Chapter2. Second-order differential Equations

Transcription

1 Chapter2. Second-order differential Equations 1. Linear Differential Equations If we can express a second order differential equation in the form y + p x y + q(x)y = r(x) it is called linear. Otherwise, it is nonlinear. Consider a linear differential equation. If r(x) = 0 it is called homogeneous, otherwise it is called nonhomogeneous. Some examples are: Linear Combination: A linear combination of y 1 ; y 2 is y = a y 1 + b y 2. For a homogeneous linear differential equation any linear combination of solutions is again a solution. The above result does NOT hold for nonhomogeneous equations. For example, both y = sin x and y = cos x are solutions to y + y = 0, so is y = 2 sin x + 5 cos x. Both y = sin x + x and y = cos x + x are solutions to y + y = x, but y = sin x + cos x + 2x is not. This is a very important property of linear homogeneous equations, called superposition. It means we can multiply a solution by any number, or add two solutions, and obtain a new solution. Linear Independence: Two functions y 1 ; y 2 are linearly independent if a y 1 + b y 2 = 0 a = 0; b = 0. Otherwise they are linearly dependent. (One is a multiple of the other). For example, e x and e 2x are linearly independent. e x and 2e x are linearly dependent. General Solution and Basis: Given a second order, linear, homogeneous differential equation, the general solution is: y = a y 1 + b y 2 where y 1 ; y 2 are linearly independent. The set y 1 ; y 2 is called a basis, or a fundamental set of the differential equation. As an illustration, consider the equation x 2 y 5xy + 8y = 0. You can easily check that y = x 2 is a solution. Therefore 2x 2 ; 7x 2 or x 2 are also solutions. But all these are linearly dependent. We expect a second, linearly independent solution, and this is y = x 4. A combination of solutions is also a solution, so y = x 2 + x 4 or y = 10x 2 5 x 4 are also solutions. Therefore the general solution is y = a x 2 + b x 4 and the basis of solutions is x 2, x Reduction of Order If we know one solution of a second order homogeneous differential equation, we can find the second solution by the method of reduction of order. Consider the differential equation y + p x y + q(x)y = 0 Suppose one solution y 1 is known, then set y 2 = uy 1 and insert in the equation. The result will be 1

2 y 1 u + 2y 1 + py 1 u + (y 1 + py 1 + qy 1) = 0 But y 1 is a solution, so the last term is canceled. So we have y 1 u + 2y 1 + py 1 u = 0 This is still second order, but if we set w = u, we will obtain first order equation: y 1 w + 2y 1 + py 1 = 0 Solving this, we can find w, then u and then y 2. Legendre's equation. The equation is known as Legendre's equation, after the French mathematician Adrien Marie Legendre (

3 3. Homogeneous Equations with Constant Coefficients Up to now we have studied the theoretical aspects of the solution of linear homogeneous differential equations. Now we will see how to solve the constant coefficient equation y + ay + by = 0 in practice. We have the sum of a function and its derivatives equal to zero, so the derivatives must have the same form as the function. Therefore we expect the function to be e x. If we insert this in the equation, we obtain: 2 + a + b = 0 This is called the characteristic equation of the homogeneous differential equation y + ay + by = 0. If we solve the characteristic equation, we will see three different possibilities: Two real roots, double real root and complex conjugate roots. Two Real Roots: The general solution is y = c 1 e 1x + c 2 e 2x 3

4 Double Real Root: One solution is ex but we know that a second order equation must have two independent solutions. Let's use the method of reduction of order to find the second solution. Let's insert y 2 = u e ax in the equation. Obviously, u = 0 therefore u = c 1 + c 2 x. The general solution is 4

5 y = c 1 e x + c 2 x e x Solve y + 2y + y = 0. y = e x. The characteristic equation is = 0. Its solution is the double root = +,-1, therefore the general solution is y = c 1 e x + c 2 e x Complex Conjugate Roots: We need the complex exponentials for this case. Euler's formula is e ix = cos x + i sin(x) This can be proved using Taylor series expansions. If the solution of the characteristic equation is 1 = a + i b, and 2 = a - i b then the general solution of the differential equation will be y = c 1 e ax (cos bx + i sin bx ) + c 2 e ax (cos bx i sin bx ) By choosing new constants A;B, we can express this as y = e ax (A cos bx + B sin bx ) 4. Cauchy-Euler Equation The equation x 2 y + axy + by = 0 is called the Cauchy-Euler equation. By inspection, we can easily see that the solution must be a power of x. Let's substitute y = x r in the equation and try to determine r. We will obtain r r 1 x r + arx r + bx r = 0 r 2 + a 1 r + b = 0 This is called the auxiliary equation. Once again, we have three different cases according to the types of roots. The general solution is given as follows: Two real roots: y = c 1 x r 1 + c 2 x r 2 Double real root: y = c 1 x r + c 2 x r lnx Complex conjugate roots where r 1, r 2 = r ± si. y = x r (c 1 cos s lnx + c 2 sin(s lnx)) 5

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13 6. Solution of Nonhomogeneous Equation Thus far, for differential equations of second-order and higher, we have studied only the homogeneous equation r[y] = 0. In this section we turn to the nonhomogeneous case r[y] = f(x) That is, this time we include a nonzero forcing function f(x). Before proceeding with solution techniques, let us reiterate that the function f(x) (that is, what's left on the right-hand side when the terms involving y and its derivatives are put on the left-side side) is, essentially, a forcing function, and we will call it that, in this text, as a reminder of its physical significance. For instance, we have already met the equation governing the displacement x(t) of a mechanical oscillator. Here, the forcing function is the applied force F(t). For the analogous electrical oscillator governed by the equations on the current i(t), and on the charge Q(t) on the capacitor, the forcing functions are the time derivative of the applied voltage E(t), and the applied voltage E(t), respectively. 7. General and Particular Solutions Consider the nonhomogeneous equation y + p x y + q(x)y = r(x) Let y p be a solution of this equation. Now consider the corresponding homogeneous equation y + p x y + q(x)y = 0 Let y h be the general solution of this one. If we add y h and y p, the result will still be a solution for the nonhomogeneous equation, and it must be the general solution because y h contains two arbitrary constants. This interesting property means that we need the homogeneous equation when we are solving the nonhomogeneous one. The general solution is of the form y = y h + y p Example 1. 13

14 Example 2. Example 3. Example 4. 14

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16 8. Method of Undetermined Coefficients To solve the constant coefficient equation d 2 y dy + a + by = r(x) dx2 dx Solve the corresponding homogeneous equation, y p and y h. Find a candidate for y p using the following table: (You don't have to memorize the table. Just note that the choice consists of r(x) and all its derivatives) If your choice for y p occurs in y h, you have to change it. Multiply it by x if the solution corresponds to a single root, by x 2 if it is a double root. Find the constants in y p by inserting it in the equation. The general solution is y = y h + y p Note that this method works only for constant coefficient equations, and only when r(x) is relatively simple. Example 5. 16

17 Example Method of Variation of Parameters Consider the linear second order nonhomogeneous differential equation y + p x y + q(x)y = r(x) If a(x); b(x) and c(x) are not constants, or if r(x) is not among the functions given in the table, we can not use the method of undetermined coefficients. In this case, the variation of parameters can be used if we know the homogeneous solution. Let y h = c 1 y 1 + c 2 y 2 be the solution of the associated homogeneous equation Let us express the particular solution as: a(x)y + b x y + c(x)y = 0 y p = v 1 (x)y 1 + v 2 (x)y 2 There are two unknowns, so we may impose an extra condition. Let's choose v 1 y 1 + v 2 y 2 = 0 for simplicity. Inserting y p in the equation, we obtain v 1 y 1 + v 2 y 2 = r a v 1 y 1 + v 2 y 2 = 0 The solution to this linear system is where W is the Wronskian v 1 = y 2r aw, v 2 = y 1r aw 17

18 Therefore the particular solution is Example 7. Example 8. References: 1. Greenberg, M.D. Advanced Engineering Mathematics, 2nd edition. Prentice Hall, O'Neil, P.V. Advanced Engineering Mathematics, 5th edition. Thomson, Ross, S.L. Introduction to Ordinary Differential Equations, 4th edition. Wiley,