation and we follow the strategy above: separate and integrate dy dx = y = e x y 2, dy = e x y 2 dx, ˆ ˆ y 2 dy = e x dx, y = e x + k, y 2 dy = e x dx (y 0) y = e x k, y = e x k, where k is an arbitrary constant Hence the general solution is y = e x k Remark 24 In the solution above, we assumed y 0 and found the general solution We observe For value of k, the general solution cannot produce y = 0 2 y = 0 is also a solution of the differential equation, because y(x) = 0 satisfies the equation y = y 2 e x for all x These two observations defines a singular solution of a differential equation, ie, the solution y(x) = 0 is called the singular solution of the equation, which cannot be deduced from the general solution Another example of the differential equation having a singular solution is Solve and check this out by yourself y = 6x(y ) 2/3 ANSWER The general solution is y = + (x 2 + C) 3 and the differential equation has a singular solution y(x) = Example 25 Solve the differential equation x 2 y = + y ANSWER It is easy to see y = + y x 2 = ( + y), x2 which is the form of a separable equation So it is a separable equation Using the strategy 22, dy dx = y = x 2 ( + y), dy = x 2 ( + y)dx, + y dy = dx, ( + y 0) x2 Page 8 of 93

9 ˆ ˆ + y dy = x 2 dx, ln + y = x +C, + y = De x, y = + De x, where C and D = e C are arbitrary constants Hence the general solution is Now we discuss the case when + y = 0, ie, y(x) = y = + De x (2) Since y(x) = does satisfy the given differential equation, it is a solution of the equation 2 If D = 0 in the general solution (2), then we have y = Because of the second observation, y = cannot be a singular solution It is just one particular solution Example 26 Solve the initial value problem y = e x y 2 with y() = 4 ANSWER From the solution in the Example 23, the general solution is y = e x k, where k is an arbitrary constant Using the given initial condition y() = 4, we determine the constant k: 4 = y() = e k, 4 = e ke, ke = e 4, k = e/4 = e e 4 Therefore, the solution of the initial value problem is y = Example 27 Solve the initial value problem e x + /4 /e ANSWER It is easy to see y = y (x )2, y(3) = y + 3 y (x )2 = y y + 3 = (x y )2 y + 3 which is the form of a separable equation So it is a separable equation Using the strategy 22, dy dx = y = (x ) 2 y y + 3, dy = (x y )2 y + 3 dx, y + 3 dy = (x ) 2 dx, (y 0) y ˆ ˆ ( y + 3 dy = + 3 ) ˆ dy = (x ) 2 (x )3 dx, y + 3ln y = + k, y y 3 where k is an arbitrary constant and the solution is implicitly defined By the given initial condition, y(3) + 3ln y(3) = (3 )3 3 + k, + 3ln = Hence, the solution of the initial value problem is y + 3ln y = (x )3 3 (3 )3 3 + k, = k, k = 3 3, (22) which is implicitly defined Since a logarithmic function is defined on positive real numbers, ie, ln y in the solution (22) is defined for y 0, thus the condition y 0 is satisfied Page 9 of 93

10 Example 28 Solve the differential equations y = + x + y + xy y = 6e 2x y with y(0) = 0 p(x)y + q(x)y = q(x), where p(x) 0 and q(x) 0 are given functions p(x)y + q(x)y 2 = q(x), where p(x) 0 and q(x) 0 are given functions e 3x y + xsin(2y) = 0 xy + y = (Final Exam of Fall 2009) 2 Extended Method: Reduction to Separable Form Certain nonseparable differential equations can be made separable by transformation that introduce for y a new unknown function, simply, a substitution We discuss this technique for equations ( y y = f, (23) x) where f is a differentiable function of y/x This form of such a differential equation suggests that we set u = y/x Then the substitution implies, by the Product Rule, Putting into the given equation (23), we get y = ux, and y = xu + u xu + u = f (u), ie, xu = f (u) u, which is a separable equation du dx = u = f (u) u = ( f (u) u), x x Example 29 Solve the differential equation: which is not in the form of a separable equation 2xyy = y 2 x 2, ANSWER When we divide the whole equation 2xy, the given equation becomes y = ( y 2 x x ) y (24) Because of the form, we try the substitution u = y/x Then we have y = ux and y = xu + u Putting them into the equation (24), we get xu + u = 2 ( u ) = u2 u 2u, xu = u2 + 2u, which is a separable equation Using the strategy 22 for a separable equation, we deduce x du ˆ ˆ dx = + u2 2u, 2u u 2 + du = x dx, 2u u 2 + du = x dx, ln u 2 + = ln x +C, ln u 2 + = ln x +C, u2 + = D x, where C and D 0 are arbitrary constants Finally, putting u = y/x back to the result, we conclude ( y x ) 2 + = D x, Page 0 of 93 y2 + x 2 = Dx,

11 which gives the implicitly defined solution A little bit more work (explicitly completing the square) shows that the equation y 2 + x 2 = Dx is equivalent to ( x D 2 ) 2 + y 2 = ( ) D 2, 2 of which graphs, ie, integral curves of the differential equation, are clearly circles centered at (D/2,0) with radius D /2 Example 20 Solve the differential equations xy = x y 2xyy = 3y 2 + x 2 with y() = 2 xyy = x 2 + 2y 2 (Final Exam of Fall 2009) 22 Some Applications of Separable Differential Equations Skip Please read the textbook Page of 93

12 3 Linear Differential Equations Definition 3 A first order differential equation is called to be linear if it has the form where p(x) and q(x) are assumed to be continuous y + p(x)y = q(x), How to solve a first order linear differential equation y + p(x)y = q(x)? Theorem 32 (STRATEGY: INTEGRATING FACTOR e p(x)dx ) Multiplying the differential equation by e p(x)dx, we get e p(x)dx y + e p(x)dx p(x)y = q(x)e p(x)dx Then the left hand side of the equation becomes the derivative of the product e p(x)dx y That is, the equation becomes ( e p(x)dx y) = q(x)e p(x)dx After integrating both sides, we obtain e p(x)dx y = ˆ ( ) q(x)e p(x)dx dx +C, where C is the constant of integration Therefore, we deduce the general solution ˆ ( ) p(x)dx y = e q(x)e p(x)dx dx +Ce p(x)dx (3) Remark 33 The function e p(x)dx is called the integrating factor of the differential equation It is not recommended memorizing the formula 3 However, it is suggested that you should remember the integrating factor e p(x)dx and the steps to deduce the general solution 2 When can a separable differential equation y = A(x)B(y) be linear? When can a linear differential equation y + p(x)y = q(x) be separable? If a differential equation is both separable and linear, then which strategy should we use to solve the differential equation? In the linear differential equation, y + p(x)y = q(x), if q(x) = 0, then the differential equation becomes separable The strategy 32 or the strategy 22 on a separable differential equation implies the general solution y = Ce p(x)dx Example 34 Solve the differential equation y + y = sinx ANSWER Since the equation has the form of a linear equation with p(x) = and q(x) = sin(x), so it is a linear differential equation and we follow the strategy 32, ie, integrating factor e p(x)dx = e dx = e x Multiplying the equation by the integrating factor e x, we have e x y + e x y = e x sinx, ie, (e x y) = e x sinx Integrating both sides implies ˆ e x y = e x sinxdx = 2 ex (sinx cosx) +C, by the Integration by Parts formula in CALCULUS I and C is the constant of integration Finally, we simplify and get the general solution y = 2 (sinx cosx) +Ce x Page 2 of 93

13 Example 35 Solve the initial value problem ANSWER We observe the equation is same as y = 3x 2 y, y() = 5 x y + x y = 3x2 (32) Since the equation has the form of a linear equation with p(x) = /x and q(x) = 3x 2, so it is a linear differential equation and we follow the strategy 32, ie, integrating factor e p(x)dx = e /xdx = e lnx = x, for x > 0 Multiplying the equation (32) by the integrating factor x, we have Integrating both sides implies ˆ xy = By the initial condition y() = 5, we have xy + y = 3x 3, ie, (xy) = 3x 3 3x 3 dx = 3 4 x4 +C, ie, y = 3 4 x3 + C x 5 = y() = C, C = 7 4 Finally, the general solution is, for x > 0, y = 3 4 x x In the solution of this example, we have chosen x > 0 in computing the integrating factor Why did we choose x > 0 rather than x < 0? The reason lies on the initial condition y() = 5 As you may recall from SECTION PRELIMINARY CONCEPTS, the initial condition restricts our concerns to certain integral curve passing through the point (x,y) = (,5) In other words, since we want to find the solution whose graph passes through the point (x,y) = (,5), we have to choose x > 0 Suppose the initial condition y( 4) = 0 is given Then the integral factor becomes x and the general solution is, for x < 0, y = 3 4 x3 232 x Can we always have the general solution in which all the integrals can be evaluated? The answer is NO For instance, the linear differential equation, y + xy = 2, has the integrating factor and the general solution is xdx e = e x2 2 ˆ y = 2e x2 2 e x2 2 dx +Ce x2 2 where e x2 2 dx cannot be evaluated explicitly in elementary terms Example 36 Solve the differential equations y xy = e 2x y + ytanx = sin(2x) with y(0) = x 2 y + xy = sinx with y() = y 0, where y 0 is a given real number xy + y = 3xy with y() = 0 y = ( y)cosx with y(π) = 2 y = + x + y + xy with y(0) = 0 xy 2y = x 2 (Final Exam of Fall 2009) 2xy 3y = 9x 3 with y() = 0 (Final Exam of Fall 2009) Page 3 of 93

14 4 Exact Differential Equations Compared to previous three sections, this section is difficult and long So please focus and try to understand all examples We will have the implicitly defined solution The chain rule for partial derivatives in CALCULUS II will be used We have seen that a general solution y(x) of a first order differential equation is often defined implicitly by an equation of the form φ(x,y(x)) = C, (4) where C is a constant On the other hand, given the equation (4), we can recover the original differential equation by differentiating each side with respect to x Provided that the equation (4) implicitly defines y as a differentiable function of x, this gives the original differential equation in the form where φ x + φ dy y dx = x C = 0, M(x,y) = φ(x,y) x M(x,y) + N(x,y)dy = 0, M(x,y)dx + N(x,y)dy = 0, (42) dx = φ x (x,y) and N(x,y) = φ(x,y) y = φ y (x,y) The last equation in (42) is called the differential form The general first order differential equation y = f (x,y) can be written in this form with M = f (x,y) and N = or M = f (x,y) and N = The preceding discussion shows that, if there exists a function φ(x,y) such that φ x = M and φ y = N, then the equation φ(x,y) = C implicitly defines a general solution of the equation (42) In this case, the equation (42) is called an exact differential equation With this background, we make the following definitions Definition 4 A function φ(x,y) is called a potential function for the differential equation M(x,y) + N(x,y)y = 0 or M(x,y)dx + N(x,y)dy = 0 on a region R of the plane if, for each (x,y) in R, φ x = M(x,y) and φ y = N(x,y) Definition 42 When a potential function exists on a region R for the differential equation this equation is said to be exact on R Remark 43 (QUESTIONS) M(x,y) + N(x,y)y = 0, How can we determine whether the differential equation (42) is exact? That is, how do we know the existence of a potential function? The answer is given in the theorem 44 below 2 If a differential equation is exact, ie, if the potential function exists, then how can we find it? That is, how can we find the function φ(x,y) such that The answer will be explained through the examples φ x = M(x,y) and φ y = N(x,y)? Page 4 of 93

15 Theorem 44 (CRITERION/TEST FOR EXACTNESS) Suppose that the functions M(x, y) and N(x, y) are continuous and have continuous first order partial derivatives on a region R of the plane Then the differential equation M(x,y) + N(x,y)y = 0 is exact in R if and only if M y = N x (43) at each point of R That is, there exists a potential function φ(x,y) defined on R with φ x = M and φ y = N if and only if the equation (43) holds on R Example 45 Solve the differential equation 6xy y 3 + (4y + 3x 2 3xy 2 )y = 0 which is neither separable nor linear or ( 6xy y 3 ) dx + ( 4y + 3x 2 3xy 2) dy = 0, ANSWER Step Test for Exactness Let M(x,y) = 6xy y 3 and N(x,y) = 4y + 3x 2 3xy 2 Then M y = 6x 3y2 = N x So by the Theorem 44, the given equation is exact Step 2 Solution Again by the Theorem 44, a potential function φ satisfies φ x = M and φ y = N, ie, φ x = 6xy y3, and φ y = 4y + 3x2 3xy 2 (44) To find φ, we integrate either one Let us integrate the first one φ x with respect to x: ˆ ( ) ˆ φ (6xy φ(x,y) = dx = y 3 ) dx = 3x 2 y xy 3 + g(y), x thinking of the function g(y) as an arbitrary constant of integration, as far as the variable x is concerned Inserting this result to the second equation in (44), we determine the function g(y): 4y + 3x 2 3xy 2 = φ y = ( 3x 2 y xy 3 + g(y) ) = 3x 2 3xy 2 + g (y), y ˆ 4y + 3x 2 3xy 2 = 3x 2 3xy 2 + g (y), g (y) = 4y, g(y) = 4ydy = 2y 2 +C Thus, finally we obtain the potential function φ(x,y): and the general solution of the differential equation φ(x,y) = 3x 2 y xy 3 + 2y 2 +C φ(x,y) = 3x 2 y xy 3 + 2y 2 +C = D, ie, 3x 2 y xy 3 + 2y 2 = E, where C, D and E = D C are arbitrary constants Step 3 Checking We check this implicitly defined solution by implicitly differentiating and see whether it leads to the given differential equation: 0 = d dx E = d ( 3x 2 y xy 3 + 2y 2) dx = 6xy + 3x 2 y y 3 3xy 2 y + 4yy = 6xy y 3 + ( 4y + 3x 2 3xy 2) y This is the given differential equation and so the implicitly defined function y in Step 2 above is really the general solution Page 5 of 93

16 In Step 2 of the solution above, we have integrated φ x with respect to x and 2 found φ having g(y) and 3 by differentiating φ with respect to y, we found g(y) and finally φ However, we can argue in the other way, ie, integrate φ y with respect to y and 2 find φ having h(x) and 3 by differentiating φ with respect to x, we can find h(x) and finally φ For the argument in detail, see the example below Example 46 Solve the differential equation cos(x + y) + (3y 2 + 2y + cos(x + y))y = 0 or cos(x + y)dx + ( 3y 2 + 2y + cos(x + y) ) dy = 0 ANSWER Step Test for Exactness Let M(x,y) = cos(x + y) and N(x,y) = 3y 2 + 2y + cos(x + y) M y = sin(x + y) = N x So by the Theorem 44, the given equation is exact Step 2 Solution Again by the Theorem 44, a potential function φ satisfies φ x = M and φ y = N, ie, φ x = cos(x + y), and φ y = 3y2 + 2y + cos(x + y) (45) To find φ, let us integrate the second one φ y with respect to y: ˆ ( ) ˆ φ (3y φ(x,y) = dy = 2 + 2y + cos(x + y) ) dy = y 3 + y 2 + sin(x + y) + h(x), y thinking of the function h(x) as an arbitrary constant of integration, as far as the variable y is concerned Inserting this result to the first equation in (45), we determine the function h(x): cos(x + y) = φ x = ( y 3 + y 2 + sin(x + y) + h(x) ) = cos(x + y) + h (x), x cos(x + y) = cos(x + y) + h (x), h (x) = 0, h(x) = C Thus, finally we obtain the potential function φ(x,y): and the general solution of the differential equation φ(x,y) = y 3 + y 2 + sin(x + y) +C φ(x,y) = y 3 + y 2 + sin(x + y) +C = D, ie, y 3 + y 2 + sin(x + y) = E, where C, D and E = D C are arbitrary constants Step 3 Checking We check this implicitly defined solution by implicitly differentiating and see whether it leads to the given differential equation: 0 = d dx E = d ( y 3 + y 2 + sin(x + y) ) dx = 3y 2 y + 2yy + cos(x + y) + cos(x + y)y = cos(x + y) + ( 3y 2 + 2y + cos(x + y) ) y This is the given differential equation and so the implicitly defined function y in Step 2 above is really the general solution Page 6 of 93

17 Example 47 Solve the differential equation 2xy (3x 2 y 2 + 8e 4y )y = 0 or ( 2xy ) dx + ( 3x 2 y 2 + 8e 4y) dy = 0 ANSWER Step Test for Exactness Let M(x,y) = 2xy and N(x,y) = 3x 2 y 2 + 8e 4y Then M y = 6xy2 = N x So by the Theorem 44, the given equation is exact Step 2 Solution Again by the Theorem 44, a potential function φ satisfies φ x = M and φ y = N, ie, φ x = 2xy3 + 2, and φ y = 3x2 y 2 + 8e 4y (46) To find φ, we integrate either one Let us integrate the first one φ x with respect to x: ˆ ( ) ˆ φ (2xy φ(x,y) = dx = ) dx = x 2 y 3 + 2x + g(y), x thinking of the function g(y) as an arbitrary constant of integration, as far as the variable x is concerned Inserting this result to the second equation in (46), we determine the function g(y): 3x 2 y 2 + 8e 4 y = φ y = ( x 2 y 3 + 2x + g(y) ) = 3x 2 y 2 + g (y), y ˆ 3x 2 y 2 + 8e 4 y = 3x 2 y 2 + g (y), g (y) = 8e 4y, g(y) = Thus, finally we obtain the potential function φ(x,y): and the general solution of the differential equation φ(x,y) = x 2 y 3 + 2x + 2e 4y +C 8e 4y dy = 2e 4y +C φ(x,y) = x 2 y 3 + 2x + 2e 4y +C = D, ie, x 2 y 3 + 2x + 2e 4y = E, where C, D and E = D C are arbitrary constants Step 3 Checking We check this implicitly defined solution by implicitly differentiating and see whether it leads to the given differential equation: 0 = d dx E = d ( x 2 y 3 + 2x + 2e 4y) dx = 2xy 3 + 3x 2 y 2 y e 4y y = 2xy ( 3x 2 y 2 + 8e 4y) y This is the given differential equation and so the implicitly defined function y in Step 2 above is really the general solution Example 48 Solve the differential equation x 2 + 3xy + (4xy + 2x)y = 0 or ( x 2 + 3xy ) dx + (4xy + 2x)dy = 0 ANSWER Step Test for Exactness Let M(x,y) = x 2 + 3xy and N(x,y) = 4xy + 2x Then M y = 3x, N x = 4y + 2 We have the equality M y = N x along the straight line 3x = 4y + 2, but the equality does not hold for every point in a region of the plane Hence the differential equation is not exact and we cannot solve the differential equation using the strategy above Page 7 of 93

18 Example 49 Solve the differential equation e x siny 2x + (e x cosy + )y = 0 or (e x siny 2x)dx + (e x cosy + )dy = 0, which is neither separable nor linear ANSWER Step Test for Exactness Let M(x,y) = e x siny 2x and N(x,y) = e x cosy + Then M y = ex cosy = N x So by the Theorem 44, the given equation is exact Step 2 Solution Again by the Theorem 44, a potential function φ satisfies φ x = M and φ y = N, ie, φ x = ex siny 2x, and φ y = ex cosy + (47) To find φ, we integrate either one Let us integrate the first one φ x with respect to x: ˆ ( ) ˆ φ φ(x,y) = dx = (e x siny 2x) dx = e x siny x 2 + g(y), x thinking of the function g(y) as an arbitrary constant of integration, as far as the variable x is concerned Inserting this result to the second equation in (47), we determine the function g(y): e x cosy + = φ y = ( e x siny x 2 + g(y) ) = e x cosy + g (y), y ˆ e x cosy + = e x cosy + g (y), g (y) =, g(y) = dy = y +C Thus, finally we obtain the potential function φ(x,y): and the general solution of the differential equation φ(x,y) = e x siny x 2 + y +C φ(x,y) = e x siny x 2 + y +C = D, ie, e x siny x 2 + y = E, where C, D and E = D C are arbitrary constants Step 3 Checking We check this implicitly defined solution by implicitly differentiating and see whether it leads to the given differential equation: 0 = d dx E = d ( e x siny x 2 + y ) dx = e x siny + e x (cosy)y 2x + y = e x siny 2x + (e x cosy + )y This is the given differential equation and so the implicitly defined function y in Step 2 above is really the general solution Before we end the section, let us bring the caution through an example Example 40 Solve the differential equation y 3 + 3xy 2 y = 0 or y 3 dx + 3xy 2 dy = 0 Page 8 of 93

19 ANSWER SOLUTION OF EXACT EQUATION Step Test for Exactness Let M(x,y) = y 3 and N(x,y) = 3xy 2 Then M y = 3y2 = N x So by the Theorem 44, the given equation is exact Step 2 Solution Again by the Theorem 44, a potential function φ satisfies φ x = M and φ y = N, ie, φ x = y3, and φ y = 3xy2 (48) To find φ, we integrate either one Let us integrate the first one φ x with respect to x: φ(x,y) = ˆ ( ) ˆ φ dx = x y 3 dx = xy 3 + g(y) Inserting this result to the second equation in (48), we determine the function g(y): 3xy 2 = φ y = ( xy 3 + g(y) ) = 3xy 2 + g (y), y Thus, finally we obtain the potential function φ(x,y): and the general solution of the differential equation 3xy 2 = 3xy 2 + g (y), g (y) = 0, g(y) = C φ(x,y) = xy 3 +C φ(x,y) = xy 3 +C = D, ie, xy 3 = E, (49) where C, D and E = D C are arbitrary constants Step 3 Checking We check this implicitly defined solution by implicitly differentiating and see whether it leads to the given differential equation: 0 = d dx E = d dx ( xy 3 ) = y 3 + 3xy 2 y This is the given differential equation and so the implicitly defined function y in Step 2 above is really the general solution Suppose that we divide each term of the differential equation in the example above by y 2 to obtain y + 3xy = 0 or ydx + 3xdy = 0 Then this equation is not exact, because, with M(x,y) = y and N(x,y) = 3x, we have M y = 3 = N x That is, it does not pass the TEST FOR EXACTNESS 43 So it is suggested that we should not modify the given differential equation by dividing by common factors However, in the example above, due to the common factors, we can solve the differential equation as follows Page 9 of 93

20 ANSWER 2 SOLUTION OF SEPARABLE EQUATION We observe y 3 + 3xy 2 y = 0, y = y3 3xy 2 = y, (x 0) 3x which means the given equation is separable So we use the strategy for the separable equation dy dx = 3x y, y dy = ˆ ˆ 3x dx, y dy = dx, (y 0) 3x ln y = 3 ln x +C, y = D x /3, xy 3 = E, where D = e C and E = D 3 are arbitrary constants and we assumed x > 0 and y > 0 As we can see, the solution is exactly same as the one (49) in ANSWER SOLUTION OF EXACT EQUATION Remark 4 (IMPORTANT CAUTION) If φ is a potential function for M +Ny = 0, then φ itself is not the solution The general solution is defined implicitly by the equation φ(x,y) = C Exercise 42 Solve the differential equations 2x + 3y + (3x + 2y)y = 0 ANSWER x 2 + 3xy + y 2 = C ( + ye xy )dx + (2y + xe xy )dy = 0 ANSWER x + e xy + y 2 = C 3x 2 + 2y 2 + (4xy + 6y 2 )y = 0 ANSWER x 3 + 2xy 2 + 2y 3 = C (cosx + lny)dx + ( xy + e y) dy = 0 (Final of Fall 2009) ANSWER sinx + xlny + e y = C 4 Appendix: Formula For an exact differential equation Mdx + Ndy = 0, we discuss an easier(?) way to find the potential function φ(x,y) and so the implicitly defined solution Personally, I do not recommend the formula(?) below Let M = M dx and N = N dy Since the potential function φ(x,y) satisfies φ x = M, we just integrate M with respect to x and add g(y), ˆ φ(x,y) = M dx + g(y) = M + g(y) (40) Because of the relations φ x = M and φ y = N, to find g(y), we differentiate the equation (40) with respect to y and compare with N, M y + g (y) = y (M + g(y)) = N, g (y) = N M y, ˆ ˆ ˆ ˆ g(y) = (N M y ) dy = N dy M y dy = N M y dy Therefore, we deduce a formula (if one wants to call it so) ˆ φ = M + N M y dy and the implicitly defined solution is ˆ M + N M y dy = C Example 43 Solve the exact differential equation ( ) x (cosx + lny)dx + y + ey dy = 0 Page 20 of 93

21 ANSWER Since the problem says that the given equation is exact So we do not have to test for exactness We just find the potential function and the implicitly defined solution Let us use the formula above Since M = cosx + lny and N = x/y + e y, so ˆ ˆ M = M dx = (cosx + lny) dx = sinx + xlny, M y = x ˆ ˆ x y, M y dy = dy = xlny, y ˆ ˆ ( ) x N = N dy = y + ey dy = xlny + e y (We do not have to add the constants of integration) The formula above implies the potential function φ and the implicitly defined solution, φ = sinx + xlny + xlny + e y xlny = sinx + xlny + e y, sinx + xlny + e y = C Page 2 of 93

22 5 Integrating Factors Let us start with the differential equation y + xy = 0, (5) which is not an exact differential equation However, if we multiply the whole equation by /x 2, then the equation becomes y x 2 + x y = 0 and we observe the left hand side of the equation is, in fact, the derivative of y/x, ie, d ( y ) = y dx x x 2 + x y = 0, ie, y x = C That is, the given differential equation (5) has the general solution y/x = C Check: Implicitly differentiating y/x = C, we get Here is another example Example 5 Solve the differential equation which is neither separable, linear, nor exact y x 2 + x y = 0, y + xy = 0 y 2 6xy + (3xy 6x 2 )y = 0, (52) ANSWER Step Multiply by µ(x,y) = y When we multiply the equation by µ(x,y) = y, the given equation becomes y(y 2 6xy) + y(3xy 6x 2 )y = 0, ie, y 3 6xy 2 + (3xy 2 6x 2 y)y = 0 (53) We observe the resulting equation is exact, because with M(x,y) = y 3 6xy 2 and N(x,y) = 3xy 2 6x 2 y, M y = 3y2 2xy = N x Step 2 Solution By the strategy for the exact differential equation, we deduce the potential function φ(x, y) and the implicitly defined solution of the equation (53) as follows: ˆ ˆ (y φ(x,y) = M dx = 3 6xy 2) dx = xy 3 3x 2 y 2 + g(y) The partial derivative of φ(x,y) with respect to y should be N, ie, 3xy 2 6x 2 y = N = φ(x,y) = ( xy 3 3x 2 y 2 + g(y) ) = 3xy 2 6x 2 y + g (y), y y ie, g (y) = 0, g(y) = C Hence, we deduce the potential function and the general solution defined implicitly, φ(x,y) = xy 3 3x 2 y 2 +C φ(x,y) = D, xy 3 3x 2 y 2 +C = D, xy 3 3x 2 y 2 = E Wherever y 0, this defines the general solution of the original non exact equation (52) given in the problem Page 22 of 93

23 Let us review the solution above The given differential equation was not exact 2 Multiplying the whole equation by µ(x,y), we made the given equation to be exact 3 By the strategy for an exact equation, we found the general solution of the modified equation (53), which was eventually same as the solution of the original equation (52) The function µ(x,y) has played a very important role in the solution above and so we should give a name to the function µ(x, y) Definition 52 Let M(x,y) and N(x,y) be defined on a region R of the plane Then µ(x,y) is an integrating factor for M(x,y) + N(x,y)y = 0 if µ(x,y) 0 for all (x,y) R and 2 µm + µny = 0 is exact on R How to find an integrating factor of the differential equation M + Ny = 0? Theorem 53 (STRATEGY) In order for a function µ(x,y) to be an integrating factor, it should make the equation µm + µny = 0 to be exact (in some region of the plane), which means µm and µn should satisfy the TEST FOR EXACT- NESS 44, ie, y (µm) = (µn) (54) x in the region Thus, we solve the differential equation (54) to find the integrating factor µ Example 54 Solve the differential equation x xy y = 0 (55) ANSWER Step Test for Exactness Let M(x, y) = x xy and N(x, y) = Then M y = x, but N x = 0 So the given equation is not exact Step 2 Integrating Factor µ(x,y) We solve the following equation for µ(x,y): y (µm) = x (µn), y (µ(x xy)) = x µ µ ( µ), (x xy) xµ = y x We choose µ(x,y) such that µ y = 0 Then µ(x,y) becomes a function of only x, ie, µ(x,y) = µ(x), and so the equation becomes simple to be solved dµ dx = xµ, dµ dx = xµ, which is a separable equation with the solution µ(x,y) = Ce x2 /2 Since we need just one integrating factor, we choose C =, ie, µ(x,y) = e x2 /2 Step 3 Solving Exact Equation Multiplying the original equation (55) by µ(x,y) = e x2 /2, (x xy)e x2 /2 e x2 /2 y = 0, (56) Page 23 of 93

24 which is an exact differential equation A potential function φ satisfies φ x = (x xy)ex2 /2, and φ y = ex2 /2 (57) To find φ, we integrate either one Let us integrate the second one φ y with respect to y: φ(x,y) = ˆ ( ) ˆ φ ( ) dy = e x2 /2 dy = e x2 /2 y + h(x) y Inserting this result to the first equation in (57), we determine the function h(x): (x xy)e x2 /2 = φ x = ( ) e x2 /2 y + h(x) = xye x2 /2 + h (x), x ˆ (x xy)e x2 /2 = xye x2 /2 + h (x), h (x) = xe x2 /2, h(x) = Thus, finally we obtain the potential function φ(x,y): φ(x,y) = ye x2 /2 + e x2 /2 +C = ( y)e x2 /2 +C, and the general solution of the differential equation (56) xe x2 /2 dx = e x2 /2 +C φ(x,y) = ( y)e x2 /2 +C = D, ie, ( y)e x2 /2 = E, (58) where C, D and E = D C are arbitrary constants Although we solved the modified equation (56), the solution (58) also satisfies the original differential equation (55) If we cannot find an integrating factor that is a function of just x or just y, then we must try something else So finding an integrating factor is the most difficult part in the solution Example 55 (DIFFICULT) Solve the differential equation 2y 2 9xy + (3xy 6x 2 )y = 0 (59) ANSWER Step Test for Exactness Let M(x,y) = 2y 2 9xy and N(x,y) = 3xy 6x 2 Then M y = 4y 9x, but N x So the given equation is not exact Step 2 Integrating Factor µ(x,y) We solve the equation for µ(x,y), y (µm) = x (µn), = 3y 2x ( µ(2y 2 9xy) ) = ( µ(3xy 6x 2 ) ), y x (2y 2 9xy) µ y + µ(4y 9x) = (3xy 6x2 ) µ + µ(3y 2x) (50) x If we choose µ(x,y) such that µ y = 0, then µ(x,y) = µ(x) and the equation becomes which is not easy to solve for µ as just a function of x µ(4y 9x) = (3xy 6x 2 ) µ + µ(3y 2x), x Page 24 of 93

25 If we choose µ(x,y) such that µ x = 0, then µ(x,y) = µ(y) and the equation becomes (2y 2 9xy) µ + µ(4y 9x) = µ(3y 2x), y which is not easy to solve for µ as just a function of y So we must try something else Let us try µ(x,y) = x a y b, where a and b are constants to be determined Putting µ(x,y) = x a y b into the equation (50) and find the constants a and b: (2y 2 9xy)(bx a y b ) + (x a y b )(4y 9x) = (3xy 6x 2 )(ax a y b ) + (x a y b )(3y 2x) Expanding the whole equation and dividing by x a y b (with assumption x 0 y), we have 2by 9bx + 4y 9x = 3ay 6ax + 3y 2x Rearranging the terms and using that x and y are independent, we get + 2b 3a = 0, and 3 + 9b 6a = 0 Solving the equations for a and b, we get a = = b, ie, the integrating factor µ(x,y) = xy Step 3 Solving Exact Equation Multiplying the original equation (59) by µ(x,y) = xy, xy(2y 2 9xy) + xy(3xy 6x 2 )y = 0, 2xy 3 9x 2 y 2 + (3x 2 y 2 6x 3 y)y = 0, (5) which is an exact differential equation A potential function φ satisfies φ x = 2xy3 9x 2 y 2, and φ y = 3x2 y 2 6x 3 y (52) To find φ, we integrate either one Let us integrate the first one φ x with respect to x: ˆ ( ) ˆ φ (2xy φ(x,y) = dx = 3 9x 2 y 2) dx = x 2 y 3 3x 3 y 2 + g(y) x Inserting this result to the second equation in (52), we determine the function g(y): 3x 2 y 2 6x 3 y = φ y = ( x 2 y 3 3x 3 y 2 + g(y) ) = 3x 2 y 2 6x 3 y + g (y), y 3x 2 y 2 6x 3 y = 3x 2 y 2 6x 3 y + g (y), g (y) = 0, g(y) = C Thus, finally we obtain the potential function φ(x,y): φ(x,y) = x 2 y 3 3x 3 y 2 +C, and the general solution of the differential equation (5) φ(x,y) = x 2 y 3 3x 3 y 2 +C = D, ie, x 2 y 3 3x 3 y 2 = E, (53) where C, D and E = D C are arbitrary constants Although we solved the modified equation (5), the solution (53) also satisfies the original differential equation (59) The method with the integrating factor may not give all the solutions It may not give a singular solution (SECTION 2 SEPARABLE EQUATIONS) which is a solution but cannot be deduced from the general solution Two examples, Example 56 and Example 57, on this case are given below It is a peculiar case, so you may skip the examples, as long as you can understand the limitation of the integrating factor technique Page 25 of 93

26 Example 56 Solve the differential equation 2xy y y = 0 (54) ANSWER Particular Solution We observe y = 0 does satisfy the given equation Hence, y = 0 is a particular solution General Solution: Strategy for Separable Equation Since it is a separable equation, one can get the general solution easily by using the strategy 22 for separable equations: y ln y = x 2 +C, where y 0 and y General Solution: Strategy with Integrating Factor for Exact Equation Let us use the integrating factor technique for an exact equation The given equation is not exact, but µ(x, y) = (y )/y is an integrating factor for y 0 (We don t have to solve a differential equation to find this integrating factor) Multiplying the whole equation (54) by the integrating factor, we have 2x y y y = 0, which is exact with the potential function φ(x,y) = x 2 y + ln y Thus the equation (54) has the general solution x 2 y + ln y = C, y 0 As we can see, the particular solution y = 0 cannot be obtained from the general solution That is, y = 0 is a singular solution Example 57 Solve the equation y 3 xy = 0 (55) ANSWER Particular Solution We observe y = 3 does satisfy the given equation Hence, y = 3 is a particular solution General Solution: Strategy for Linear Equation We modify the given equation, xy y = 3, y x y = 3 x, x 0, which is a linear equation So, one can get the general solution easily by using the strategy 32 for linear equations with the integrating factor e ( /x)dx = /x: ( y x ) = 3 x 2, y x = 3 x +C, y = 3 +Cx, where x 0 With C = 0, we can get the particular solution y = 3, ie, y = 3 is not a singular solution General Solution: Strategy with Integrating Factor for Exact Equation Let us use the integrating factor technique for an exact equation The given equation is not exact, but µ(x,y) = /[x(y 3)] is an integrating factor for x 0 and y 3 (We don t have to solve a differential equation to find this integrating factor) Multiplying the whole equation (55) by the integrating factor, we have x y 3 y = 0, which is exact with the potential function φ(x,y) = ln x ln y 3 Thus the equation (55) has the general solution ln x ln y 3 = C, Page 26 of 93

27 where x 0 and y 3 The solution implicitly defined above does not give the particular solution y = 3 However, we can simplify the equation and find y explicitly: y = 3 + Dx, where D is a constant As we can see, with D = 0, we can get the particular solution y = 3 That is, y = 3 is not a singular solution 5 Separable Equations and Integrating Factors The separable equation y = A(x)B(y), A(x)B(y) y = 0 (56) is not exact, because with M(x,y) = A(x)B(y) and N(x,y) =, the partial derivatives M y = A(x)B (y), and N x = 0 are not equal in general Multiplying the whole equation (56) by the integrating factor µ(x, y) = /B(y) for B(y) 0, the given equation (56) becomes A(x) B(y) y = 0, which is exact and has the general solution ˆ ˆ A(x)dx dy = C B(y) The act of separating the variables is the same as multiplying by the integrating factor /B(y) 52 Linear Equations and Integrating Factors The linear equation y + p(x)y = q(x), p(x)y q(x) + y = 0 (57) is not exact, because with M(x,y) = p(x)y q(x) and N(x,y) =, the partial derivatives M y = p(x), and N x = 0 are not equal in general unless p(x) is identically zero Multiplying the whole equation (57) by the integrating factor µ(x,y) = e p(x)dx, the equation (57) becomes which is exact and has the general solution e p(x)dx (p(x)y q(x)) + e p(x)dx y = 0, e p(x)dx y = ˆ ( ) q(x)e p(x)dx dx +C, ˆ ( ) p(x)dx y = e q(x)e p(x)dx dx +Ce p(x)dx See the formula (3) in SECTION 3 LINEAR DIFFERENTIAL EQUATIONS Page 27 of 93

28 53 Appendix: Formula We deduce a formula on the integrating factor µ(x,y) for an exact differential equation Mdx + Ndy = 0 Personally, I do not recommend that you should memorize the formulas We start with the condition for the integrating factor µ, y (µm) = x (µn), µ ym + µm y = µ x N + µn x (58) Case µ(x,y) = µ(x) In this case, µ y = 0 and the equation (58) becomes µm y = dµ dx N + µn x, N dµ dx = µm y µn x = (M y N x ) µ, µ dµ = M y N x dx N If (M y N x )/N is a function of only x, then we can integrate both sides and get ln µ = ˆ ˆ µ dµ = My N x dx, N {ˆ } My N x µ(x) = exp dx N Case 2 µ(x,y) = µ(y) In this case, µ x = 0 and the equation (58) becomes dµ dy M + µm y = µn x, dµ dy M = µn x µm y = (N x M y ) µ, µ dµ = N x M y M dy If (N x M y )/M is a function of only y, then we can integrate both sides and get ˆ ˆ {ˆ } ln µ = µ dµ = Nx M y Nx M y dy, µ(y) = exp M M dy Example 58 Find an integrating factor of the form either a function of x, µ(x), or a function of y, µ(y), which makes the differential equation 2xy + 3y 2x 3 = 0 to be exact ANSWER Let us use the formula above The equation is rewritten as (3y 2x 3 )dx + 2xdy = 0 and so letting M = 3y 2x 3 and N = 2x, we have M x = 6x 2, M y = 3, N x = 2 and N y = 0 Since M y N x N = 3 2 2x = 2x Since N x M y M formula, then we have is a function of only x, the formula implies {ˆ My N x µ(x) = exp N = 2 3 = 6x 2 6x 2 µ(y) = exp } { ˆ } { } dx = exp 2 x dx = exp 2 lnx = x /2 is not a function of only y, we cannot use the formula If we force to use the {ˆ Nx M y M } { ˆ } { y } dy = exp 6 x 2 dy = exp 6x 2, which is wrong, because it should be the function of only y (the left hand side is µ(y)) Hence, the integrating factor is µ(x) = x /2 Page 28 of 93

29 6 Homogeneous, Bernoulli, and Riccati Equations 6 Homogeneous Differential Equations Definition 6 A first order differential equation is homogeneous if it has the form ( y y = f (6) x) Example 62 The following differential equations are homogeneous, y = x y sin ( y x ), y = y x + x y, y = y2 x 2 e y x Theorem 63 (STRATEGY) A homogeneous equation is always transformed into a separable one by the change of variable u = y, ie, y = ux x Differentiating it, y = u x + ux = u x + u Putting it into the equation, the given equation (6) becomes u x + u = f (u), u x = f (u) u, x du = f (u) u, dx which is separable Using the strategy 22 for a separable equation, we get x du dx = f (u) u, f (u) u du = ˆ ˆ x dx, f (u) u du = dx = ln x +C, x where f (u) u 0 x are assumed We compute the integral and put u = y/x back to the result and get the general solution to the equation (6) Example 64 Solve the differential equation ANSWER Step Type of Equation xy = y2 x + y, x 0 (62) It is easy to see that the equation (62) is neither separable nor linear 2 The equation (62) is equivalent to y 2 x + y xy = 0, which is not exact, because with M(x,y) = y 2 /x + y and N(x,y) = x, M y = 2y N + = x x, on a region in the plane 3 Is it easy to find an integrating factor µ(x,y) which makes the given equation to be exact? µ y ( y 2 ) x + y ( 2y + µ x + (µm) y ) = µ x ( x) + µ( ) = (µn), µ y M + µm y = µ x N + µn x, x As we can see, it is not easy to find µ satisfying the equation above Page 29 of 93

30 4 We observe the given equation is equivalent to y = y2 x 2 + y ( y ) 2 x = y + x x, (63) of which the right hand side is a function of y/x Hence, the equation (62) is homogeneous Step 2 Substitution u = y/x Since the equation is homogeneous, we use the substitution u = y/x Putting it into the equation (63), it becomes y = ux, y = u x + ux = u x + u u x + u = u 2 + u, u x = u 2, which is separable Using the strategy 22 for a separable equation, we have x du dx = u2, u 2 du = x dx, ˆ u 2 du = ˆ x dx, u = ln x +C, u = ln x +C Hence, by putting u = y/x back to the result, we deduce the solution of the equation (62) y x = ln x +C, ie, y = x ln x +C Example 65 Solve the differential equation ANSWER Step Type of Equation The equation is equivalent to y = x2 + 3y 2, x 0 y (64) 2xy y = x 2 y y x, (65) which is homogeneous Step 2 Substitution u = y/x Since the equation is homogeneous, we use the substitution u = y/x Putting it into the equation (65), it becomes y = ux, y = u x + ux = u x + u u x + u = 2 u u, u x = 2u + u 2 = + u2 2u, which is separable Using the strategy for a separable equation, we have x du dx = + u2 2u, 2u + u 2 du = x dx, + u 2 = ln x +C, u2 = ˆ 2u + u 2 du = ( ln x +C, u = ± ˆ x dx, ln x +C Hence, by putting u = y/x back to the result, we deduce the solution of the equation (64) ( y x = ± ln x +C ) /2, ie, y = ±x( Page 30 of 93 ) /2 ) /2 ln x +C

31 Before we end this subsection, let us consider the equation y = which is not homogeneous But the equation can be written by y = y x + y, (66) y x + y = y/x + y/x, ie, y = y/x, x 0, (67) + y/x which is homogeneous The solution of (67) also satisfies the equation (66) But the equation (66) may have other solutions as well When we perform manipulations on a differential equation, we should be careful Exercise 66 Solve the homogeneous equations y 2 + 2xy x 2 y = 0 xyy = x 2 + 2y 2 (Final Exam of Fall 2009) 62 Bernoulli Differential Equation Definition 67 A Bernoulli equation is a first order equation in which α is a real number We observe: When α = 0, the Bernoulli equation becomes which is linear 2 When α =, the Bernoulli equation becomes y + p(x)y = q(x)y α, y + p(x)y = q(x), y + p(x)y = q(x)y, y + (p(x) q(x))y = 0, which is both separable and linear 3 For 0 α, the Bernoulli equation is nonlinear Theorem 68 (STRATEGY) We use the substitution (ie, change of variable) v = y α How does it work? We will discuss it through examples For the rigorous and general argument, please see the APPENDIX 63 Example 69 Solve the differential equation which is nonlinear y + x y = 3x2 y 3, (68) ANSWER Step Substitution v = y α Since the equation is of Bernoulli type with α = 3, we use the change of variable v = y 3 = y 2, ( v = 2y 3 y = 2y 3 ) x y + 3x2 y 3 = 2 x y 2 6x 2 = 2 x v 6x2 Page 3 of 93

32 That is, we get v 2 x v = 6x2, (69) which is linear Step 2 Solve Transformed Equation We introduce the integrating factor I(x) = e 2 x dx = e 2ln x = x 2 = x 2 Multiplying the whole equation (69) by the integrating factor I(x) = /x 2, we get x 2 v 2 x 2 x v = ( 6x 2 ) x 2 = 6, ( ) x 2 v = 6, x 2 v = 6x +C, v = Cx2 6x 3 Since we are looking for the solution to the equation (68), we should change the equation back in terms of y by using v = y 2, y 2 = Cx 2 6x 3, y = ( Cx 2 6x 3) /2 = Cx 2 6x 3 Example 60 Solve the differential equation which is nonlinear x 2 y + 2xy y 3 = 0, x > 0, (60) ANSWER Step Substitution v = y α Since the equation is of Bernoulli type with α = 3, we use the change of variable y = v 3 = y 2 with y = 2 x y + x 2 y 3, v = 2y 3 y = 2y 3 ( 2 x y + x 2 y3 ) = 4 x y 2 2 x 2 = 4 x v 2 x 2 That is, we get v 4 x v = 2 x 2, (6) which is linear Step 2 Solve Transformed Equation We introduce the integrating factor I(x) = e 4 x dx = e 4ln x = x 4 = x 4 Multiplying the whole equation (6) by the integrating factor I(x) = /x 4, we get x 4 v 4 x 4 x v = ( 2x ) x 4 2 = 2 ( ) x 6, x 4 v = 2 x 6, x 4 v = 2 5x 5 +C, v = 2 5x +Cx4 Since we are looking for the solution to the equation (60), we should change the equation back in terms of y by using v = y 2, y 2 = 2 5x +Cx4, y = ( ) 2 /2 5x +Cx4 = 2/(5x) +Cx 4 Exercise 6 Solve the Bernoulli equation, y + (x + )y = e x2 y 3 Page 32 of 93

33 63 Appendix: Formula The Bernoulli equation can be solved by the change of variable v = y α y + p(x)y = q(x)y α (62) PROOF We assume α 0 and α, because when α = 0 or α =, the given equation becomes linear and it is easy to solve (For the convenience, we use p and q for p(x) and q(x)) Step Substitution v = y α The change of variable implies v = ( α)y α y = ( α)y α ( py + qy α ) = ( α) ( py α + q ) = ( α)( pv + q), That is, the given equation (62) turns to be v + ( α)pv = ( α)q, (63) which is linear Step 2 Solve Transformed Equation We introduce the integrating factor I(x) = e ( α) pdx Multiplying the whole equation (63) by the integrating factor I(x), we get ˆ Iv + I( α)pv = ( α)qi, (Iv) = ( α)qi, Iv = ( α) (qi) dx Hence, the solution of the equation (63) is given by v = α ˆ ( α) ˆ ( pdx ( α) ) pdx (qi) dx = ( α)e qe dx I Since we look for the solution to the equation (62), we should express it in terms of y by using v = y α, y α ( α) ˆ ( pdx ( α) ) pdx = ( α)e qe dx, y = { ( α) ˆ ( p(x)dx ( α) ) } α p(x)dx ( α)e q(x)e dx 7 Applications to Mechanics, Electrical Circuits, and Orthogonal Trajectories Skip Please read the textbook 8 Existence and Uniqueness for Solutions of Initial Value Problems Skip Please read the textbook Page 33 of 93

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35 Chapter 2 Second Order Differential Equations 2 Preliminary Concepts Definition 2 A second order differential equation is an equation which contains a second derivative, but no higher derivative Generally, it has the form F(x,y,y,y ) = 0, where a term involving y should be appeared A solution of F(x,y,y,y ) = 0 on an interval I is a function φ(x) satisfying the differential equation at each point of I: F(x,φ(x),φ (x),φ (x)) = 0 for all x in I Example 22 Second order differential equations The followings are second order differential equations: y = x 3, xy cosy = e x, y 4xy + y = 2, y =, y = y, x 3 y = 5xy + 6x 0 y y + 6y = 0 has a solution φ(x) = 6cos(4x) 7sin(4x) Why? Check by putting the solution into the differential equation! 3 x 2 y 5xy +0y = 0 for x > 0 has a solution φ(x) = x 3 cos(lnx) Why? Check by putting the solution into the differential equation! Definition 23 The linear second order differential equation has the form R(x)y + P(x)y + Q(x)y = S(x), simply, y + p(x)y + q(x)y = f (x), (2) by dividing both sides of the first equation by R(x) In this chapter, we have two main goals: Study on the existence and uniqueness of the solution of the equation (2) (Throughout this course, we study only differential equations having a solution So we do not focus on the existence) 2 How to solve the equation (2) 35

36 22 Theory of Solutions of y + p(x)y + q(x)y = f (x) Let us start with the following problem from CALCULUS I or GENERAL PHYSICS Example 22 A distance function s(x) has the acceleration a(x) = 2x with the initial velocity v(0) = 27 and the initial position s(0) = 4 Find the distance function s(x) ANSWER One should be able to solve this kind of problem The problem is, in fact, a second order initial value problem: find s(x) such that Integrating it, we get ˆ ˆ s (x) = s (x)dx = s (x) = 2x, s(0) = 4, s (0) = 27 (22) ˆ 2xdx = x 2 +C, s(x) = s (x)dx = ˆ (x 2 +C ) dx = x3 3 where C and D are constants of integration Applying the initial conditions to the results above, 27 = s (0) = 0 2 +C, C = 27, 4 = s(0) = C(0) + D = D, D = 4 Hence, we deduce the solution of the given second order initial value problem (22), s(x) = x3 3 The example says that for a second order differential equation, +Cx + D, + 27x + 4 The solution s(x) = x3 3 +Cx + D has two constants of integration C and D 2 To determine those constants, we need two initial conditions s(0) = 4 and s (0) = 27 3 When we have two initial conditions, we deduce only one solution s(x) = x3 3 It is summarized in the following Theorem Theorem 222 (UNIQUENESS OF SOLUTION) A second order initial value problem, y + p(x)y + q(x)y = f (x), y(x 0 ) = A, y (x 0 ) = B, has a unique (ie, only one) solution, where A and B are given constants Definition x + 4 A second order differential equation y + p(x)y + q(x)y = 0 is called homogeneous 2 A second order differential equation y + p(x)y + q(x)y = f (x) with f (x) 0 is called nonhomogeneous 3 A linear combination of two functions f (x) and g(x) is a function of form a f (x)+bg(x), where a and b are any constants One should not be confused: Homogeneous first order differential equation is of form, ( y y = f x) Homogeneous second order differential equation is of form, y + p(x)y + q(x)y = 0 From SECTION 22 to SECTION 25, we mainly focus on the homogeneous second order equation For a nonhomogeneous equation, we will discuss in SECTION 26 Page 36 of 93

37 Theorem 224 (LINEAR COMBINATION) Let y (x) and y 2 (x) be solutions of the second order differential equation y + p(x)y + q(x)y = 0 (222) on an interval I Then, any linear combination of these functions is also a solution of the equation (222), that is, c y (x) + c 2 y 2 (x) is also a solution, where c and c 2 are any constants PROOF Let φ(x) = c y (x) + c 2 y 2 (x) Then, φ = c y + c 2 y 2, φ = c y + c 2 y 2 So when putting them into the given equation (222), the given equation becomes φ + p(x)φ + q(x)φ = (c y + c 2 y 2 ) + p(x)(c y + c 2 y 2 ) + q(x)(c y + c 2 y 2 ) = c y + c 2 y 2 + p(x) ( c y + c 2 y 2) + q(x)(c y + c 2 y 2 ) ( = c y + p(x)y ) ( + q(x)y + c2 y 2 + p(x)y ) 2 + q(x)y 2 = c (0) + c 2 (0) = 0 Hence, φ(x) = c y (x) + c 2 y 2 (x) is also a solution of (222) Remark 225 We point out two remarks on the Theorem 224 The Theorem does not hold for a nonhomogeneous equation 2 The Theorem is about producing a new solution c y + c 2 y 2 from already found two solutions y and y 2 How can we find those two solutions y and y 2? It will be studied in following sections In this section, we deal with solutions already found Definition 226 Two functions f (x) and g(x) are said to be linearly dependent on an open interval I, if for some constant c, either f (x) = cg(x) or g(x) = c f (x) for all x in I If f and g are not linearly dependent, then we say they are linearly independent Example 227 The functions y (x) = cosx and y 2 (x) = sinx are solutions of the homogeneous second order differential equation y + y = 0 () Verify that y and y 2 are really solutions of the given differential equation (2) The Theorem 224 says that y = c y + c 2 y 2 = c cosx + c 2 sinx with any constants c and c 2 is also a solution of the given differential equation Verify that y = c cosx + c 2 sinx is really a solution (2) Are y = cosx and y 2 = sinx linearly dependent or linearly independent? Justify your answer ANSWER () We observe that for y = cosx, y = (cosx) = sinx, y = (cosx) = ( sinx) = cosx, y + y = cosx + cosx = 0 Since y = cosx satisfies the given differential equation, y = cosx is really a solution Similarly, for y 2 = sinx, y 2 = (sinx) = cosx, y 2 = (sinx) = (cosx) = sinx, y 2 + y 2 = sinx + sinx = 0 Since y 2 = sinx satisfies the given differential equation, y 2 = sinx is really a solution (2) For y = c cosx + c 2 sinx with any constants c and c 2, y = (c cosx + c 2 sinx) = c sinx + c 2 cosx, y = (c cosx + c 2 sinx) = ( c sinx + c 2 cosx) = c cosx c 2 sinx, Page 37 of 93

38 y + y = c cosx c 2 sinx + c cosx + c 2 sinx = 0 Since y = c cosx + c 2 sinx satisfies the given differential equation, y = c cosx + c 2 sinx is really a solution (3) Suppose y = cosx and y 2 = sinx are linear dependent Then for some constant k, y = ky 2, cosx = k sinx, = k tanx, tanx = k for all x It means that for any x, the tangent function gives only one value /k That is, the tangent function is a constant function This is obviously wrong, because the tangent function is not a constant function Hence, the assumption is wrong and we conclude that y = cosx and y 2 = sinx must be linearly independent As one can see, when we need to know the linear dependency/independency of given two functions, it is very painful if we use the definition Is there a better way or method telling us whether two functions are linearly dependent or independent? Yes, it is: Wronskian Test Since the test is very useful, it is very important, ie, one must memorize it Definition 228 For two functions f (x) and g(x), the function W(x) = W( f (x),g(x)) = is called Wronskian of f (x) and g(x) f (x) f (x) g(x) g (x) = f (x)g (x) f (x)g(x) Theorem 229 (WRONSKIAN TEST) Let y (x) and y 2 (x) be two solutions of the differential equation y + p(x)y + q(x)y = 0 on an interval I Then for the Wronskian W(x) = y (x)y 2 (x) y (x)y 2(x) of y and y 2, we have either W(x) = 0 for all x in I or W(x) 0 for all x in I 2 W(x) 0 on I if and only if y and y 2 are linearly independent Simply speaking, the Theorem 229 says that the Wronskian is either a zero function or not If the Wronskian is a zero function, those two functions y and y 2 are linearly dependent If the Wronskian is not a zero function, those two functions y and y 2 are not linearly dependent, ie, linearly independent Example 220 From Example 227, we recall that y = cosx and y 2 = sinx are linearly independent Let us use the Wronskian Test 229 to determine the linear independency: W(x) = y (x)y 2(x) y (x)y 2 (x) = cosx(sinx) (cosx) sinx = cos 2 x + sin 2 x = 0 for all x Since the Wronskian W(x) is not a zero function, by the Test, we conclude that y = cosx and y 2 = sinx are linearly independent Theorem 22 (GENERAL SOLUTION) Let y and y 2 be linearly independent solutions of the differential equation y + p(x)y + q(x)y = 0 (223) on an open interval I Then every solution of the differential equation is a linear combination of y and y 2, ie, the general solution of the differential equation (223) is where c and c 2 are any constants y(x) = c y (x) + c 2 y 2 (x), Page 38 of 93

39 Definition 222 Let y and y 2 be solutions of the differential equation y + p(x)y + q(x)y = 0 on an open interval I If y and y 2 are linearly independent, y and y 2 forms a fundamental set of solution on I 2 If y and y 2 forms a fundamental set of solution on I, then c y + c 2 y 2 is called the general solution of the given differential equation on I One will see that this definition goes to the course, LINEAR ALGEBRA (also called as CALCULUS 4 in this university) That is, one must understand and memorize it We end this section with the proof of the Theorem 22 PROOF OF THEOREM 22 Let φ be any solution of the given differential equation on I Then we have to show that there exist some numbers c and c 2 such that φ = c y + c 2 y 2 We choose x 0 I Let φ(x 0 ) = A and φ (x 0 ) = B Then by Theorem 222, φ is a unique solution of the second order initial value problem, Now we consider the system of equations, Solving the equations for c and c 2, we get y + p(x)y + q(x)y = 0, y(x 0 ) = A, y (x 0 ) = B c y (x 0 ) + c 2 y 2 (x 0 ) = A, c y (x 0 ) + c 2 y 2(x 0 ) = B c = Ay 2 (x 0) By 2 (x 0 ) W(x 0 ) where W(x) is the Wronskian of y and y 2 Then, we observe c y + c 2 y 2 = Ay 2 (x 0) By 2 (x 0 ) W(x 0 ), c 2 = By (x 0 ) Ay (x 0), (224) W(x 0 ) y + By (x 0 ) Ay (x 0) W(x 0 ) is a solution of the given differential equation By the uniqueness of the solution in Theorem 222, we conclude φ = c y + c 2 y 2 = Ay 2 (x 0) By 2 (x 0 ) y + By (x 0 ) Ay (x 0) y 2 W(x 0 ) W(x 0 ) y 2 Page 39 of 93

40 23 Reduction of Order Throughout this section, we consider only the homogeneous second order differential equation, y + p(x)y + q(x)y = 0 (23) In the previous section, we have argued on generating a general solution c y + c 2 y 2 of the homogeneous second order differential equation y + p(x)y + q(x)y = 0, when linearly independent two solutions y and y 2 are known In this section, we develop a method of finding another solution y 2 (which is linearly independent to y ) of the same differential equation, when only one solution y is known So in this way, for a given one solution y, we can deduce a general solution Specifically, we look for the solution y 2 (x) of form u(x)y (x), ie, y 2 (x) = u(x)y (x) Be careful! If u(x) is a constant, then y and y 2 are linearly dependent, which is not what we want So u(x) should not be a constant First, we describe the strategy and then see how to use it through examples Please do not memorize it as a formula, but try to understand how to do it Goal When a solution y (x) 0 is given, we want to find another solution y 2 (x) = u(x)y (x) such that y and y 2 are linearly independent Theorem 23 (STRATEGY) We follow the steps called the method of reduction of order Step Inserting y 2 (x) = u(x)y (x) into Equation We put it into the equation (23) and simplify it Then we have a first order equation, u + p(x)u = 0, v + g(x)v = 0, which is both separable and linear by the substitution v = u Step 2 Solving Resulting Equation v + g(x)v = 0 It s easy to solve the resulting equation u ˆ = v = e g(x)dx, u = e g(x)dx dx Step 3 Linear Independency (Wronskian Test: Optional) From Step 2, we have found another solution (ˆ ) y 2 (x) = u(x)y (x) = e g(x)dx dx y (x) Using the Wronskian Test 229, we check the linear independency between y and y 2 This Step is optional Step 4 General Solution Since we have two solutions y (x) and y 2 (x) = u(x)y (x) which are linearly independent, the general solution of the equation (23) is obtained by Let us study each step above y = c y + c 2 y 2 = c y + c 2 uy = (c + c 2 u)y, y = (c + c 2 u)y FULL EXPLANATION Step Inserting y 2 (x) = u(x)y (x) into Equation Since we want y 2 (x) = u(x)y (x) to be a solution, it should satisfy the equation, y 2 + p(x)y 2 + q(x)y 2 = 0, [u(x)y (x)] + p(x)[u(x)y (x)] + q(x)[u(x)y (x)] = 0 Expanding and simplifying it, we get 0 = ( u y + uy ) + p ( u y + uy ) + quy = u y + u y + u y + uy + p ( u y + uy ) + quy Page 40 of 93

41 = u y + 2u y + uy + pu y + puy + quy = u ( y + py + qy ) + u y + 2u y + pu y = u ( y + py + qy ) + u y + ( 2y + py ) u Since y is a solution, we have y + py + qy = 0, and so the result above becomes 0 = u y + ( 2y + py ) u Since we want to find u, we solve the differential equation by the change of variable v = u, 0 = v y + ( 2y + py ) v, v + 2y + py y v = 0 (232) Step 2 Solving Resulting Equation We simplify the equation (232) by letting g(x) = 2y (x) + p(x)y (x) y (x) Then equation (232) becomes a first order differential equation, which is both separable and linear Solving the equation, we have v + g(x)v = 0, (233) v(x) = e g(x)dx, where we ignored the constant Since we are looking for u but not v, so we solve the equation u ˆ = v = e g(x)dx, u(x) = e g(x)dx dx (234) and finally we deduce another solution y 2 = uy, (ˆ ) y 2 (x) = e g(x)dx dx y (x), g(x) = 2y (x) + p(x)y (x) y (x) Step 3 Linear Independency (Wronskian Test) Let us check whether or not y and y 2 = uy are linearly independent We use the Wronskian Test 229, W(x) = y y 2 y y 2 = y (uy ) y ( ( (uy ) = y u y + uy ) uy y = u y 2 ) g(x)dx = e y 2 0 where the result (234) is used and the exponential function never vanishes Since W(x) 0, the Wronskian Test implies that y and y 2 = uy are linearly independent Step 4 General Solution Since y and y 2 = uy are linearly independent solutions, by the Theorem 22, the general solution of the differential equation (23) is given by y(x) = c y + c 2 y 2 = c y + c 2 uy = (c + c 2 u)y, y(x) = (c + c 2 u(x))y (x), where c and c 2 are any constants Remark 232 Why is this method called the Reduction of Order? The big picture is as follows: for a given solution y, we want to find another solution u(x)y (x) which is linearly independent to y, ie, u(x) satisfying the given requirements In order to find such a u(x), we have to solve the equation (233), which is the first order differential equation For this reason, the problem on the second order differential equation (23) turns to be the problem on the first order differential equation (233), ie, we reduce the order of the differential equation from order 2 to order Page 4 of 93

42 Let us see how to use the method through examples Example 233 Suppose the homogeneous second order differential equation y +4y +4y = 0 has a solution y (x) = e 2x () Find another solution y 2 (x) which is linearly independent to y (x) (2) Find the general solution of the given differential equation ANSWER Step Inserting y 2 (x) = u(x)y (x) into Equation Let y 2 (x) = u(x)y (x) = u(x)e 2x be a solution Then y 2 (x) should satisfy the given differential equation, ie, 0 = y 2 + 4y 2 + 4y 2 = ( ue 2x) + 4 ( ue 2x ) + 4ue 2x = ( u e 2x 2ue 2x) + 4 ( u e 2x 2ue 2x) + 4ue 2x = u e 2x 4u e 2x + 4ue 2x + 4 ( u e 2x 2ue 2x) + 4ue 2x = u e 2x Since an exponential function never vanishes, ie, e 2x 0 for all x, so we get u (x) = 0 Step 2 Solving Resulting Equation It is easy to solve the equation u (x) = 0, u (x) = C, u(x) = Cx + D, where C and D are constants of integration Since we need only one more solution, it is fine even if we choose C = and D = 0 (Be careful! u(x) should not be constant So we cannot choose C = 0) Hence, we have u(x) = x, y 2 (x) = u(x)y (x) = xe 2x Step 3 Linear Independency (Wronskian Test) Let us check whether or not y and y 2 = uy are linearly independent We use the Wronskian Test, W(x) = y y 2 y y 2 = e 2x ( xe 2x) ( e 2x ) xe 2x = e 4x 0, for all x Since W(x) 0 for all x, the Wronskian Test implies that y (x) = e 2x and y 2 = u(x)y (x) = xe 2x are linearly independent Step 4 General Solution Since y (x) = e 2x and y 2 (x) = u(x)y (x) = xe 2x are linearly independent solutions, by the Theorem 22, a general solution of the given differential equation y(x) = c y (x) + c 2 y 2 (x) = c e 2x + c 2 xe 2x = (c + c 2 x)e 2x where c and c 2 are any constants Example 234 Suppose the homogeneous second order differential equation y 3 x y + 4 x 2 y = 0, x > 0, has a solution y (x) = x 2 () Find another solution y 2 (x) which is linearly independent to y (x) (2) Find the general solution of the given differential equation ANSWER Step Inserting y 2 (x) = u(x)y (x) into Equation Let y 2 (x) = u(x)y (x) = u(x)x 2 be a solution Then y 2 (x) should satisfy the given differential equation, ie, That is, we have the differential equation, 0 = y 2 3 x y x 2 y 2 = ( ux 2) 3 ( ux 2 ) 4 ( + ux 2 ) x x 2 = ( u x 2 + 2xu ) 3 ( u x 2 + 2xu ) + 4 ( ux 2 ) x x 2 = ( u x 2 + 4xu + 2u ) 3 ( u x + 2u ) + 4u = u x 2 + xu x 2 u + xu = 0, u + x u = 0 Page 42 of 93

43 The change of variable v = u implies v + x v = 0, which is both separable and linear Step 2 Solving Resulting Equation Solving the equation, we get ˆ v = e dx x = e lnx = x, u = v = x, u(x) = x dx = lnx, where we ignored the constants (Because of the given condition x > 0 in the problem, ln x = lnx) Hence, we have deduced u(x) = lnx, y 2 (x) = u(x)y (x) = x 2 lnx Step 3 Linear Independency (Wronskian Test) Let us check whether or not y and y 2 = uy are linearly independent We use the Wronskian Test 229, y y 2 W(x) = y y = x 2 x 2 lnx 2 2x 2xlnx + x = x2 (2xlnx + x) x 2 lnx(2x) = x 3 0, for all x > 0 (Recall the given condition in the problem) Since W(x) 0 for all x > 0, the Wronskian Test implies that y (x) = x 2 and y 2 = u(x)y (x) = x 2 lnx are linearly independent Step 4 General Solution Since y (x) = x 2 and y 2 (x) = u(x)y (x) = x 2 lnx are linearly independent solutions, by the Theorem 22, a general solution of the given differential equation y(x) = c y (x) + c 2 y 2 (x) = c x 2 + c 2 x 2 lnx = (c + c 2 lnx)x 2 where c and c 2 are any constants Exercise 235 Suppose the given homogeneous second order differential equation has a solution y () Find another solution y 2 which is linearly independent to y (2) Find the general solution of the differential equation y y + 4 y = 0 with y (x) = e x/2 2t 2 y + 3ty y = 0, t > 0, with y (t) = t ( x 2 x ) y xy + y = 0 with y (x) = x x 2 y + xy 4y = 0 with y (x) = x 2 y 6y + 9y = 0 with y (x) = e 3x ( x 2) y 2xy + 2y = 0 with y (x) = x 2x 2 y + xy 3y = 0 with y (x) = /x (Final Exam of Fall 2009) Page 43 of 93

44 24 Constant Coefficient Homogeneous Linear Equation Throughout this section, we consider only the homogeneous second order differential equation with constant coefficient, y + Ay + By = 0, (24) where A and B are real numbers We observe that the derivatives of an exponential function e λx with constant λ become the constant multiples of the exponential function itself, ie, (e λx ) = λe λx and (e λx ) = λ 2 e λx So it could be the solution of the equation (24) For this reason, we look for the solution y(x) = e λx Putting it into the equation, we have 0 = λ 2 e λx + Aλe λx + Be λx = e λx ( λ 2 + Aλ + B ) Since an exponential function can never be zero, it implies λ 2 + Aλ + B = 0, (242) which is called the characteristic equation of the differential equation (24) Solving the characteristic equation for λ, we get λ = A ± A 2 4B 2 Depending on the value of A 2 4B, we have three cases: two different real roots, repeated root, and complex roots Before we study each case, let us compare two equations (24) and (242) The characteristic equation is obtained by replacing y = y (2), y = y () and y = y (0) in the equation (24) with λ 2, λ and λ 0 =, respectively 24 Case A 2 4B > 0 In this case, the characteristic equation has two distinct real roots: a = A + A 2 4B 2, b = A A 2 4B 2 and each gives the solution of the equation (24), y (x) = e ax and y 2 (x) = e bx, respectively In order for y and y 2 to form the fundamental set of solutions of (24), they should be linearly independent, ie, Wronskian of y and y 2 should not be zero Let us run the Wronskian Test: W(x) = y (x)y 2(x) y (x)y 2 (x) = be ax e bx ae ax e bx = e (a+b)x (b a) 0, because a b Thus, y and y 2 are linearly independent, ie, they forms the fundamental set of solutions, ie, the general solution of the equation (24) is given by where c and c 2 are arbitrary constants Example 24 Solve the differential equation, ANSWER It has the characteristic equation y(x) = c y (x) + c 2 y 2 (x) = c e ax + c 2 e bx, (243) y y 6y = 0 0 = λ 2 λ 6 = (λ + 2)(λ 3), of which roots are λ = 2 and λ = 3 Hence, by the formula (243), the general solution is where c and c 2 are arbitrary constants y(x) = c e 2x + c 2 e 3x, Page 44 of 93

45 242 Case 2 A 2 4B = 0 The characteristic equation (242) has the repeated root λ = A 2 So one solution is y (x) = e A 2 x Since we know one solution, we get another solution y 2 (x) in the form of y 2 (x) = u(x)y (x) = u(x)e A 2 x by using the method of Reduction of Order in the previous section Then again by the result in the previous Section 23 Reduction of Order, the general solution of the equation (24) is y(x) = c y (x) + c 2 y 2 (x) = c e A 2 x + c 2 u(x)e A 2 x = (c + c 2 u(x))e A 2 x We substitute y 2 (x) = u(x)e A 2 x into the differential equation (24) to find u(x): 0 = [u(x)e A x] 2 + A [u(x)e A x] 2 + B [u(x)e A x] 2 [ = u (x)e A 2 x A ] 2 u(x)e A 2 x + A [u (x)e A2 x A2 ] u(x)e A2 x + Bu(x)e A 2 x = u (x)e A 2 x Au (x)e A 2 x + A2 4 u(x)e A 2 x + A [u (x)e A2 x A2 ] u(x)e A2 x + Bu(x)e A 2 x = u (x)e A 2 x A2 4 u(x)e A 2 x + Bu(x)e A 2 x [ ) ] = u (x) + (B A2 u(x) e A 2 x 4 Since an exponential function can never be zero, we deduce u + ) (B A2 u = 0 4 However, we are discussing the case A 2 4B = 0, so the equation is in fact u = 0, u (x) = C, u(x) = Cx + D, and we choose C = and D = 0 (Why? See the Section 23 Reduction of Order) That is, u(x) = x and we get y 2 (x) = xe A 2 x and the general solution of the equation (24), y(x) = c y (x) + c 2 y 2 (x) = (c + c 2 u(x))e A 2 x = (c + c 2 x)e A 2 x (244) Example 242 Solve the differential equation, ANSWER It has the characteristic equation y 6y + 9y = 0 0 = λ 2 6λ + 9 = (λ 3) 2, which has the repeated root, λ = 3 Hence, by the formula (244), the general solution is y(x) = (c + c 2 x)e 3x, where c and c 2 are arbitrary constants Page 45 of 93

46 243 Case 3 A 2 4B < 0 A complex number i is defined by i = so that i 2 = If A 2 4B < 0, then 4B A 2 > 0 and we have A 2 4B = ( )(4B A 2 ) = i 2 (4B A 2 ) = i 4B A 2 So in the case of A 2 4B < 0, the characteristic equation has the complex roots λ = A ± i 4B A 2 2 For convenience, we write Then, the roots become p = A 4B A 2, and q = 2 2 λ = A ± i 4B A 2 2 = A 2 ± i 4B A 2 2 = p ± iq which yields two solutions y (x) = e (p+iq)x, y 2 (x) = e (p iq)x Now we check the linear independency of y and y 2 via the Wronskian Test: W(x) = y (x)y 2(x) y y (x) y 2 (x) (x)y 2 (x) = y (x) y 2 (x) = e (p+iq)x e (p iq)x (p + iq)e (p+iq)x (p iq)e (p iq)x = (p iq)e (p+iq)x e (p iq)x (p + iq)e (p+iq)x e (p iq)x = e 2px ( 2iq) The condition A 2 4B < 0 implies q = 4B A 2 0 That is, W(x) 0 for all x and so y and y 2 are linearly independent and thus the general solution of the equation (24) is given by Example 243 Solve the differential equation, ANSWER It has the characteristic equation y(x) = c y (x) + c 2 y 2 (x) = c e (p+iq)x + c 2 e (p iq)x (245) y + 2y + 6y = 0 λ 2 + 2λ + 6 = 0, which has the complex roots, λ = ± i 5 Hence, by the formula (245), the general solution is y(x) = c e ( +i 5)x + c 2 e ( i 5)x, where c and c 2 are arbitrary constants Page 46 of 93

47 244 Alternative General Solution in Case of Complex Roots Let us introduce Euler s Formula: e ix = cosx + isinx It implies e ix = cos( x) + isin( x) = cosx isinx When we use the formula, the general solution (245) becomes y(x) = c e (p+iq)x + c 2 e (p iq)x = c e px+iqx + c 2 e px iqx = c e px e iqx + c 2 e px e iqx = e px ( c e iqx + c 2 e iqx) = e px {c [cos(qx) + isin(qx)] + c 2 [cos(qx) isin(qx)]} = e px [(c + c 2 )cos(qx) + i(c c 2 )sin(qx)] = (c + c 2 )e px cos(qx) + i(c c 2 )e px sin(qx) If we choose c = c 2 = /2, we obtain one solution y 3 (x) = e px cos(qx) If we choose c = /(2i) and c 2 = /(2i), we obtain anther solution y 4 (x) = e px sin(qx) The Wronskian Test on y 3 and y 4 implies W(x) = y 3 y 4 y 3y 4 = e 2px 0, ie, they are linearly independent, ie, the general solution of the equation (24) is their linear combination, y(x) = c 3 y 3 (x) + c 4 y 4 (x) = c 3 e px cos(qx) + c 4 e px sin(qx) = e px (c 3 cos(qx) + c 4 sin(qx)), (246) where c 3 and c 4 are arbitrary constants Example 244 Solve the initial value problem, y 4y + 53y = 0, y(π) = 3, y (π) = 2 ANSWER USING (246) The differential equation has the characteristic equation, λ 2 4λ + 53 = 0, which has the complex roots λ = 2 ± 7i = p ± iq So by the formula (246), we have the general solution of the differential equation, The initial conditions say y(x) = e px (c cos(qx) + c 2 sin(qx)) = e 2x (c cos(7x) + c 2 sin(7x)) 3 = y(π) = e 2π (c cos(7π) + c 2 sin(7π)) = e 2π (c ( ) + c 2 (0)) = c e 2π, where Euler s formula is used, 2 = y (π) = 2e 2π (c cos(7π) + c 2 sin(7π)) + e 2π ( 7c sin(7π) + 7c 2 cos(7π)) = 2e 2π (c ( ) + c 2 (0)) + e 2π ( 7c (0) + 7c 2 ( )) = e 2π (2c + 7c 2 ), e 7πi = cos(7π) + isin(7π) = cosπ + isinπ =, e 7πi = cos(7π) isin(7π) = cosπ isinπ = That is, e (2+7i)π = e 2π = e (2 7i)π Solving two equation coming from the initial conditions, we get c = 3e 2π and 2 = e 2π [ 2 ( 3e 2π) + 7c 2 ], 2 = 6 7c2 e 2π, c 2 = 8 7 e 2π Page 47 of 93

48 Therefore, we conclude that the solution of the initial value problem is y(x) = e 2x (c cos(7x) + c 2 sin(7x)) = e (3e 2x 2π cos(7x) 8 ) 7 e 2π sin(7x) = e 2x 2π ( 3cos(7x) 8 7 sin(7x) ) = e 2(x π) ( 3cos(7x) 8 7 sin(7x) ) ANSWER 2 USING (245) The differential equation has the characteristic equation, λ 2 4λ + 53 = 0, which has the complex roots λ = 2 ± 7i = p ± iq So by the formula (245), we have the general solution of the differential equation, The initial conditions say y(x) = c e (p+iq)x + c 2 e (p iq)x = c e (2+7i)x + c 2 e (2 7i)x 3 = y(π) = c e (2+7i)π + c 2 e (2 7i)π, 2 = y (π) = c (2 + 7i)e (2+7i)π + c 2 (2 7i)e (2 7i)π Using Euler s formula, e 7πi = cos(7π) + isin(7π) = cosπ + isinπ =, e 7πi = cos(7π) isin(7π) = cosπ isinπ = That is, e (2+7i)π = e 2π = e (2 7i)π It allows us to simplify the equations coming from the initial conditions, 3 = c e (2+7i)π + c 2 e (2 7i)π = e 2π (c + c 2 ), ie, c + c 2 = 3e 2π, 2 = c (2 + 7i)e (2+7i)π + c 2 (2 7i)e (2 7i)π = e 2π [c (2 + 7i) + c 2 (2 7i)] = e 2π [2(c + c 2 ) + 7i(c c 2 )] We have two equations and two unknowns c and c 2 So we can solve the system of equations for the unknowns c and c 2 Solving the system of equations, we have 2 = 6 7ie 2π (c c 2 ), c c 2 = 8 7i e 2π = i 8 7 e 2π, where the identity /i = i is used Finally, we deduce c = 2 (3e 2π + i 87 e 2π ) = e 2π 2 Therefore, the solution of the initial value problem is y(x) = c e (2+7i)x + c 2 e (2 7i)x = e 2π 2 [( = e 2π 3 + i 8 ) e (2+7i)x ( 3 + i 8 ), c 2 = (3e 2π i 87 ) ( 7 2 e 2π = e 2π 3 i 8 ) 2 7 ( 3 + i 8 )e (2+7i)x + e 2π 7 2 ( 3 i 8 7 We have given two answers above Are those two answers same? That is, Answer = e 2(x π) ( 3cos(7x) 8 7 sin(7x) ) ( 3 i 8 ) e (2 7i)x 7 )e (2 7i)x ] Page 48 of 93

49 [( Same? = e 2π 3 + i 8 ) ( e (2+7i)x + 3 i 8 ] )e (2 7i)x = Answer (Please see the Appendix 245) The example above suggests the following preference for a differential equation in the complex root case If the problem has the initial conditions, ie, for the initial value problem in the complex roots case, it is recommended that the general solution (246) should be used If the problem does not have a initial condition, ie, when we are asked to find only the general solution, one may use either (245) or (246) However, personally, I would recommend the second one (246) Exercise 245 Solve the differential equations y + y 2y = 0 with y(0) = 4 and y (0) = 5 y + y + 4 y = 0 with y(0) = 3 and y (0) = 7/2 y + 04y + 904y = 0 with y(0) = 0 and y (0) = 3 y 6y 7y = 0 4y 20y + 25y = 0 y + 2y + 5y = 0 y + 5y + 6y = 0 with y(0) = 2 and y (0) = 3 (Final Exam of Fall 2009) Remark 246 (SUMMARY) Solve the constant coefficient second order differential equation y +Ay +By = 0, where A and B are real numbers We find and solve the characteristic equation λ 2 + Aλ + B = 0, λ = A ± A 2 4B 2 Case If the characteristic equation has two distinct real roots λ = a and λ = b, then the general solution of the differential equation is y = c e ax + c 2 e bx Case 2 If the characteristic equation has repeated roots λ = a, then the general solution of the differential equation is y = c e ax + c 2 xe ax = (c + c 2 x)e ax Case 3 If the characteristic equation has complex roots λ = p ± qi, then the general solution of the differential equation is y = e px (c cos(qx) + c 2 sin(qx)) or y = c e (p+iq)x + c 2 e (p iq)x 245 Appendix Let us answer to the following question: Answer = e 2(x π) ( 3cos(7x) 8 7 sin(7x) ) [( Same? = e 2π 3 + i 8 ) ( e (2+7i)x + 3 i 8 ] )e (2 7i)x = Answer We will expand and simplify the right hand side and obtain the left hand side [( Answer 2 = e 2π 3 + i 8 ) ( e (2+7i)x + 3 i 8 ] )e (2 7i)x x e 2π = e 2 2x e 2π = e 2 [( 3 + i 8 ) ( e 7ix + 3 i 8 ] )e 7ix 7 7 [ 3 ( e 7ix + e 7ix) + i 8 7 ( e 7ix e 7ix)] Page 49 of 93

50 Now we develop amazing identities from Euler s formula: e 7ix = cos(7x) + isin(7x), e 7ix = cos(7x) isin(7x) Adding/Substracting those two equations, we get e 7ix + e 7ix = 2cos(7x), e 7ix e 7ix = 2isin(7x) Theorem 247 (IDENTITIES ON EXPONENTIAL AND TRIGONOMETRIC FUNCTIONS) For any real x, By the Theorem, we have Answer 2 = e 2π 2 cosx = exi + e xi, sinx = exi e xi 2 2i 2x e 2π = e 2 2x e 2π = e 2 = e2x 2π 2 [( 3 + i 8 ) ( e (2+7i)x + 3 i 8 ] )e (2 7i)x 7 7 [ 3 ( e 7ix + e 7ix) + i 8 ( e 7ix e 7ix)] 7 [6cos(7x) + i 87 ] 2isin(7x) [6cos(7x) 67 sin(7x) ] = e 2(x π) [ 3cos(7x) 8 7 sin(7x) ] = Answer ( because i 2 = ) Page 50 of 93

51 25 Euler s Equation Throughout this section, we consider only the homogeneous second order differential equation with certain coefficient, y + A x y + B x 2 y = 0, or x2 y + Axy + By = 0 (x > 0) (25) where A and B are real numbers The equation is called Euler s equation Since the domain of y(x) is positive real numbers (x > 0), we can use the change of variable, x(t) = e t, or equivalently t(x) = lnx, and so y(x) = y(x(t)) = (y x)(t) Let Y (t) = (y x)(t) Then by the Chain Rule in CALCULUS I, Y dy (t) (t) = dt Y (t) = dy (t) dt = d dt (y x)(t) = d (y x)(t)dx dx dt = d dx = d ( y (x)x ) = d ( y (x)x ) dx dt dx dt (y(x)) dx dt = y (x) dx dt = y (x)e t = y (x)x, = ( y (x)x + y (x) ) dx dt = ( y (x)x + y (x) ) e t = ( y (x)x + y (x) ) x = y (x)x 2 + y (x)x From those results, we deduce xy (x) = Y (t), x 2 y (x) = Y (t) xy (x) = Y (t) Y (t) Hence, Euler s equation (25) becomes in terms of Y (t), Y (t) Y (t) + AY (t) + BY (t) = 0, Y (t) + (A )Y (t) + BY (t) = 0, (252) which is a constant coefficient differential equation for Y (t) In the previous Section 24 Constant Coefficient Homogeneous Linear Equation, we have studied on how to solve the equation (252) We solve the equation (252) and then later we put x = e t or t = lnx back to get the solution of (252) in terms of x In carrying out the strategy, it is useful to recall that for x > 0 and any real number b, x b = e blnx Remark 25 What should we memorize? We should memorize the equation (252) Please do not forget that we have to express the solution of the equation (252) in terms of x, because the original equation (25) is given in terms of x Remark 252 (ANOTHER STRATEGY) There is another strategy for Euler s equation Please see the Appendix 25, which I personally recommend, although we will follow the strategy developed above, because it is given in the textbook Example 253 Solve the Euler equation y + 2 x y 6 x 2 y = 0, ie, x2 y + 2xy 6y = 0 ANSWER Let x = e t Then by the change of variable x = e t or t = lnx, the equation becomes Y (t) + (2 )Y (t) 6Y (t) = 0, ie, Y (t) +Y (t) 6Y (t) = 0, (253) which is a constant coefficient homogeneous linear differential equation We use the technique in the previous section 24 It has the characteristic equation 0 = λ 2 + λ 6 = (λ 2)(λ + 3), Page 5 of 93

52 which has two distinct real roots λ = 2 and λ = 3 So the equation (253) has the general solution Y (t) = c e 2t + c 2 e 3t, where c and c 2 are arbitrary constants Now using the change of variable (x = e t or t = lnx) again, we deduce the solution of the original equation y(x) = c x 2 + c 2 x 3, (x > 0) because e 2t = (e t ) 2 = x 2 and e 3t = (e t ) 3 = x 3 or e 2t = e 2lnx = x 2 and e 3t = e 3lnx = x 3 Example 254 Solve the Euler equation y 5 x y + 9 x 2 y = 0, ie, x2 y 5xy + 9y = 0 ANSWER Let x = e t Then by the change of variable x = e t or t = lnx, the equation becomes Y (t) + ( 5 )Y (t) + 9Y (t) = 0, ie, Y (t) 6Y (t) + 9Y (t) = 0, (254) which is a constant coefficient homogeneous linear differential equation We use the technique in the previous section 24 It has the characteristic equation 0 = λ 2 6λ + 9 = (λ 3) 2, which has repeated root λ = 3 So the equation (254) has the general solution Y (t) = (c + c 2 t)e 3t, where c and c 2 are arbitrary constants Now using the change of variable (x = e t or t = lnx) again, we deduce the solution of the original equation y(x) = (c + c 2 lnx)x 3, (x > 0) because e 3t = (e t ) 3 = x 3 or e 3t = e 3lnx = x 3 Example 255 Solve the Euler equation y + 3 x y + 0 x 2 y = 0, ie, x2 y + 3xy + 0y = 0 ANSWER Let x = e t Then by the change of variable x = e t or t = lnx, the equation becomes Y (t) + (3 )Y (t) + 0Y (t) = 0, ie, Y (t) + 2Y (t) + 0Y (t) = 0, (255) which is a constant coefficient homogeneous linear differential equation We use the technique in the previous section 24 It has the characteristic equation which has complex roots So the equation (255) has the general solution λ 2 + 2λ + 0 = 0, λ = ± i 3 Y (t) = e t (c cos(3t) + c 2 sin(3t)), where c and c 2 are arbitrary constants Now using the change of variable (x = e t or t = lnx) again, we deduce the solution of the original equation y(x) = x (c cos(3lnx) + c 2 sin(3lnx)), (x > 0) because e t = (e t ) = x or e t = e lnx = x Page 52 of 93

53 As usual, we can solve an initial value problem by finding the general solution of the differential equation and then finding the constants satisfying the initial conditions Example 256 Solve the initial value problem x 2 y 5xy + 0y = 0, y() = 4, y () = 6 ANSWER We observe that the given differential equation is Euler s equation Let x = e t Then by the change of variable x = e t or t = lnx, the equation becomes Y (t) + ( 5 )Y (t) + 0Y (t) = 0, ie, Y (t) 6Y (t) + 0Y (t) = 0, (256) which is a constant coefficient homogeneous linear differential equation We use the technique in the previous section 24 It has the characteristic equation λ 2 6λ + 0 = 0, which has complex roots So the equation (256) has the general solution λ = 3 ± i Y (t) = e 3t (c cost + c 2 sint), where c and c 2 are arbitrary constants Now using the change of variable (x = e t or t = lnx) again, we deduce the solution of the original equation y(x) = x 3 (c cos(lnx) + c 2 sin(lnx)), (x > 0) because e 3t = (e t ) 3 = x 3 or e 3t = e 3lnx = x 3 Now we use the initial conditions to determine the constants c and c 2 y() = 4 implies The solution becomes 4 = y() = 3 (c cos(ln) + c 2 sin(ln)) = c cos0 + c 2 sin0 = c, c = 4 y(x) = x 3 (4cos(lnx) + c 2 sin(lnx)) ( y (x) = 3x 2 (4cos(lnx) + c 2 sin(lnx)) + x 3 4sin(lnx) x + c 2 cos(lnx) ) x = 3x 2 (4cos(lnx) + c 2 sin(lnx)) + x 2 ( 4sin(lnx) + c 2 cos(lnx)) = (2 + c 2 )x 2 cos(lnx) + (3c 2 4)x 2 sin(lnx) The other initial condition y () = 6 yields 6 = y () = (2 + c 2 ) 2 cos(ln) + (3c 2 4) 2 sin(ln) = 2 + c 2, c 2 = 8 Therefore, finally, we get the solution of the initial value problem y(x) = x 3 (4cos(lnx) 8sin(lnx)) = 2x 3 (2cos(lnx) 9sin(lnx)) (x > 0) Remark 257 (IMPORTANT OBSERVATION) Let us observe the structure of the solutions Constant coefficient linear equation (Section 24) has the solution of form y + Ay + By = 0 (257) e αx, xe αx, e αx cos(βx), e αx sin(βx) Page 53 of 93

54 2 Euler s equation (This Section) y + A x y + B x 2 y = 0 or x2 y + Axy + By = 0 (258) has the solution of form x r, x r lnx, x p cos(qlnx), x p sin(qlnx) By the observation above, we can say x 3 cannot be a solution of the equation (257) and e 6x cannot be a solution of Euler s equation (258) Exercise 258 Solve the differential equation x 2 y 2y = 0, x > 0 (Final Exam of Fall 2009) Remark 259 (SUMMARY) Solve the Euler equation y + A x y + B x 2 y = 0, or x2 y + Axy + By = 0 (x > 0) where A and B are real numbers By using the technique in the previous section 24, we solve the constant coefficient equation Y (t) + (A )Y (t) + BY (t) = 0 Once we have the general solution, we express it in terms of x by using x = e t or t = lnx 25 Appendix Part I Basic Transformation We introduce another strategy for Euler s equation y + A x y + B x 2 y = 0, or x2 y + Axy + By = 0 (x > 0) (259) where A and B are real numbers We look for the solution in the form of y(x) = x λ, where λ is a constant to be determined Putting it into the equation (259), we get 0 = x 2 ( x λ ) + Ax ( x λ ) + Bx λ = x 2 ( λ (λ )x λ 2) + Axλx λ + Bx λ = x λ (λ (λ ) + Aλ + B) = x λ ( λ 2 + (A )λ + B ) Since x > 0, so x λ > 0 and thus we have the auxiliary equation λ 2 + (A )λ + B = 0, (250) which has roots λ = A + ± (A ) 2 4B 2 = A 2 ± 4 ( A)2 B Page 54 of 93

55 Case Two Real Distinct Roots Suppose the equation (250) has two real distinct roots a = A ( A)2 B, b = A 2 4 ( A)2 B In this case, solutions of (259) are y (x) = x a, y 2 (x) = x b They are linearly independent through the Wronskian Test Thus, the general solution of (259) is where c and c 2 are arbitrary constants Example 250 Solve the differential equation y(x) = c y (x) + c 2 y 2 (x) = c x a + c 2 x b, (25) y + 3 2x y 2x 2 y = 0, or x2 y xy y = 0 (x > 0) 2 ANSWER From the given equation, we deduce the equation on the power of x λ like (250) ( ) 3 λ λ 2 = 0, 0 = 2λ 2 + λ = (λ + )(2λ ), λ =, λ = 2 By the formula (25), we have the general solution where c and c 2 are arbitrary constants Case 2 Repeated Root y(x) = c x + c 2 x /2, Suppose ( A) 2 4B = 0 in the equation (250) Then the equation has repeated root In this case, one solution of (259) is λ = A 2 y (x) = x A 2 To get another solution y 2 (x) = u(x)y (x) = u(x)x A 2 which is linearly dependent to y (x), we use the method of Reduction of Order in Section 23 Then we obtain another solution y 2 (x) = u(x)x A 2 = (lnx)x A 2 (How? Why? Check/Deduce y 2 (x) by yourselves) Therefore, the general solution of (259) is y(x) = c y (x) + c 2 y 2 (x) = c x A 2 + c 2 (lnx)x A 2 = (c + c 2 lnx)x A 2 (252) where c and c 2 are arbitrary constants Example 25 Solve the differential equation y 5 x y + 9 x 2 y = 0, or x2 y 5xy + 9y = 0 (x > 0) ANSWER From the given equation, we deduce the equation on the power of x λ like (250) λ 2 + ( 5 )λ + 9 = 0, 0 = λ 2 6λ + 9 = (λ 3) 2, λ = 3 By the formula (252), we have the general solution where c and c 2 are arbitrary constants y(x) = (c + c 2 lnx)x A 2 = (c + c 2 lnx)x 3, Page 55 of 93

56 Case 3 Complex Roots Suppose ( A) 2 4B < 0 in the equation (250) Then the equation has complex roots λ = A ± i 2 4 ( A)2 B When we let λ = p ± iq, p = A 2, q = 4 ( A)2 B, we observe y (x) = x p cos(qlnx) and y 2 (x) = x p sin(qlnx) are solutions of the original equation (259) and also they are linearly independent Therefore, the general solution of (259) is y(x) = c x p cos(qlnx) + c 2 x p sin(qlnx) = x p (c cos(qlnx) + c 2 sin(qlnx)), (253) where c and c 2 are arbitrary constants Example 252 Solve the differential equation y x y + 4 x 2 y = 0, or x2 y xy + 4y = 0 (x > 0) ANSWER From the given equation, we deduce the equation on the power of x λ like (250) λ 2 + ( )λ + 4 = 0, 0 = λ 2 2λ + 4, λ = ± i 3 By the formula (253), we have the general solution y(x) = x p (c cos(qlnx) + c 2 sin(qlnx)) = x ( ( ) ( )) c cos 3 lnx + c 2 sin 3 lnx, where c and c 2 are arbitrary constants Remark 253 (SUMMARY) Solve the Euler equation where A and B are real numbers We find and solve its auxiliary equation y + A x y + B x 2 y = 0, or x2 y + Axy + By = 0 (x > 0) λ 2 + (A )λ + B = 0 Depending on the roots, we have three different general solutions of the Euler equation, y(x) = c x a + c 2 x b, y(x) = (c + c 2 lnx)x a, y(x) = x p (c cos(qlnx) + c 2 sin(qlnx)) Part II Generalization (Purely Optional) We consider a little bit generalized Euler equation y + A B ax + b y + (ax + b) 2 y = 0, x > b a, where A, B, a and b are constants If a = and b = 0, then we have Euler s equation which we have seen We introduce a change of variable to solve the equation Note that the strategy introduced in the textbook/lecture note is also based on a change of variable Page 56 of 93

57 We take the change of variable u = ax + b, x = u b a, du dx = a, dx du = a and so y(x) = y(x(u)) = (y x)(u) Let Y (u) = (y x)(u) Then by the Chain Rule, Y (u) = dy du = d Y (u) = dy (u) du du (y x)(u) = d dx (y x)(u) dx du = d dx (y(x)) dx du = y (x) dx du = y (x) a, = d ( ) du a y (x) = d ( y (x) ) = d a du a dx ( y (x) ) dx du = a y (x) dx du = a 2 y (x) Thus we have y (x) = a 2 Y (u) and y (x) = ay (u) and so the Euler equation for y(x) becomes a 2 Y (u) + A u ay (u) + B u 2 Y (u) = 0, ie, Y (u) + A/a u Y (u) + B/a2 u 2 Y (u) = 0 The resulting differential equation has the characteristic equation ( ) A λ 2 + a λ + B a 2 = 0, ie, a2 λ 2 + a(a a)λ + B = 0, which has the roots a(a a) ± a 2 (A a) 2 4a 2 B (A a) 2 4B λ = 2a 2 = A a 2a ± 2a Just like other strategies, we have three cases: two distinct real roots, repeated root, and two complex roots Case Two Distinct Real Roots: Suppose the characteristic equation has two distinct real roots λ = α and λ = β Then we have two solutions and the general solution is Y (u) = u α, Y 2 (u) = u β Y (u) = c Y (u) + c 2 Y 2 (u) = c u α + c 2 u β, where c and c 2 are arbitrary constants Therefore, the general solution y(x) of the original Euler equation is Example 254 Solve the differential equation y(x) = c (ax + b) α + c 2 (ax + b) β ( + 2x) 2 y + 8( + 2x)y 6y = 0, x > 2 ANSWER We use the change of variable u = 2x + and let Y (u) = y(x) Then we have y (x) = Y (u) du dx = 2Y (u), and the equation is transformed It has the characteristic equation y (x) = 4Y (u) 4u 2 Y (u) + 6uY (u) 6Y (u) = 0, ie, Y + 4 u Y 4 u 2 Y = 0 0 = λ 2 + 3λ 4 = (λ + 4)(λ ), λ = 4, λ = Thus, we have two solutions Y (u) = u 4 and Y 2 (u) = u and the general solution Y (u) = c Y (u) + c 2 Y 2 (u) = c u 4 + c 2 u Therefore, finally the general solution of the original differential equation is obtained by y(x) = c (2x + ) 4 + c 2 (2x + ) For the Case 2 Repeated Root and Case 3 Complex Roots, we can argue similarly as we did for other strategies Page 57 of 93

58 26 Nonhomogeneous Equation y + p(x)y + q(x)y = f (x) Suppose that we can find the general solution y h of the linear homogeneous equation Then the linear nonhomogeneous equation y + p(x)y + q(x)y = 0 (26) y + p(x)y + q(x)y = f (x) (262) has the general solution y = y h + y p, where y p is any particular solution of the nonhomogeneous equation (262) That is, the strategy for solving (262) is as follows: Step Find the general solution y h of its homogeneous equation (26) Step 2 Find a particular solution y p of the nonhomogeneous equation (262) Step 3 We conclude the general solution of the equation (262) is y = y h + y p In Sections 23, 24 and 25, we have studied on how to find the general solution of a homogeneous equation So the work on the Step can be done by the strategies in those previous sections In this section, we mainly focus on the Step 2, ie, how to find a particular solution y p of a nonhomogeneous equation (262) We will discuss two methods: Method of Variation of Parameters and Method of Undetermined Coefficients 26 Method of Variation of Parameters We recall that if y and y 2 forms the fundamental set of solutions of the homogeneous equation (26), then the general solution of (26) is given by where c and c 2 are arbitrary constants We look for a particular solution y p of the equation (262) in the form y h (x) = c y (x) + c 2 y 2 (x), (263) y p (x) = u(x)y (x) + v(x)y 2 (x), (264) where u(x) and v(x) will be determined and it is formed by replacing constants c and c 2 in (263) with functions u(x) and v(x) This technique is called the method of variation of parameters Let us find u(x) and v(x) explicitly Since y p in (264) should be a particular solution of (262), it should satisfy f (x) = y p + p(x)y p + q(x)y p = (uy + vy 2 ) + p(uy + vy 2 ) + q(uy + vy 2 ) = ( u y + v y 2 + uy + vy 2) + p ( u y + v y 2 + uy + vy 2) + q(uy + vy 2 ) = ( u y + v y 2 ) + ( uy + vy 2) + p ( u y + v y 2 ) + p ( uy + vy 2) + q(uy + vy 2 ) = ( u y + v ) ( y 2 + u y + uy + v y 2 + vy ( 2) + p u y + v ) ( y 2 + p uy + vy 2) + q(uy + vy 2 ) = u ( y + py ) ( + qy + v y 2 + py ) ( 2 + qy 2 + u y + v ) ( y 2 + u y + v y ) ( 2 + p u y + v ) y 2 = u(0) + v(0) + ( u y + v ) ( y 2 + u y + v y ) ( 2 + p u y + v ) y 2 = ( u y + v y 2 ) + ( u y + v y 2) + p ( u y + v y 2 ), where y + py +qy = 0 and y 2 + py 2 +qy 2 = 0, because y and y 2 are solutions of (26) Suppose we give a condition on u and v as follows u y + v y 2 = 0 Then the equation above is simplified by f (x) = ( u y + v ) ( y 2 + u y + v y ) ( 2 + p u y + v ) y 2 Page 58 of 93

59 = (0) + ( u y + v y 2) + p(0) = u y + v y 2 Therefore, we can deduce two equations for u and v: u y + v y 2 = 0 and u y + v y 2 = f (x) Since we have two equations and two unknowns u and v, we can solve for u and v, in fact, u = y 2 f W, v = y f W, u = ˆ y2 f W dx, v = ˆ y f W dx, (265) where W(x) is the Wronskian of y and y 2, ie, W(x) = y (x)y 2 (x) y (x)y 2(x) (Since y and y 2 forms the fundamental set of solutions, we should have W(x) 0 for all x) Finally, we obtain a particular solution y p (x) of the nonhomogeneous equation (262), y p (x) = u(x)y (x) + v(x)y 2 (x) = [ ˆ ] [ˆ ] y2 (x) f (x) y (x) f (x) dx y (x) + dx y 2 (x) (266) W(x) W(x) The general solution of the nonhomogeneous equation (262) is eventually obtained by [ y(x) = y h (x) + y p (x) = c y (x) + c 2 y 2 (x) + [ ˆ ] y2 (x) f (x) = c dx W(x) Remark 26 ˆ y2 (x) f (x) y (x) + W(x) [ c 2 + ] dx ˆ y (x) f (x) [ˆ y (x) f (x) y (x) + W(x) W(x) ] dx y 2 (x) ] dx y 2 (x) Suppose f (x) = 0 Then the nonhomogeneous equation (262) becomes the homogeneous equation (26) and the particular solution y p (x) of the nonhomogeneous equation in (266) becomes [ ˆ ] [ˆ ] y2 (x)0 y p (x) = W(x) dx y (x)0 y (x) + W(x) dx y 2 (x) [ ˆ ] [ˆ ] = 0dx y (x) + 0dx y 2 (x) = c y (x) + c 2 y 2 (x) = y h (x), which is the general solution of the homogeneous equation 2 Please do not try to memorize all of them In the method of variation of parameters, the results (265) are worthwhile to be memorized 3 One should not be confused In the formula (266), the coefficient of y (x) has y 2 (x) in the integral, while the coefficient of y 2 (x) has y (x) in the integral When we have two solutions, it is up to us to choose which one is y and which one is y 2 But, once we fixed them, they should be carefully plugged into the formula (266) We summarize the steps of the method Theorem 262 (METHOD OF VARIATION OF PARAMETERS) A particular solution of y + py +qy = f can be found through the following steps Step General Solution We find the general solution y h = c y +c 2 y 2 of the equation y + py +qy = 0 Step 2 Method of Variation of Parameters We set y p = uy + vy 2 and find u and v satisfying u y + v y 2 = 0 and u y + v y 2 = f Page 59 of 93

60 In fact, those u and v are obtained by ˆ ˆ y2 (x) f (x) y (x) f (x) u(x) = dx, v(x) = dx, W(x) W(x) where W(x) is the Wronskian of y and y 2 Hence, the particular solution y p is obtained by ( ˆ ) (ˆ ) y2 (x) f (x) y (x) f (x) y p = uy + vy 2 = dx y + dx y 2 W(x) W(x) Now let us see how to use the results above through the examples Example 263 A differential equation is given by y + 4y = secx, π 4 < x < π 4 Step y h Find two solutions y and y 2 and the general solution y h (x) = c y (x) + y 2 (x) of its homogeneous equation y + 4y = 0, π/4 < x < π/4 Step 2 y p Find a particular solution y p = uy + vy 2 of the given nonhomogeneous equation Step 3 y = y h + y p Find the general solution y = y h + y p of the given nonhomogeneous equation ANSWER Step y h The homogeneous equation has the characteristic equation λ = 0, λ = ±2i Thus, we have two solutions y (x) = e 0x cos(2x) = cos(2x), and y 2 (x) = e 0x sin(2x) = sin(2x), and the general solution of the homogeneous equation is y h (x) = c y (x) + c 2 y 2 (x) = c cos(2x) + c 2 sin(2x) (267) Step 2 y p We find a particular solution y p (x) = u(x)y (x) + v(x)y 2 (x), precisely, u(x) and v(x): y p (x) = u(x)cos(2x) + v(x)sin(2x), by using the formula (266) The Wronskian of y (x) = cos(2x) and y 2 (x) = sin(2x) is found by W(x) = y y 2 y y 2 = 2cos(2x)cos(2x) + 2sin(2x)sin(2x) = 2 ( cos 2 (2x) + sin 2 (2x) ) = 2 The formula (266) implies u (x) = 2 sin(2x)secx = 2sinxcosxsecx = sinx, 2 v (x) = 2 cos(2x)secx = ( 2cos 2 x ) secx = cosx 2 2 secx Integrating them, we get ˆ ˆ u(x) = sinxdx = cosx, v(x) = (cosx 2 ) secx dx = sinx ln secx + tanx 2 (We do not have to put the constants of integration, because this is the work for finding a particular solution y p We can take the constants to be zero) Therefore, we deduce a particular solution of the nonhomogeneous equation y p (x) = u(x)cos(2x) + v(x)sin(2x) = cosxcos(2x) + (sinx 2 ) ln secx + tanx sin(2x) (268) Page 60 of 93

61 Step 3 y = y h + y p Since we have the general solution y h in (267) of the homogeneous equation and a particular solution y p in (268) of the nonhomogeneous equation, therefore, the general solution of the nonhomogeneous equation is given by y(x) = y h (x) + y p (x) = c cos(2x) + c 2 sin(2x) + cosxcos(2x) + (sinx 2 ) ln secx + tanx sin(2x) = (c + cosx)cos(2x) + (c 2 + sinx 2 ) ln secx + tanx sin(2x), where c and c 2 are arbitrary constants Example 264 A differential equation is given by y 4 x y + 4 x 2 y = x2 +, (x > 0) Step y h Find two solutions y and y 2 and the general solution y h (x) = c y (x) + y 2 (x) of its homogeneous equation y (4/x)y + (4/x 2 )y = 0, x > 0 Step 2 y p Find a particular solution y p = uy + vy 2 of the given nonhomogeneous equation Step 3 y = y h + y p Find the general solution y = y h + y p of the given nonhomogeneous equation ANSWER Step y h The homogeneous equation has the auxiliary equation Thus, we have two solutions 0 = λ 2 + ( 4 )λ + 4 = λ 2 5λ + 4 = (λ )(λ 4), λ =, λ = 4 y (x) = x, and y 2 (x) = x 4, and the general solution of the homogeneous equation is y h (x) = c y (x) + c 2 y 2 (x) = c x + c 2 x 4 (269) (For the method used in Step, please read the Appendix 25 in the lecture note on Section 25 Euler s Equation) Step 2 y p We find a particular solution y p (x) = u(x)y (x) + v(x)y 2 (x), precisely, u(x) and v(x): y p (x) = u(x)x + v(x)x 4, by using the formula (266) The Wronskian of y (x) = x and y 2 (x) = x 4 is found by W(x) = y y 2 y y 2 = x ( 4x 3) ( x 4) = 3x 4, which is not zero for x > 0 The formula (266) implies ( u (x) = x4 x 2 + ) 3x 4 = ( x 2 + ), v (x) = x( x 2 + ) 3 3x 4 = 3 Integrating them, we get u(x) = 3 ˆ (x 2 + ) dx = 3 ( x + ) x 3 ( ) 3 x3 + x, v(x) = ˆ ( 3 x + ) x 3 dx = ( lnx ) 3 2x 2 (We do not have to put the constants of integration, because this is the work for finding a particular solution y p We can take the constants to be zero) Therefore, we deduce a particular solution of the nonhomogeneous equation y p (x) = u(x)x + v(x)x 4 = ( ) 3 3 x3 + x x + ( lnx ) 3 2x 2 x 4 = 9 x4 2 x2 + 3 x4 lnx (260) Page 6 of 93

62 Step 3 y = y h + y p Since we have the general solution y h in (269) of the homogeneous equation and a particular solution y p in (260) of the nonhomogeneous equation, therefore, the general solution of the nonhomogeneous equation is given by y(x) = y h (x) + y p (x) = c x + c 2 x 4 9 x4 2 x2 + 3 x4 lnx ( = c x + c 2 ) x x2 + 3 x4 lnx, (x > 0) where c and c 2 are arbitrary constants 262 Method of Undetermined Coefficients The method in this topic is only applicable for constant coefficient nonhomogeneous equation y + Ay + By = f (x), (26) where A and B are constants The goal is to find a particular solution y p of (26) without finding the solution y h of its homogeneous part Theorem 265 (METHOD OF UNDETERMINED COEFFICIENTS) Step From f (x), make a first conjecture for the form of y p (How? See the table below) Step 2 Solve its homogeneous equation y + Ay + By = 0 If a solution of the homogeneous equation appears in any term of the conjectured form for y p, then modify this form by multiplying it by x 2 If the modified function still occurs in a solution of the homogenous equation, then multiply it by x again (so the original y p is multiplied by x 2 in this case) Step 3 Substitute the final proposed y p into the nonhomogeneous equation y + Ay + By = f (x) and solve for its coefficients Here is the list of functions which we may try in the Step of formulating y p P(x), Q(x) and R(x) represent polynomials of the same degree (Please memorize the following table) f (x) P(x) ce ax α cos(bx) or β sin(bx) P(x)e ax P(x)cos(bx) or P(x)sin(bx) P(x)e ax cos(bx) or P(x)e ax sin(bx) Initial Guess for y p Q(x) de ax ccos(bx) + d sin(bx) Q(x)e ax Q(x)cos(bx) + R(x)sin(bx) Q(x)e ax cos(bx) + R(x)e ax sin(bx) Let us see how the method above works through examples Example 266 (POLYNOMIAL f ) Use the method of undetermined coefficients to find a particular solution of the constant coefficient nonhomogeneous equation y 4y = 8x 2 2x (262) Page 62 of 93

63 ANSWER Step Since f (x) = 8x 2 2x is a polynomial of degree 2, we guess that a particular solution y p may be a polynomial of degree 2, and so we set where a, b and c are constants to be determined later Step 2 The homogeneous equation y 4y = 0 y p (x) = ax 2 + bx + c, (263) has two solutions y (x) = e 2x and y 2 (x) = e 2x and its general solution y h (x) = c e 2x + c 2 e 2x (Why? See the Section 24) Since any solution y or y 2 of the homogeneous equation does not appear in the conjectured form (263) for y p, so we move to next step Step 3 Putting the guess (263) into the given equation (262), we have 8x 2 2x = y p 4y p = 2a 4ax 2 4bx 4c, 4(a + 2)x 2 + 2(2b )x + 2(2c a) = 0, for all x In general, a n x n + a n x n + + a 2 x 2 + a x + a 0 = 0 if and only if a n = 0 = a n = = a 2 = a = a 0 for all x So we should get a + 2 = 0, and 2b = 0, and 2c a = 0, ie, a = 2, b = /2 and c = and the particular solution is y p (x) = 2x x One can verify by substituting it into the given nonhomogeneous equation General Solution of (262) The general solution y(x) = y h (x) + y p (x) of the nonhomogeneous equation (262) is given by y(x) = y h (x) + y p (x) = c e 2x + c 2 e 2x 2x x Example 267 (EXPONENTIAL FUNCTION f ) Use the method of undetermined coefficients to find a particular solution of the constant coefficient nonhomogeneous equation y + 2y 3y = 4e 2x (264) ANSWER Step Since f (x) = 4e 2x is an exponential function, we guess that a particular solution y p may be an exponential function, and so we set y p (x) = ae 2x, (265) where a is a constant to be determined later Step 2 The homogeneous equation y + 2y 3y = 0 has two solutions y (x) = e 3x and y 2 (x) = e x and its general solution y h (x) = c e 3x + c 2 e x (Why? See the Section 24) Since any solution y or y 2 of the homogeneous equation does not appear in the conjectured form (265) for y p, so we move to next step Step 3 Putting the guess (265) into the given equation (264), we have 4e 2x = y p + 2y p 3y p = 4ae 2x + 4ae 2x 3ae 2x = 5ae 2x, ie, e 2x (5a 4) = 0 Page 63 of 93

64 An exponential function can never be zero So we should get and the particular solution is 5a 4 = 0, a = 4 5, y p (x) = 4 5 e2x One can verify by substituting it into the given nonhomogeneous equation General Solution of (264) The general solution y(x) = y h (x) + y p (x) of the nonhomogeneous equation (264) is given by y(x) = y h (x) + y p (x) = c e 3x + c 2 e x e2x Example 268 (TRIGONOMETRIC FUNCTION f ) Use the method of undetermined coefficients to find a particular solution of the constant coefficient nonhomogeneous equation y 5y + 6y = 3sin(2x) (266) ANSWER Step Since f (x) = 3 sin(2x) is a trigonometric function, we guess that a particular solution y p may be a combination of trigonometric functions, and so we set where a and b are constants to be determined later Step 2 The homogeneous equation y 5y + 6y = 0 y p (x) = acos(2x) + bsin(2x), (267) has two solutions y (x) = e 3x and y 2 (x) = e 2x and its general solution y h (x) = c e 3x + c 2 e 2x (Why? See the Section 24) Since any solution y or y 2 of the homogeneous equation does not appear in the conjectured form (267) for y p, so we move to next step Step 3 Putting the guess (267) into the given equation (266), we have 3sin(2x) = y p 5y p + 6y p In general, = 4acos(2x) 4bsin(2x) 5( 2asin(2x) + 2bcos(2x)) + 6(acos(2x) + bsin(2x)) ie, 0 = 2(a 5b)cos(2x) + (0a + 2b + 3)sin(2x), for all x Acosx + Bsinx = 0 for all x if and only if A = 0 = B In LINEAR ALGEBRA or maybe CALCULUS 4, we have the following theorem: Suppose two functions p(x) and q(x) are linearly independent Then, Ap(x) + Bq(x) = 0 for all x if and only if A = 0 = B Since cos(2x) and sin(2x) are linearly independent (the Wronskian Test), so we should get a 5b = 0 and 0a + 2b + 3 = 0 Solving the equations for a and b, we get a = 5/52 and b = 3/52 and the particular solution is y p (x) = 5 52 cos(2x) 3 52 sin(2x) One can verify by substituting it into the given nonhomogeneous equation General Solution of (266) The general solution y(x) = y h (x) + y p (x) of the nonhomogeneous equation (266) is given by y(x) = y h (x) + y p (x) = c e 3x + c 2 e 2x 5 52 cos(2x) 3 52 sin(2x) Page 64 of 93

65 There are problems that the main ideas in the method of undetermined coefficients may not work, however, still we can solve the problem with a slight modification of the method Let us see an example Example 269 Use the method of undetermined coefficients to find a particular solution of the differential equation y + 2y 3y = 8e x (268) ANSWER Step Since f (x) = 8e x is an exponential function, we guess that a particular solution y p may be an exponential function, and so we set y p (x) = ae x, (269) where a is a constant to be determined later Step 2 The homogeneous equation y + 2y 3y = 0 has two solutions y (x) = e 3x and y 2 (x) = e x and its general solution y h (x) = c e 3x + c 2 e x (Why? See the Section 24) Since the solution y 2 = e x of the homogeneous equation does appear in the conjectured form (269) for y p, so we modify our guess by multiplying by x and try a new guess y p (x) = axe x (2620) Step 3 Putting the new guess (2620) into the given equation (268), we have 8e x = y p + 2y p 3y p = 2ae x + axe x + 2(ae x + axe x ) 3axe x = 4ae x, 8e x = 4ae x, ie, a = 2 Hence, we deduce a particular solution y p (x) = 2xe x One can verify by substituting it into the given nonhomogeneous equation General Solution of (268) The general solution y(x) = y h (x) + y p (x) of the nonhomogeneous equation (268) is given by y(x) = y h (x) + y p (x) = c e 3x + c 2 e x + 2xe x Exercise 260 Use the method of undetermined coefficients to find a particular solution of the differential equation 2y y + y = x + e x (Final Exam of Fall 2009) If we use the method of undetermined coefficients to find a particular solution y p and eventually the general solution y = y h + y p, then it is suggested to follow the steps in the theorem Theorem 26 (GENERAL SOLUTION OF NONHOMOGENEOUS EQUATION) Step General Solution y h of Homogeneous Equation Step 2 Particular Solution y p Step 3 General Solution y = y h + y p of Nonhomogeneous Equation Example 262 Solve the nonhomogeneous equation y 6y + 9y = 5e 3x ANSWER Step General Solution y h To avoid a useless guess on a particular solution, we first find the general solution y h of its homogeneous equation y 6y + 9y = 0 It has the characteristic equation 0 = λ 2 6λ + 9 = (λ 3) 2, Page 65 of 93

66 which has repeated root λ = 3 Hence, the general solution y h is obtained by y h (x) = (c + c 2 x)e 3x Step 2 Particular Solution y p Since the general solution y h has terms e 3x and xe 3x, so we try a particular solution y p (x) = ax 2 e 3x, where e 3x in y p comes from the function f (x) = 5e 3x Putting y p (x) = ax 2 e 3x into the nonhomogeneous equation, we have 5e 3x = ( ax 2 e 3x) 6 ( ax 2 e 3x) + 9y = 2ae 3x + 2ae 3x + 9ax 2 e 3x 6 ( 2axe 3x + 3ax 2 e 3x) + 9ax 2 e 3x = 2ae 3x, ie, a = 5 2 Thus, we have a particular solution y p (x) = 2 5 x2 e 3x Step 3 General Solution From Step and Step 2, we conclude the general solution of the nonhomogeneous equation is y(x) = y h (x) + y p (x) = (c + c 2 x)e 3x x2 e 3x = (c + c 2 x + 52 ) x2 e 3x Example 263 Solve the nonhomogeneous equation y + 9y = 4xsin(3x) ANSWER Step General Solution y h To avoid a useless guess on a particular solution, we first find the general solution y h of its homogeneous equation It has the characteristic equation y + 9y = 0 λ = 0, which has two complex roots λ = ±3i Hence, the general solution y h is obtained by y h (x) = c cos(3x) + c 2 sin(3x) Step 2 Particular Solution y p For the function f (x) = 4xsin(3x), we follow the suggested form in the table above So we try a particular solution y p (x) = (ax + b)cos(3x) + (cx + d)sin(3x) But unfortunately, it does not work So we modify by multiplying it by x, ie, we try y p (x) = x(ax + b)cos(3x) + x(cx + d)sin(3x) = ( ax 2 + bx ) cos(3x) + ( cx 2 + dx ) sin(3x) Putting it into the nonhomogeneous equation, we have 4xsin(3x) = [( ax 2 + bx ) cos(3x) + ( cx 2 + dx ) sin(3x) ] ie, + 9 [( ax 2 + bx ) cos(3x) + ( cx 2 + dx ) sin(3x) ], 0 = (2a + 6d)cos(3x) + ( 6b + 2c)sin(3x) + 2cxcos(3x) + ( 2a + 4)xsin(3x), for all x In LINEAR ALGEBRA or CALCULUS 4 ON BASIS, we have the following theorem: Suppose functions p(x), q(x), r(x) and s(x) are linearly independent Then, Ap(x) + Bq(x) +Cr(x) + Ds(x) = 0 for all x if and only if A = 0, B = 0, C = 0 and D = 0 Page 66 of 93

67 Since cos(3x), sin(3x), xcos(3x) and xsin(3x) are linearly independent (Wronskian Test), we should get 2a + 6d = 0, and 6d + 2c = 0, and 2c = 0, and 2a + 4 = 0 Solving them for a, b, c and d, we have Thus, we have a particular solution a = 3, b = 0 = c, d = 9 y p (x) = 3 x2 cos(3x) 9 xsin(3x) Step 3 General Solution From Step and Step 2, we conclude the general solution of the nonhomogeneous equation is y(x) = y h (x) + y p (x) = c cos(3x) + c 2 sin(3x) + 3 x2 cos(3x) 9 xsin(3x) = (c + 3 ) x2 cos(3x) + (c 2 9 ) x sin(3x) There are certain nonconstant coefficient nonhomogeneous equation which can be transformed constant coefficient nonhomogeneous equation For this case, we may use the method of undetermined coefficients on the transformed constant coefficient equation and transform the solution back to the original equation See the example below Example 264 Solve the nonconstant coefficient nonhomogeneous differential equation y 5 x y + 8 x 2 y = 2lnx x 2 or x 2 y 5xy + 8y = 2lnx, (x > 0) of which homogeneous equation is Euler s equation ANSWER Step Transform to Constant Coefficient Equation Since the given equation has Euler s equation as its homogeneous equation This observation makes us to use the change of variable x = e t or equivalently t = lnx The change of variable implies that the given nonhomogeneous equation turns to be Y (t) 6Y (t) + 8Y (t) = 2t (How to convert? See the Section 25 Euler s Equation) Since the transformed equation has constant coefficients, we use the method of undetermined coefficients Step 2 General Solution Y h From the Section 25 Euler s Equation, the homogeneous equation Y (t) 6Y (t) + 8Y (t) = 0 has the general solution Y h (t) = c e 2t + c 2 e 4t Step 3 Particular Solution Y p By the method of undetermined coefficients, the transformed equation has the particular solution Y p (t) = 4 t Step 4 General Solution From Steps above, we conclude the transformed equation has the general solution Y (t) = Y h (t) +Y p (t) = c e 2t + c 2 e 4t + 4 t Our purpose is the general solution of the original equation Thus we use the change of variable again to deduce the general solution of the original equation y(x) = c x 2 + c 2 x lnx Page 67 of 93

68 263 Principle of Superposition We consider the equation Suppose y pi is a solution of the equation y + p(x)y + q(x)y = f (x) + f 2 (x) + + f N (x) (262) y + p(x)y + q(x)y = f i (x), where i =,2,,N Then, y p + y p2 + + y pn Principle of Superposition is a solution of the equation (262) This is called the PROOF We put y p + y p2 + + y pn into the equation (262) (y p + y p2 + + y pn ) + p(x)(y p + y p2 + + y pn ) + q(y p + y p2 + + y pn ) = ( y p + py p + qy p ) + ( y p2 + py p 2 + qy p2 ) + + ( y pn + py p N + qy pn ) = f + f f N Thus, the combination is really the solution of the given differential equation This principle allows us to separate the differential equation into smaller problems Example 265 Solve the differential equation y + 4y = x + 2e 2x ANSWER General Solution y h We first solve the homogeneous equation y + 4y = 0 It has the general solution y h (x) = c cos(2x) + c 2 sin(2x) Particular Solution y p For a particular solution, we separate the differential equation into two pieces y + 4y = x and y + 4y = e 2x and solve each of them separately Problem y + 4y = x Method of undetermined coefficients gives a solution y p (x) = x 4 Problem 2 y + 4y = e 2x Method of undetermined coefficients gives a solution y p2 (x) = e 2x 4 By the principle of superposition, thus, the original equation has a particular solution y p (x) = y p (x) + y p2 (x) = x 4 + e 2x 4 = ( x + ) 4 e 2x General Solution From Steps above, the general solution of the given differential equation is y(x) = y h (x) + y p (x) = c cos(2x) + c 2 sin(2x) + 4 ( x + ) e 2x Page 68 of 93

69 264 Higher Order Differential Equations As the last topic in this section, we consider a differential equation having order more than 2 Let us consider, for example, a 6 th order differential equation y (6) 4y (4) + 2y + 5y = 0 Since it has constant coefficients, if we look for a solution of form e λx, then we have the characteristic equation λ 6 4λ 4 + 2λ + 5 = 0, which has 6 complex roots, say, p ± iq, p 2 ± iq 2, and p 3 ± iq 3 Hence, the general solution is y(x) = e p (a cos(q x) + a 2 sin(q x)) + e p 2 (b cos(q 2 x) + b 2 sin(q 2 x)) + e p 3 (c cos(q 3 x) + c 2 sin(q 3 x)), where a i, b i and c i, i =,2, are arbitrary constants Is there a better way to solve the given higher order differential equation? We use the series of change of variable as follows: z = y, z 2 = y, z 3 = y, z 4 = y (3), z 5 = y (4), z 6 = y (5) Then, the given differential equation is expressed by a system z = z 2, z 2 = z 3, z 3 = z 4, z 4 = z 5, z 5 = z 6, z 6 = 4z 5 2z 2 5z Remark 266 (ASIDE) Using the Matrix notation (which will be discussed in LINEAR ALGEBRA or CALCULUS 4), the system is formed as follows: z z 2 ( z,z 2,,z ) 6 = z 3 or z = Az t, z z where A is the matrix and z = (z,,z 6 ) and z = ( z,,z 6) are vectors and z t is the transpose of z So simply speaking, the problem looks like dz dx = Az, which is a separable equation with the general solution, z = e Ax How can we define and compute the matrix power of an exponential function? References for Graduate Study in Math: Richard Bellman s Introduction to Matrix Analysis, Vladimir Arnold s Ordinary Differential Equations Far and Far Beyond the level of this course 27 Application of Second Order Differential Equations to a Mechanical System Skip Please read the textbook z 6 Page 69 of 93

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71 Chapter 3 Laplace Transform So far through Calculus courses and previous Chapters, we have seen many techniques on change of variable In this chapter, we introduce a very special and powerful tool for solving a differential equation, which transforms an initial value problem itself to an algebra problem The process is as follows Initial Value Problem Step Laplace Transform Algebra Problem Difficult Step 2 Algebraic Methods Solution of Initial Value Problem Step 3 Inverse Laplace Transform Solution of Algebra Problem 3 Definition and Basic Properties Definition 3 The Laplace Transform L [ f ] of f is a function defined by for all s such that this integral converges L [ f ](s) = ˆ 0 e st f (t)dt, The Laplace transform converts a function f into a new function L [ f ] Often we use t as the independent variable of f and s as the independent variable of the function L [ f ] It is convenient to use a capital letter for the Laplace transform, eg, and so on F = L [ f ], G = L [g], H = L [h], Example 32 For f (t) = e at, with a any real number, find its Laplace transform L [ f ](s) ANSWER By definition, L [ f ](s) = F(s) = ˆ 0 = lim b a s e st e at dt = ˆ 0 [e (a s)t] b e (a s)t dt = lim 0 = lim b a s b ˆ b 0 [ e (a s)b e (a s)t dt ] = a s [ ] = s a, provided by a s < 0, s > a Thus the Laplace transform of f (t) = e at is F(s) = s a defined for s > a Example 33 For g(t) = sint, find its Laplace transform L [g](s) ANSWER By definition, L [g](s) = G(s) = Thus G(s) = defined for all real s s 2 + ˆ 0 e st sint dt = lim = lim b s 2 + b ˆ b 0 e st sint dt [ e bs cosb + se bs sinb ] = s 2 + 7

72 Whenever we need a Laplace transform of a function, should we compute and find it? Clearly, it is not efficient So we use the table of the Laplace transform See the Appendix 3 Theorem 34 (LINEARITY OF LAPLACE TRANSFORM) Suppose L [ f ](s) and L [g](s) are defined for s > a and α and β are real numbers Then, for s > a, (Constant Multiple) L [α f ](s) = αl [ f ](s) 2 (Sum) L [ f + g](s) = L [ f ](s) + L [g](s) The two formulas can be compressed into the following one formula, L [α f + βg](s) = αl [ f ](s) + βl [g](s), which is called the linearity property of Laplace transform PROOF Since L [ f ](s) and L [g](s) are well defined for s > a, the integrals 0 e st f (t)dt and 0 e st g(t)dt converge for s > a It implies that for s > a, L [α f + βg](s) = = α ˆ 0ˆ 0 e st (α f (t) + βg(t)) dt e st f (t)dt + β ˆ 0 e st g(t)dt = αl [ f ](s) + βl [g](s) We can generalize the formula to finitely many functions Under the appropriate circumstances, we have [ n L α k f k ](s) = L [α f + α 2 f α n f n ](s) k= n = α L [ f ](s) + α 2 L [ f 2 ](s) + + α n L [ f n ](s) = α k L [ f k ](s) k= Now we focus on functions Let us refresh our memory on CALCULUS I & II On differentiation, can any function be differentiable? No! Then, what kind of function can be differentiable? We studied on the differentiable function To be a differentiable function, it should be at least continuous 2 On integration, can any function be integrable? No! Then, what kind of function can be integrable? We studied on the integrable function To be an integrable function, it should be continuous (roughly speaking, because we can even integrate a function having finite number of jump discontinuities) As we can see, the continuity is very important in differentiation and integration Moreover, it goes to our subject, Laplace transform Can any function have the Laplace transform? No! Then, what kind of function can have the Laplace transform? What is the requirement on the function? To have the Laplace transform, the function should be at least piecewise continuous Definition 35 (PIECEWISE CONTINUITY) A function f is said to be piecewise continuous on the interval [a,b] if there are points a = t 0 < t < t 2 < < t n < t n = b such that f is continuous on each open interval (t j,t j ), j =,2,,n and 2 all of the following one sided limits are finite, lim t t + 0 f (t), lim t t j f (t), lim t t + j f (t), lim t t n f (t), j =,2,,n Page 72 of 93

73 Differential Equations and Engineering Applications Fall, 200 In other words, f is piecewise continuous on [a,b] if it is continuous there except for a finite number of jump discontinuities If f is piecewise continuous on [a,b] for every b > a, then f is said to be piecewise continuous on [a, ) See the figures 3 and 32 The figure 32 is the graph of the piecewise continuous function f (t) defined by t 2 for 0 t < 2 2 at t = 2 f (t) = for 2 < t 3 for 3 < t 4 a = t 0 t t 2 t 3 b = t 4 t Figure 3: Example of piecewise continuous function on [a,b] y(t) t Figure 32: Example of piecewise continuous function f (t) on [0,4] If f is piecewise continuous on [0,k] for every positive k, then does its Laplace transform exist? The answer is no Under the given condition on f, e st f (t) is piecewise continuous on [0,k] and so k 0 e st f (t)dt does k exist However, lim k 0 e st f (t)dt may not converge and the Laplace transform of f may not exist For instance, f (t) = e t2 is continuous everywhere but 0 e st e t2 dt does not converge for every real value of s For this reason, we need one more condition on f to guarantee the existence of its Laplace transform The following theorem gives the conditions for the existence of the Laplace transform Page 73 of 93

74 Theorem 36 (EXISTENCE OF L [ f ]) Suppose f is piecewise continuous on [0,k] for every positive k 2 Suppose there are numbers M and b such that f (t) Me bt for t 0 Then 0 e st f (t)dt converges for s > b and hence the Laplace transform L [ f ](s) of f is defined for s > b ROUGH PROOF From the first condition, we deduce that e st f (t) is piecewise continuous on [0,k] and so k 0 e st f (t)dt does exist Now we use the second condition, which implies, for s b, but, lim k ˆ k 0 f (t) Me bt, e st f (t) Me (b s)t, Me (b s)t dt = { M s b diverges So by comparison, for s > b, ˆ ˆ e st f (t)dt e st f (t) dt 0 0 for b s < 0, ie, s > b for b s 0, ie, s b ˆ 0 Me (b s)t dt = M s b, ie, for s > b, the Laplace transform L [ f ](s) = 0 e st f (t)dt exists Example 37 Elementary functions having Laplace transforms are polynomials 2 trigonometric functions sin(at) and cos(at) 3 exponential functions e at where a is a constant Let us consider the converse of the Theorem 36 If a function has the Laplace transform, then should it always satisfy both conditions in the Theorem 36? The answer is no Even though a function has a Laplace transform, it may not satisfy both conditions For instance, the function f (t) = t /2, for t > 0, has the Laplace transform, L [ f ](s) = ˆ 0 e st t /2 dt Subst:x=t/2 = 2 ˆ 0 e sx2 dx Subst:z=x s = ˆ 2 e z2 dz = s where the fact 0 e z2 dz = π 2 is used However, the function f (t) = t /2 is not piecewise continuous on [0,k] for every positive k, precisely, lim t 0 + t /2 = and so it does not satisfy the first condition in the Theorem 36 Next, we introduce the inverse Laplace transform Definition 38 (INVERSE LAPLACE TRANSFORM) Given a function G, a function g such that L [g] = G is called an inverse Laplace transform of G In this event, we write g = L [G] Example 39 Since f (t) = e at has the Laplace transform L [ f ](s) = s a, so the inverse Laplace transform of F(s) = s a is f (t) = eat, ie, L [ e at] (s) = s a, Since f (t) = sint has the Laplace transform L [ f ](s) = is f (t) = sint, ie, L [sint](s) = s 2 +, [ ] eat = L (t) s a s π s, so the inverse Laplace transform of F(s) = s 2 + [ sint = L s 2 + ] (t) Page 74 of 93

75 To find the inverse Laplace transform, we use the same table as the one for the Laplace transform Let us consider two functions, { e f (t) = e t t for t 3 and g(t) = 0 for t = 3 We observe L [ f ](s) = s+ = L [g](s) for s >, that is, f and g have the same Laplace transform Then? That is, which one of f and g will be the inverse Laplace transform of s+ [ ] [ L (t) = f (t)? or L s + s + The following theorem gives the answer to this question ] (t) = g(t)? Theorem 30 (LERCH S THEOREM) Let f and g be continuous on [0, ) and suppose L [ f ] = L [g] Then f = g That is, if two continuous functions have the same Laplace transform, then they should be same Thus, we can deduce the uniqueness of the inverse Laplace transform, provided the inverse Laplace transform is continuous Notice that the function g(t) presented before the theorem is not continuous Because of the linearity of the Laplace transform, its inverse is also linear Theorem 3 (LINEARITY OF INVERSE LAPLACE TRANSFORM) If L [F] = f and L [G] = g and α and β are real numbers, then L [αf + βg] = αl [F] + βl [G] = α f + βg Page 75 of 93

76 3 Appendix (Table of Laplace Transform) f (t) F(s) = L [ f ](s) t s s 2 t n n! s n+ e at s a te at (s a) 2 sin(at) cos(at) t sin(at) t cos(at) e at sin(bt) e at cos(bt) a s 2 + a 2 s s 2 + a 2 2as (s 2 + a 2 ) 2 s 2 a 2 (s 2 + a 2 ) 2 b (s a) 2 + b 2 s a (s a) 2 + b 2 Here n! = n(n )(n 2) 3(2)() and 0! = Page 76 of 93

77 32 Solution of Initial Value Problem Using the Laplace Transform We use the Laplace transform to solve an initial value problem Since a derivative y (t) of a function y(t) is also a function, with appropriate conditions, we can find the Laplace transform L [y ](s) of the derivative y (t) and thus we can change the differential equation of y(t) into the equation of its Laplace transform L [y](s) For this purpose, we need the following theorem Theorem 32 (LAPLACE TRANSFORM OF DERIVATIVE) Suppose f is continuous on [0, ) and has the limit lim k e sk f (k) = 0 for s > 0 and f is piecewise continuous on [0,k) for every positive k Then, with F(s) = L [ f ](s), L [ f ] (s) = sf(s) f (0) PROOF The Integration by Parts formula with u = e st and dv = f (t)dt implies, for k > 0, lim k ˆ k 0 ˆ k 0 e st f (t)dt = ˆ k e st f (t)dt = lim k 0 ˆ k udv = [uv] k 0 vdu 0 = [ e st f (t) ] sˆ k t=k t=0 + e st f (t)dt = e ks f (k) f (0) + s 0 [ ] e ks f (k) f (0) + s ˆ k L [ f ] (s) = sl [ f ](s) f (0) = sf(s) f (0) 0 e st f (t)dt ˆ k 0 = f (0) + s lim k e st f (t)dt ˆ k 0 e st f (t)dt In the case that f has a jump discontinuity at t = 0, the formula becomes L [ f ](s) = sf(s) f (0+), where f (0+) = lim t 0 + f (t) Now we apply the Laplace transform to the higher derivatives Recall that f ( j) means the j th derivative of f and f (0) = f Theorem 322 (LAPLACE TRANSFORM OF HIGHER DERIVATIVE) Suppose f, f,, f (n ) are continuous on [0, ), lim k e sk f ( j) (k) = 0 for s > 0 and j =,2,,n and f (n) is piecewise continuous on [0, k) for every positive k Then, with F(s) = L [ f ](s), [ L f (n)] (s) = s n F(s) s n f (0) s n 2 f (0) s f (n 2) (0) f (n ) (0) n = s n F(s) s n i f (i ) (0), The Theorems 32 and 322 imply the following corollary Corollary 323 i= L [ f ] (s) = sf(s) f (0) L [ f ] (s) = s 2 F(s) s f (0) f (0) [ L f (3)] (s) = s 3 F(s) s 2 f (0) s f (0) f (0) Now we are ready to solve an initial value problem using the Laplace transform Example 324 Solve the initial value problem y (t) = 3 with y(0) = 5 ANSWER Applying the Laplace transform to the differential equation y (t) = 3, we get L [ y ] (s) = L [3](s) = 3L [](s), sy (s) y(0) = 3 s, sy (s) 5 = 3 s, Y (s) = 3 s s Applying the inverse Laplace transform to the resulting function Y (s) = s 2 s for s > 0, we get [ 3 y(t) = L [Y ](t) = L s [ ] [ ] s ](t) = 3L s 2 (t) + 5L (t) = 3t + 5 s Page 77 of 93

78 Example 325 Solve the initial value problem y 4y = 0, y(0) = ANSWER Applying the Laplace transform to the differential equation, we get L [ y 4y ] (s) = L [0](s), L [ y ] (s) 4L [y](s) = L [0](s), sy (s) y(0) 4Y (s) = 0, (s 4)Y (s) =, Y (s) = (s > 4) s 4 Applying the inverse Laplace transform to the resulting function Y (s) = s 4 [ ] y(t) = L [Y ](t) = L (t) = e 4t s 4 for s > 4, Example 326 Solve the initial value problem y + 4y + 3y = e t, y(0) = 0, y (0) = 2 ANSWER Applying the Laplace transform to the differential equation, we get L [ e t] (s) = L [ y + 4y + 3y ] (s) = L [ y ] (s) + 4L [ y ] (s) + 3L [y](s) s = s2 Y (s) sy(0) y (0) + 4[sY (s) y(0)] + 3Y (s) = s 2 Y s(0) 2 + 4[sY 0] + 3Y 2s s = s + 2 = ( s 2 + 4s + 3 ) Y Y (s) = 2s (s )(s 2 + 4s + 3) (s > ) Before we apply the inverse Laplace transform, we need to separate the fraction partially, ie, Partial Fraction Technique in CALCULUS I should be used, because the table on the Laplace transform does not provide 2s the (inverse) Laplace transform for the fraction, which one can check by yourself (s )(s 2 +4s+3) We separate the fraction as follows: 2s (s )(s 2 + 4s + 3) = 2s (s )(s + )(s + 3) = A s + B s + + C s + 3, where A, B and C are constants to be determined Multiplying the equation by (s )(s + )(s + 3), 2s = A(s + )(s + 3) + B(s )(s + 3) +C(s )(s + ) Putting s =, s = and s = 3, we have respectively 2() = A(2)(4) + B(0) +C(0), = 8A, A = 8 2( ) = A(0) + B( 2)(2) +C(0), 3 = 4B, B = 3 4 2( 3) = A(0) + B(0) +C( 4)( 2), 7 = 8C, C = 7 8 2s Y (s) = (s )(s 2 + 4s + 3) = A s + B s + + C s + 3 = 8 s s s + 3 Now applying the inverse Laplace transform to the resulting function for s >, we get [ y(t) = L [Y ](t) = L 8 s s + 7 ] (t) 8 s + 3 = [ ] 8 L (t) + 3 [ ] s 4 L (t) 7 [ ] s + 8 L (t) = s et e t 7 8 e 3t Page 78 of 93

79 Example 327 Solve the initial value problem y + y = t, y(0) =, y (0) = 0 ANSWER Applying the Laplace transform to the differential equation, we get L [ y + y ] (s) = L [t](s), L [ y ] (s) + L [y](s) = L [t](s), s 2 Y (s) sy(0) y (0) +Y (s) = s 2, s2 Y s 0 +Y = s 2, ( s 2 + ) Y = s 2 + s, Y (s) = s 2 (s 2 + ) + s s 2 + (s > 0) Applying the inverse Laplace transform to the resulting function for s > 0, we get [ y(t) = L [Y ](t) = L s 2 (s 2 + ) + s ] s 2 (t) + [ [ ] = L ](t) s 2 (s 2 + L s + ) s 2 (t) = t sint + cost + An interesting consequence of the Laplace transform can be found as follows We recall Suppose f (t) is defined by Then f (t) = g(t) and f (0) = 0 So we observe L [ f ] = sl [ f ] f (0) f (t) = ˆ t 0 g(τ) dτ L [g] = L [ f ] = sl [ f ] f (0) = sl [ f ] 0 = sl [ f ], L [ f ] = s L [g], L [ˆ t It enables us to take the Laplace transform of a function defined by an integral 0 ] g(τ) dτ = s L [g] Page 79 of 93

80 33 Shifting Theorems and the Heaviside Function In this section, we discuss four topics: (i) the first shifting theorem, (ii) the heaviside function and pulses, (iii) the first shifting theorem, and (iv) analysis of electrical circuits Let us compare two graphs of y = 2t and y = 2(t 5) We observe the graph of y = 2(t 5) is obtained by shifting the graph of y = 2t by 5 units along the positive t axis, ie, by moving parallelly the graph of y = 2t by 5 units along the positive t axis See the figure 33 y(t) y = 2t y = 2(t 5) 5 units t Figure 33: Graphs of y = 2t and y = 2(t 5) In general, when we shift the graph y = f (t) by a units along the positive t axis, we obtain the graph of y = f (t a) Thanks to the shifting relation, once we know the properties of y = f (t), we can deduce the properties of y = f (t a) easily We will see how greatly this shifting technique works in the Laplace transform 33 The First Shifting Theorem Theorem 33 (FIRST SHIFTING THEOREM) Let L [ f ](s) = F(s) for s > b 0 Let a be any number Then L [ e at f (t) ] (s) = F(s a) for s > a + b Moreover, the inverse Laplace transform yields L [F(s a)] = e at f (t), ie, L [F(s a)] = e at L [F(s)] The theorem says that the Laplace transform of e at f (t) is exactly same as the Laplace transform of f (t) shifted a units to the positive s axis PROOF A simple computation shows L [ e at f (t) ] (s) = ˆ 0 e at e st f (t)dt = Example 332 Since L [cos(bt)] = ˆ 0 e (s a)t f (t)dt = F(s a) (s a > b, s > a + b) s s 2 +b 2, so L [e at cos(bt)] = Example 333 Since L [ t 3] = 6 s 4, so L [ t 3 e 7t] = 6 (s 7) 4 s a (s a) 2 +b 2 Exercise 334 Find the Laplace transform of f (t) = e t cos(2t) 2sin(3t) (Final Exam of Fall 2009) Page 80 of 93

81 Remark 335 (VERTEX FORM OF QUADRATIC FUNCTION FROM CALCULUS I) The complete square technique for a quadratic function f (s) = as 2 + bs + c with constants a, b and c, is very useful to find the inverse Laplace transform of the shifted function The question is how to convert f (s) in the form of f (s) = a(s ) 2 +? From CALCULUS I, f (s) = as 2 + bs + c has the critical number s = 2a b (To refresh your memory, f (s) = 2as + b = 0 at s = b 2a It gives the following equation, which is the critical number) At the critical number, f has the value ( f b ) ( = a b ) 2 + b( b ) + c = b2 4ac 2a 2a 2a 4a ( f (s) = as 2 + bs + c = a s + b ) 2 b2 4ac, 2a 4a ( which is called the vertex form of the quadratic function f with the vertex point Example 336 Find L [ ] 4 s 2 +4s+20 b 2a, b2 4ac 4a ANSWER First we find the vertex form of the denominator s 2 + 4s + 20 Using the technique above, we deduce s 2 + 4s + 20 = (s + 2) , s 2 + 4s + 20 = 4 (s + 2) It enables us to think of a function 4 F(s + 2) = (s + 2) , F(s) = 4 s Now we apply the shifting theorem, [ ] L [F(s + 2)] = e 2t L [F(s)] = e 2t sin(4t), ie, L 4 s 2 = e 2t sin(4t) + 4s + 20 Example 337 Compute L [ ] 3s s 2 6s+2 ANSWER First we find the vertex form of the denominator s 2 6s + 2 By the technique above, we deduce s 2 6s + 2 = (s 3) 2 7, 3s s 2 6s + 2 = 3s 3(s 3) (s 3) 2 = 7 (s 3) (s 3) 2 = G(s 3) + K(s 3), 7 where G(s) = 3s s 2 7, K(s) = 8 s 2 7 Now we apply the shifting theorem, [ ] 3s L s 2 = L [G(s 3) + K(s 3)] = L [G(s 3)] + L [K(s 3)] 6s + 2 = e 3t L [G(s)] + e 3t L [K(s)] = e 3t ( L [G(s)] + L [K(s)] ) ( [ ] [ ]) [ 3s 8 = e 3t L s 2 + L 7 s 2 = e (3L 3t s 7 s 2 7 = e 3t (3cosh ) ] [ ]) + 8L s 2 7 ( 7t ) sinh( 7t ) ) (in mathematics) the point where two lines meet to form an angle, or the point that is opposite the base of a shape Page 8 of 93

82 Differential Equations and Engineering Applications Fall, The Heaviside Function and Pulses We introduce a special function to deal with a nonhomogeneous differential equation whose forcing function has finite number of jump discontinuities We recall that f has a jump discontinuity at a if both lim t a f (t) and lim t a + f (t) exist and are finite but unequal The magnitude of the jump discontinuity is defined by the width of the gap, ie, the width is lim f (t) lim f (t) t a t a + See the figure 34 y(t) Magnitude of the jump discontinuity at a a t Figure 34: Jump Discontinuity at a Definition 338 The Heaviside function H is defined by { 0 for t < 0 H(t) = for t 0 See the figure 35 y(t) 0 t Figure 35: The Heaviside function H(t) For any number a, the Heaviside function shifted a units to the right is easily obtained by { 0 for t < a H(t a) = for t a See the figure 36 By the definition of the shifted Heaviside function, we can obtain { 0 for t < a H(t a)g(t) = g(t) for t a Page 82 of 93

83 Differential Equations and Engineering Applications Fall, 200 y(t) 0 a t Figure 36: A shifted Heaviside function H(t a) Example 339 For g(t) = cost for all t and with a = π, we have See the figure 37 H(t π)g(t) = H(t π)cost = { 0 for t < π cost for t π y(t) = cost y(t) = H(t π)cost 0 π 2π 3π t 0 π 2π 3π t Figure 37: Comparison of y = cost and y = H(t π)cost Definition 330 A pulse is a function of the form 0 for t < a H(t a) H(t b) = for a t < b 0 for t b See the figure 38 (a < b) y(t) 0 a b t Figure 38: Pulse function H(t a) H(t b) Example 33 For g(t) = e t for all t with a = and b = 2, we have 0 for t < [H(t a) H(t b)]g(t) = [H(t ) H(t 2)]e t = e t for t < 2 0 for t 2 See the figure 39 Page 83 of 93

84 Differential Equations and Engineering Applications Fall, 200 y(t) t Figure 39: Graph of f (t) = [H(t ) H(t 2)]e t Next we consider the function of the form H(t a)g(t a), explicitly, { 0 for t < a H(t a)g(t a) = g(t a) for t a So we observe the graph of H(t a)g(t a) is obtained by the graph of g(t) shifted a units for t a and zero for t < a Example 332 For g(t) = t 2 and a = 2, See the figure 30 H(t a)g(t a) = H(t 2)(t 2) 2 = { 0 for t < 2 (t 2) 2 for t 2 y(t) 6 y(t) t t Figure 30: Comparison of y = t 2 and y = (t 2) 2 H(t 2) It is important to understand the difference between g(t), H(t a)g(t) and H(t a)g(t a) See the figure 3 which is about the graphs of y = t 2, y = t 2 H(t 3) and y = (t 3) 2 H(t 3) 333 The Second Shifting Theorem Sometimes H(t a)g(t a) is referred to as a shifting function and the second shifting theorem deals with the Laplace transform of the shifting function Theorem 333 (SECOND SHIFTING THEOREM) Let L [ f ](s) = F(s) Then, L [H(t a) f (t a)](s) = e as F(s) for s > b, ie, L [H(t a) f (t a)](s) = e as L [ f ](s) Page 84 of 93

85 Differential Equations and Engineering Applications Fall, y(t) = t 2 36 y(t) = t 2 H(t 3) 36 y(t) = (t 3) 2 H(t 3) t t t Figure 3: Comparison of y = t 2, y = t 2 H(t 3) and y = (t 3) 2 H(t 3) Moreover, the inverse Laplace transform yields L [e as F(s)](t) = H(t a) f (t a), ie, PROOF A simple computation shows L [ e as F(s) ] (t) = H(t a)l [F](t a) L [H(t a) f (t a)](s) = ˆ e st H(t a) f (t a)dt = 0 ˆ Def: H(t a) = a ˆ e st f (t a)dt Subst: = w = t a ˆ = e as e sw f (w)dw = e as L [ f ](s) Example 334 Find the Laplace transform of H(t a) 0 a e st f (t a)dt ˆ 0 e s(a+w) f (w)dw ANSWER Let f (t) = for all t Then, f (t a) = for any a and t and so we have H(t a) = H(t a) = H(t a) f (t a) By the Theorem 333, we get L [H(t a)](s) = e as L [ f ](s) = e as L [](s) = s e as Example 335 Compute L [g] where g(t) = { 0 for 0 t < 2 t 2 + for t 2 ANSWER We observe that g(t) = H(t 2)(t 2 + ) Since the right hand side has the Heaviside function shifted 2 units, we change the multiple t 2 + as follows: t 2 + = (t 2 + 2) 2 + = (t 2) 2 + 4(t 2) + 5 Then we get [ ] g(t) = H(t 2)(t 2 + ) = H(t 2) (t 2) 2 + 4(t 2) + 5 = H(t 2)(t 2) 2 + 4H(t 2)(t 2) + 5H(t 2) By the Second Shifting Theorem 333, we deduce [ [ ]] L [g] = L H(t 2) (t 2) 2 + 4(t 2) + 5 Page 85 of 93

86 [ = L H(t 2)(t 2) 2] + 4L [H(t 2)(t 2)] + 5L [H(t 2)] = e 2s L [ t 2] + 4e 2s L [t] + 5 s e 2s = e 2s 2 s 3 + 4e 2s s s e 2s = e 2s [ 2 s s s Exercise 336 Find the Laplace transform of the function { t for 0 t < 2 g(t) = 2e 3t for t 2 Example 337 Compute L [ ] se 3s s 2 +4 ] ANSWER By the Second Shifting Theorem 333, we have [ se L 3s ] [ ] s 2 = H(t 3)L s + 4 s 2 (t 3) = H(t 3)cos(2(t 3)), + 4 because L [ ] s = cos(2t) by table of Laplace transforms s 2 +4 Exercise 338 Find the inverse Laplace transform of the function F(s) = e 2s s 2 +3s+2 Now we are ready to solve certain initial value problem involving discontinuous forcing functions Example 339 Solve the initial value problem y + 4y = f (t), y(0) = 0 = y (0), where { 0 for t < 3 f (t) = t for t 3 ANSWER The problem can be rewritten by y + 4y = H(t 3)t, y(0) = 0 = y (0) We apply the Laplace transform, L [ y + 4y ] = L [H(t 3)t], s 2 Y sy(0) y (0) + 4Y = L [H(t 3)t], To determine L [H(t 3)t], we use the Second Shifting Theorem 333, L [ y ] + 4L [y] = L [H(t 3)t], ( s ) Y = L [H(t 3)t], Y = s 2 L [H(t 3)t] + 4 H(t 3)t = H(t 3)[t 3 + 3] = H(t 3)(t 3) + 3H(t 3) L [H(t 3)t] = L [H(t 3)(t 3) + 3H(t 3)] = L [H(t 3)(t 3)] + 3L [H(t 3)] ( = e 3s L [t] + 3e 3t L [] = e 3s (L [t] + 3L []) = e 3s s ) s Hence, we have Y (s) = [ s 2 L [H(t 3)t] = + 4 s e 3s s ] = 3s + s s 2 (s 2 + 4) e 3s To get the solution of the initial value problem, we need to apply the inverse Laplace transform, [ ] 3s + y(t) = L [Y (s)](t) = L s 2 (s 2 + 4) e 3s (t) Page 86 of 93

87 In order to use the table of the Laplace transform, we separate the function by using the partial fraction technique, 3s + s 2 (s 2 + 4) e 3s = 3 4 s e 3s 3 s 4 s e 3s + 4 s 2 e 3s 4 s e 3s Applying the inverse Laplace transform, we deduce [ ] 3s + y(t) = L [Y (s)](t) = L s 2 (s 2 + 4) e 3s (t) = 3 [ ] 4 L s e 3s 3 [ ] 4 L s s e 3s + 4 [ ] s L 2 e 3s [ ] 4 L s e 3s = 3 4 H(t 3) 3 4 H(t 3)cos(2(t 3)) + 4 H(t 3)(t 3) 4 H(t 3) sin(2(t 3)) 2 = 4 H(t 3)t H(t 3)[6cos(2(t 3)) + sin(2(t 3))] 8 = See the figure 32 { 0 for t < 3 t 4 8 [6cos(2(t 3)) + sin(2(t 3))] for t 3 y(t) t Figure 32: Graph of the solution y(t) = 4 H(t 3)t 8 H(t 3)[6cos(2(t 3)) + sin(2(t 3))] We observe that the solution is differentiable everywhere even though the function f in the differential equation had a jump discontinuity at 3 This behavior is typical of the initial value problem having a discontinuous forcing function In general, if a differential equation has order n and ϕ is a solution Then ϕ and its first n derivatives will be continuous, while the n th derivative will have the jump discontinuity wherever f does and these jump discontinuities will agree in magnitude with the corresponding jump discontinuities of f Exercise 3320 Use the Laplace transform to solve y 4y = 5, y(0) = and y (0) = 0 As the last example under the topic, let us consider a function having two jump discontinuities Example 332 Express the given function f in terms of the pulse: 0 for t < 2 f (t) = t for 2 t < 3 4 for t 3 ANSWER First, we consider the function f only on [2, 3) On that interval, f is expressed by the pulse H(t 2) H(t 3) multiplied by t, ie, f (t) = [H(t 2) H(t 3)](t ) on [2,3) On the interval [3, ), the function f is expressed by f (t) = 4H(t 3) Therefore, we conclude f (t) = [H(t 2) H(t 3)](t ) 4H(t 3) Page 87 of 93

88 334 Analysis of Electrical Circuits Skip Please read the textbook Page 88 of 93

89 34 Convolution So far in the knowledge on mathematics, we have studied operations on functions such as addition ( f + g), subtraction ( f g), multiplication (c f and f g), division ( f /g) and composition f g In this section, we introduce another special operation on functions called the convolution ( f g) Why do we study the convolution in this chapter Laplace Transform? One of the reasons is the Convolution Theorem Definition 34 If f and g are defined on [0, ), then the convolution f g of f with g is the function defined by ( f g)(t) = ˆ t 0 f (t τ)g(τ)dτ (t 0) Is the convolution commutative? That is, f g = g f? The answer is yes, which will be discussed soon More important property is the following theorem Theorem 342 (CONVOLUTION THEOREM) If f g is defined, then PROOF Let F = L [ f ] and G = L [g] Then We recall It implies F(s)G(s) = F(s)G(s) = F(s) ˆ 0 L [ f g] = L [ f ]L [g] ˆ 0 e sτ g(τ)dτ = ˆ e sτ F(s) = L [H(t τ) f (t τ)](s) F(s)e sτ g(τ)dτ = The definition of the Laplace transform gives So the function F(s)G(s) becomes F(s)G(s) = ˆ τ= τ=0 L [H(t τ) f (t τ)](s) = ˆ 0 ˆ L [H(t τ) f (t τ)](s)g(τ)dτ = 0 0 F(s)e sτ g(τ)dτ L [H(t τ) f (t τ)](s)g(τ)dτ e st H(t τ) f (t τ)dt ˆ τ= ˆ t= τ=0 t=0 e st H(t τ) f (t τ)g(τ)dtdτ The definition of the Heaviside function, H(t τ) = 0 for t < τ, H(t τ) = for t τ, changes the interval of the integral F(s)G(s) = ˆ τ= ˆ t= τ=0 t=0 e st H(t τ) f (t τ)g(τ)dtdτ = ˆ τ= ˆ t= τ=0 t=τ e st f (t τ)g(τ)dtdτ Now we recall the techniques studied in CALCULUS II (Multiple Integrals) The region defining the last double integral is shaded as in the figure 33, explicitly, the region is R = {(t,τ) : 0 τ t} We reverse the order of integration, ie, So the double integral becomes R = {(t,τ) : 0 τ, τ t} (Horizontal Cut) = {(t,τ) : 0 t, 0 τ t} (Vertical Cut) F(s)G(s) = ˆ τ= ˆ t= τ=0 t=τ e st f (t τ)g(τ)dtdτ (Horizontal Cut) Page 89 of 93

90 = = = = e st f (t τ)g(τ)da R ˆ t= ˆ τ=t t=0 ˆ t= t=0 ˆ t= t=0 e st f (t τ)g(τ)dτdt (Vertical Cut) τ=0 [ˆ τ=t ] e st f (t τ)g(τ)dτ dt τ=0 e st ( f g)(t)dt = L [ f g](s) τ τ = t t Figure 33: Region defining double integral The inverse version of the convolution theorem is also very useful when we want to find the inverse transform of a function that is a production and we know the inverse transform of each factor Theorem 343 (INVERSE OF CONVOLUTION THEOREM) Let L [F] = f and L [G] = g Then we have L [FG] = f g, ie, L [FG] = L [F] L [G] Example 344 Compute L [ s(s 4) 2 ] There are several ways to solve this problem Let us see some of them ANSWER TABLE Let f (t) = t a ( e at ), a 0 The table of the Laplace transform gives F(s) = L [ f ](s) = a s 2 (s + a) So with a = 4, we observe F(s 4) 4 = s(s 4) 2, [ ] L s(s 4) 2 = 4 L [F(s 4)] The first shifting theorem implies [ ] L s(s 4) 2 = 4 L [F(s 4)] = 4 e4t L [F](t) = [ 4 e4t t ( e 4t )] = 4 4 e4t t 6 e4t + 6 ANSWER 2 PROGRAM Mathematica See the figure Page 90 of 93

91 ANSWER 3 PARTIAL FRACTION We can separate the quotient as follows: s(s 4) 2 = 6 s 6 s (s 4) 2 Applying the inverse Laplace transform and using the table of the Laplace transform, we get [ ] L s(s 4) 2 = [ ] 6 L [ ] s 6 L + 4 [ ] s 4 L (s 4) 2 = 6 6 e4t + 4 te4t ANSWER 4 CONVOLUTION Let us use the convolution theorem s(s 4) 2 = [ ] [ ] [ ] s (s 4) 2, L s(s 4) 2 = L L s (s 4) 2 By the table of the Laplace transform, we get [ ] [ ] [ ] L s(s 4) 2 = L L s (s 4) 2 = te 4t = Exercise 345 Use convolution to find the inverse Laplace transform of F(s) = 2 s 3 +4s The convolution operation is commutative Theorem 346 If f g is defined, then g f is also defined and f g = g f PROOF The substitution z = t τ implies ˆ t 0 τe 4τ dτ = 6 6 e4t + 4 te4t ( f g)(t) = ˆ τ=t τ=0 ˆ z=0 ˆ z=t f (t τ)g(τ)dτ = f (z)g(t z)dz = f (z)g(t z)dz = (g f )(t) z=t z=0 Now we use the convolution to solve an initial value problem Example 347 Solve the initial value problem y 2y 8y = f (t), y(0) = and y (0) = 0 Here f (t) is any function having the Laplace transform F(s) Page 9 of 93