# 4 Linear Dierential Equations

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1 Dierential Equations (part 2): Linear Dierential Equations (by Evan Dummit, 2012, v. 1.00) Contents 4 Linear Dierential Equations Terminology General Theory of Linear Dierential Equations Homogeneous Linear Equations with Constant Coecients Non-Homogeneous Linear Equations with Constant Coecients Undetermined Coecients Variation of Parameters Second-Order Equations: Applications to Newtonian Mechanics Spring Problems and Damping Resonance and Forcing Linear Dierential Equations The general nth-order linear dierential equation is of the form y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = Q(x), for some functions P n (x),, P 2 (x), P 1 (x), and Q(x). (Note that y (n) denotes the nth derivative of y.) The goal is to study the behavior of the solutions to these equations. Solving general linear dierential equations explicitly is generally hard, unless we are lucky and the equation has a particularly nice form. We can only give a method for writing down the full set of solutions for a small classes of linear equations: namely, linear dierential equations with constant coecients. There are a few equation types (e.g., Euler equations like x 2 y + xy + y = 0) which can be reduced to constant-coecient equations via substitution. Thus, we will spend most of our eort discussing how to solve linear equations with constant coecients, because we can always solve them these are equations of the form y (n) +a n 1 y (n 1) + +a 1 y +a 0 y = Q(x) for some constants a n 1,, a 0 and some function Q(x). Even harder than the general linear dierential equation are non-linear equations of higher order, such as y = e y y or y y 2 = y + 1. We will not discuss non-linear equations at all. Compare with the example of rst-order equations: we can solve any rst-order linear equation, but very few types of rst-order nonlinear equations. 4.1 Terminology The standard form of a dierential equation is when it is written with all terms involving y or higher derivatives on one side, and functions of the variable on the other side. Example: The equation y + y + y = 0 is in standard form. Example: The equation y = 3x 2 xy is not in standard form.

2 An equation is homogeneous if, when it is put into standard form, the x-side is zero. nonhomogeneous otherwise. An equation is Example: The equation y + y + y = 0 is homogeneous. Example: The equation y + xy = 3x 2 is nonhomogeneous. An nth order dierential equation is an equation in which the highest derivative is the nth derivative. Example: The equations y + xy = 3x 2 and y y = 2 are rst-order. Example: The equation y + y + y = 0 is second-order. A dierential equation is linear if it is a linear combination of y and its derivatives. (Note that coecients are allowed to be functions of x.) In other words, if there are no terms like y 2, or (y ) 3, or y y, or e y. Example: The equations y + xy = 3x 2 and y + y + y = 0 are linear. Example: The equations y y = 3x 2 and y + e y = 0 are not linear. We say a linear dierential equation has constant coecients if the coecients of y, y, y,... are all constants. Example: The equation y + y + y = 0 has constant coecients. Example: The equation y + xy = 3x 2 does not have constant coecients. For n functions y 1, y 2,, y n which are each dierentiable (n 1) times, the Wronskian W (y 1, y 2,, y n ) of y 1 y 2 y n y 1 y 2 y n the functions is dened to be the determinant of the matrix y (n 1) 1 y (n 1) 2 y n (n 1) Note that the Wronskian will also be a function of x. The purpose of the Wronskian is to provide a way to show that functions are linearly independent. Theorem: A collection of n, n-times dierentiable functions y 1, y 2,, y n is linearly independent (in the vector space of n-times dierentiable functions) if their Wronskian is not the zero function. To prove the theorem, note that if the functions are linearly dependent with dierentiating the appropriate number of times we see that n j=1 a j y (i) j n a j y j = 0, then by j=1 = 0 for any 0 i n. Hence, in particular, the rows of the matrix are linearly dependent (as vectors), and so the determinant of the matrix is zero. Therefore, if the determinant is not zero, the functions cannot be dependent. Remark: The theorem becomes an if-and-only if statement (i.e., the functions are linearly independent if and only if the Wronskian is nonzero) if we know that the functions y 1, y 2,, y n are innitely dierentiable. The proof of the other direction is signicantly more dicult. Example: The functions 1 and x are linearly independent, because we can compute W (1, x) = 1 x 0 1 = 1. Example: The functions sin(x) and cos(x) are linearly independent, as W (sin(x), cos(x)) = sin(x) cos(x) cos(x) sin(x) 1. Example: The functions 1, x, and 1+x are (rather clearly) linearly dependent. We have W (1, x, 1+x) = 1 x x = 0, by expanding along the bottom row =

3 4.2 General Theory of Linear Dierential Equations Theorem (Homogeneous Linear Equations): If P n (x),, P 1 (x) are continuous functions, then the set of solutions to the homogeneous nth order equation y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = 0 is an n-dimensional vector space. The fact that the set of solutions is a vector space is not so hard to show using the subspace criteria. The real result of this theorem, which is analogous to the existence-uniqueness theorem for rst-order equations, is that the set of solutions is n-dimensional. Example: Consider the homogeneous equation y (x) = 0. We can just integrate twice to see that the solutions are y(x) = Ax + B, for any constants A and B. Indeed, as the theorem dictates, the solution functions form a two-dimensional space, spanned by the two basis elements 1 and x. Theorem (Existence-Uniqueness for Linear Equations): If P n (x),, P 1 (x) and Q(x) are functions continuous on an interval containing a, then there is a unique solution (possibly on a smaller interval) to the initial value problem y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = Q(x), for any initial conditions y(a) = b 1, y (a) = b 2,, and y (n 1) (a) = b n. Additionally, every solution y gen to the general nth order equation may be written as y gen = y par + y hom, where y par is any one particular solution to the equation, and y hom is a solution to the homogeneous equation y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = 0. What this theorem says is: in order to solve the general equation y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = Q(x), it is enough to nd one solution to this equation along with the general solution to the homogeneous equation. The existence-uniqueness part of the theorem is hard, but the second part is fairly simple to show: y 1 and y 2 are solutions to the general equation, then their dierence y 1 y 2 is a solution to the homogeneous equation to see this, just subtract the resulting equations and apply derivatives rules. A more advanced way to see the same thing is to use the fact that the map L sending y to y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y is a linear transformation. Then L(y 1 ) = L(y 2 ) says L(y 1 y 2 ) = 0, so that y 1 y 2 is a solution to the homogeneous equation. Example: In order to solve the equation y (x) = e x, the theorem says we only need to nd one function which is a solution, and then solve the homogeneous equation y (x) = 0. We can just try simple functions until we discover that y(x) = e x has y (x) = e x. Then we need only solve the homogeneous equation y (x) = 0, whose solutions we know are Ax + B. Thus the general solution to the general equation y (x) = e x is y(x) = e x + Ax + B. We can also verify that if we impose the initial conditions y(0) = c 1 and y (0) = c 2, then (as the theorem dictates) there is the unique solution y = e x + (c 2 1)x + (c 1 1). 4.3 Homogeneous Linear Equations with Constant Coecients The general linear homogeneous dierential equation with constant coecients is y (n) + a n 1 y (n 1) + + a 1 y + a 0 y = 0, where a n 1,, a 0 are some constants. From the existence-uniqueness theorem we know that the set of solutions is an n-dimensional vector space. Based on solving rst-order linear homogeneous equations (i.e., y + ky = 0), we might expect the solutions to involve exponentials. If we try setting y = e rx then after some arithmetic we end up with r n e rx + a n 1 r n 1 e rx + + a 1 re rx + a 0 e rx = 0. Multiplying both sides by e rx and cancelling yields the characteristic equation r n + a n 1 r n a 1 r + a 0 = 0.

4 If we can nd n values of r satisfying this nth-degree polynomial i.e., if we can factor the polynomial and see that it has n distinct roots, then the theorem tells us we will have found all of the solutions. If we are unlucky and the polynomial has a repeated root, then we need to try something else. If there are non-real roots (note they will come in complex conjugate pairs!) r 1 = α + βi and r 2 = α βi then we would end up with e r1x and e r2x as our solutions. But we really want real-valued solutions, and e r1x and e r2x have complex numbers in the exponents. However we can just write out the real and imaginary parts using Euler's Theorem, and take linear combinations to obtain the two real-valued solutions e αx sin(βx) = 1 2i [er1x e r2x ] and e αx cos(βx) = 1 2 [er1x + e r2x ]. Taking motivation from the case of y (k) = 0, whose characteristic equation is r k = 0 (with the k-fold repeated root 0) and whose solutions are y(x) = A 1 + A 2 x + A 3 x A k x k 1, we guess wildly that if other roots are repeated, we want to multiply the corresponding exponentials e rx by a power of x. If we put all of these ideas together we can prove that this general outline will, in fact, give us n linearly independent functions, and hence gives the general solution to any homogeneous linear dierential equation with constant coecients. Advanced Remark: Another way to organize this information is to think of everything in terms of linear transformations. If D represents the linear transformation sending a function to its derivative, then we are looking for functions y with the property that L(y) = 0, where L is the linear operator D n + a n 1 D n a 1 D + a 0. Note that D 2, for instance, means taking the derivative twice in a row i.e., D 2 is the second derivative. Then what we are secretly doing is factoring the polynomial D n +a n 1 D n 1 + +a 1 D +a 0 into a product of linear terms (D r 1 ) k1 (D r j ) kj, solving each of the simpler equations (D r i ) ki y = 0, and adding up the resulting terms. To solve a linear homogeneous dierential equation with constant coecients, follow these steps: Step 1: Rewrite the dierential equation in the standard form y (n) + a n 1 y (n 1) + + a 1 y + a 0 y = 0 (if necessary). Step 2: Factor the characteristic equation r n + a n 1 r n a 1 r + a 0 = 0. Step 3: For each irreducible factor in the characteristic equation, write down the corresponding terms in the solution: For terms (r α) k where is a real number then add the terms of the form e αx, xe αx,, x k 1 e αx. For irreducible terms (r 2 + cx + d) k with roots r = α ± βi, then add the terms of the form e αx sin(βx), xe αx sin(βx),, x k 1 e αx sin(βx) and e αx cos(βx), xe αx cos(βx),, x k 1 e αx cos(βx). Step 4: If given additional conditions on the solution, solve for the coecients (if necessary). Example: Find all functions y such that y + y 6 = 0. Step 2: The characteristic equation is r 2 + r 6 = 0 which has roots r = 2 and r = 3. Step 3: We have two distinct real roots, so our terms are e 2x and e 3x. y = C 1 e 2x + C 2 e 3x. Example: Find all functions y such that y 2y + 1 = 0, with y(0) = 1 and y (0) = 2. Step 2: The characteristic equation is r 2 2r + 1 = 0 which has only the solution r = 1. So the general solution is Step 3: There is a double root at r = 1, so our terms are e x and xe x. Hence the general solution is y = C 1 e x + C 2 xe x. Step 4: Plugging in the two conditions gives 1 = C 1 e 0 + C 2 0, and 2 = C 1 e 0 + C 2 [ (0 + 1)e 0 ] from which C 1 = 1 and C 2 = 1. Hence the particular solution requested is y = e x + xe x. Example: Find all real-valued functions y such that y = 4y. Step 1: The standard form here is y + 4y = 0.

5 Step 2: The characteristic equation is r = 0 which has roots r = 2i and r = 2i. Step 3: We have two complex-conjugate roots. Since the problem asks for real-valued functions we write e r1x = cos(2x) + i sin(2x) and e r2x = cos(2x) i sin(2x) to see that the general solution is y = C 1 cos(2x) + C 2 sin(2x). Example: Find all real-valued functions y such that y (5) + 5y (4) + 10y + 10y + 5y + y = 0. Step 2: The characteristic equation is r 5 + 5r r r 2 + 5r + 1 = 0 which factors as (r + 1) 5 = 0. Step 3: We have a 5-fold repeated root r = 1. Thus the terms are e x, xe x, x 2 e x, x 3 e x, and x 4 e x. Hence the general solution is y = C 1 e x + C 2 xe x + C 3 x 2 e x + C 4 x 3 e x + C 5 x 4 e x. Example: Find all real-valued functions y whose fourth derivative is the same as y. Step 1: This is the equation y = y, or in standard form, y y = 0. Step 2: The characteristic equation is r 4 1 = 0 which factors as (r + 1)(r 1)(r + i)(r i) = 0. Step 3: We have the four roots 1, 1, i, i. Thus the terms are e x, e x, sin(x), and cos(x). Hence the general solution is y = C 1 e x + C 2 e x + C 3 sin(x) + C 4 cos(x). 4.4 Non-Homogeneous Linear Equations with Constant Coecients The general linear dierential equation with constant coecients is of the form y (n) +a n 1 y (n 1) + +a 1 y + a 0 y = Q(x), where a n 1,, a 0 are some constants and Q(x) is some function of x. From the general theory, all we need to do is nd one solution to the general equation, and nd all solutions to the homogeneous equation. Since we know how to solve the homogeneous equation in full generality, we just need to develop some techniques for nding one solution to the general equation. There are essentially two ways of doing this. The Method of Undetermined Coecients is just a fancy way of making an an educated guess about what the form of the solution will be and then checking if it works. It will work whenever the function Q(x) is linear combination of terms of the form x k e αx (where k is an integer and α is a complex number): thus, for example, we could use the method for something like Q(x) = x 3 e 8x cos(x) 4 sin(x) + x 10 but not something like Q(x) = tan(x). Variation of Parameters is a more complicated method which uses some linear algebra and cleverness to use the solutions of the homogeneous equation to nd a solution to the non-homogeneous equation. It will always work, for any function Q(x), but generally requires more setup and computation Undetermined Coecients The idea behind the method of undetermined coecients is that we can 'guess' what our solution should look like (up to some coecients we have to solve for), if Q(x) involves sums and products of polynomials, exponentials, and trigonometric functions. Specically, we try a solution y = [stu], where the 'stu' is a sum of things similar to the terms in Q(x). Here is the procedure for generating the trial solution: Step 1: Generate the rst guess for the trial solution as follows: Replace all numerical coecients of terms in Q(x) with variable coecients. If there is a sine (or cosine) term, add in the companion cosine (or sine) terms, if they are missing. Then group terms of Q(x) into blocks of terms which are the same up to a power of x, and add in any missing lower-degree terms in each block. Thus, if a term of the form x n e rx appears in Q(x), ll in the terms of the form e rx [A 0 + A 1 x + + A n x n ], and if a term of the form x n e αx sin(βx) or x n e αx cos(βx) appears in Q(x), ll in the terms of the form e ax cos(bx) [D 0 + D 1 x + + D n x n ] + e ax sin(bx) [E 0 + E 1 x + + E n x n ].

6 Step 2: Solve the homogeneous equation, and write down the general solution. Step 3: Compare the rst guess for the trial solution with the solutions to the homogeneous equation. If any terms overlap, multiply all terms in the overlapping block by the appropriate power of x which will remove the duplication. Here is a series of examples demonstrating the procedure for generating the trial solution: Example: y y = x. Step 1: We ll in the missing constant term in Q(x) to get D 0 + D 1 x. Step 2: General solution is A 1 e x + A 2 e x. Step 3: There is no overlap, so the trial solution is D 0 + D 1 x. Example: y + y = x 2. Step 1: We have D 0 + D 1 x. Step 2: General homogeneous solution is A + Be x. Step 3: There is an overlap (the solution D 0 ) so we multiply the corresponding trial solution terms by x, to get D 0 x + D 1 x 2. Now there is no overlap, so D 0 x + D 1 x 2 Example: y y = e x. is the trial solution. Step 1: We have D 0 e x. Step 2: General homogeneous solution is Ae x + Be x. Step 3: There is an overlap (the solution D 0 e x ) so we multiply the trial solution term by x, to get D 0 xe x. Now there is no overlap, so D 0 xe x is the trial solution. Example: y 2y + y = 3e x. Step 1: We have D 0 e x. Step 2: General homogeneous solution is Ae x + Bxe x. Step 3: There is an overlap (the solution D 0 e x ) so we multiply the trial solution term by x 2, to get rid of the overlap, giving us the trial solution D 0 x 2 e x. Example: y 2y + y = x 3 e x. Step 1: We ll in the lower-degree terms to get D 0 e x + D 1 xe x + D 2 x 2 e x + D 3 x 3 e x. Step 2: The general homogeneous solution is A 0 e x + A 1 xe x. Step 3: There is an overlap (namely D 0 e x + D 1 xe x ) so we multiply the trial solution terms by x 2 to get D 0 x 2 e x + D 1 x 3 e x + D 2 x 4 e x + D 3 x 5 e x Example: y + y = sin(x). as the trial solution. Step 1: We ll in the missing cosine term to get D 0 cos(x) + E 0 sin(x). Step 2: The general homogeneous solution is A cos(x) + B sin(x). Step 3: There is an overlap (all of D 0 cos(x) + E 0 sin(x)) so we multiply the trial solution terms by x to get D 0 x cos(x) + E 0 x sin(x). There is now no overlap so D 0 x cos(x) + E 0 x sin(x) is the trial solution. Example: y + y = x sin(x). Step 1: We ll in the missing cosine term and then all the lower-degree terms to get D 0 cos(x) + E 0 sin(x) + D 1 x cos(x) + E 1 x sin(x). Step 2: The general homogeneous solution is A cos(x) + B sin(x). Step 3: There is an overlap (all of D 0 cos(x) + E 0 sin(x)) so we multiply the trial solution terms in that group by x to get D 0 x cos(x) + E 0 x sin(x) + D 1 x 2 cos(x) + E 1 x 2 sin(x), which is the trial solution since now there is no overlap. Example: y y = x + xe x. Step 1: We ll in the lower-degree term for xe x and the lower-degree term for x, to get A 0 + A 1 x + B 0 e x + B 1 xe x.

7 Step 2: The general homogeneous solution is C 0 + C 1 x + De x. Step 3: There are overlaps in both groups of terms: A 0 + A 1 x and B 0 e x each overlap, so we multiply the x group by x 2 and the e x group by x to get rid of the overlaps. There are now no additional overlapping terms, so the trial solution is A 0 x 2 + A 1 x 3 + B 0 xe x + B 1 x 2 e x. Example: y + 2y + y = xe x + x cos(x). Step 1: We ll in the lower-degree term for xe x, then the missing sine term for x cos(x), and then the lower-degree terms for x cos(x) and x sin(x), to get A 0 e x + A 1 xe x + D 0 cos(x) + E 0 sin(x) + D 1 x cos(x) + E 1 x sin(x). Step 2: The general homogeneous solution is B 0 cos(x) + C 0 sin(x) + B 1 x cos(x) + C 1 x sin(x). Step 3: There is an overlap (namely, all of D 0 cos(x) + E 0 sin(x) + D 1 x cos(x) + D 1 x sin(x)) so we multiply that group by x 2 to get rid of the overlap. There are no additional overlapping terms, so the trial solution is A 0 e x + A 1 xe x + D 0 x 2 cos(x) + E 0 x 2 sin(x) + D 1 x 3 cos(x) + E 1 x 3 sin(x). Here is a series of examples nding the general trial solution and then solving for the coecients: Example: Find a function y such that y + y + y = x. The procedure produces our trial solution as y = D 0 + D 1 x, because there is no overlap with the solutions to the homogeneous equation. We plug in and get 0 + (D 1 ) + (D 1 x + D 0 ) = x, so that D 1 = 1 and D 0 = 1. So our solution is y = x 1. Example: Find a function y such that y y = 2e x. The procedure gives the trial solution as y = D 0 xe x, since D 0 e x overlaps with the solution to the homogeneous equation. If y = D 0 xe x then y = D 0 (x + 2)e x so plugging in yields y y = [D 0 (x + 2)e x ] [D 1 x e x ] = 2e x. Solving yields D 0 = 1, so our solution is y = xe x. Example: Find a function y such that y 2y + y = x + sin(x). The procedure gives the trial solution as y = (D 0 + D 1 x) + (D 2 cos(x) + D 3 sin(x)), by lling in the missing constant term and cosine term, and because there is no overlap with the solutions to the homogeneous equation. Then we have y = D 2 cos(x) D 3 sin(x) and y = D 1 D 2 sin(x) + D 3 cos(x) so plugging in yields y 2y +y = [ D 2 cos(x) D 3 sin(x)] 2 [D 1 D 2 sin(x) + D 3 cos(x)]+[d 0 + D 1 x + D 2 cos(x) + D 3 sin(x)] and setting this equal to x + sin(x) then requires D 0 2D 1 = 0, D 1 = 1, D 2 + 2D 3 D 2 = 1, D 3 2D 2 D 3 = 0, so our solution is y = x cos(x). Example: Find all functions y such that y + y = sin(x). The solutions to the homogeneous system y + y = 0 are y = C 1 cos(x) + C 2 sin(x). Then the procedure gives the trial solution for the non-homogeneous equation as y = D 0 x cos(x) + D 1 x sin(x), by lling in the missing cosine term and then multiplying both by x due to the overlap with the solutions to the homogeneous equation. We can compute (eventually) that y = D 0 x cos(x) 2D 0 sin(x) D 1 x sin(x) + 2D 1 cos(x). Plugging in yields y +y = ( D 0 x cos(x) 2D 0 sin(x) D 1 x sin(x) + 2D 1 cos(x))+(d 0 x sin(x) + D 1 x cos(x)), and so setting this equal to sin(x), we obtain D 0 = 0 and D 1 = 1 2. Therefore the set of solutions is y = 1 2 x cos(x) + C 1 cos(x) + C 2 sin(x), for constants C 1 and C 2. Advanced Remark: The formal idea behind the method of undetermined coecients is that the terms on the right-hand side are themselves solutions of dierential equations and hence are sent to zero by a polynomial in the dierential operator D sending a function to its derivative. (Thus, for example, D(x 2 ) = 2x and D(e x ) = e x.)

8 Then if we apply that polynomial in D which sends Q(x) to zero, to both sides of the original equation, we will end up with a homogeneous equation whose characteristic polynomial is the product of the original characteristic polynomial and the characteristic polynomial for Q(x). Example: Find the form of a solution to y y = x 2. If we dierentiate both sides 3 times (i.e., apply the dierential operator D 3 to both sides), in order to kill o the x 2 term on the right-hand side then we get y (5) y (3) = 0, which has characteristic polynomial r 5 r 3 = (r 2 1) (r 3 ). Observe that this polynomial is the product of the characteristic polynomial r 2 1 of the original equation and the polynomial r 3 corresponding to D 3. Now y (5) y (3) = 0 is homogeneous, so we can write down the general solution to obtain y = C 1 + C 2 x + C 3 x 2 + C 4 e x + C 5 e x. This is the same solution form that the method of undetermined coecients gives, though with some extra lower-degree terms (which are solutions to the homogeneous equation y y = 0). Example: Find the form of a solution to y + y = x + sin(x). This is the same as the equation (D 2 + 1) y = x + sin(x) We want to apply the operator D 2 to kill the x term, and the operator D 2 +1 to kill the sin(x) term. The new dierential equation, after we are done, is (D 2 + 1)(D 2 )(D 2 + 1) y = 0. The characteristic polynomial is (r 2 + 1) 2 r 2, which has a double root at each of r = 0 and ±i. Solving the resulting homogeneous equation gives the solutions as y = C 1 sin(x) + C 2 x sin(x) + C 3 cos(x) + C 4 x cos(x) + C 5 + C 6 x, which is the same thing that the method of undetermined coecients gives, up to some extra terms Variation of Parameters Variation of Parameters will solve any non-homogeneous linear equation provided that the solutions to the homogeneous equation are known. However, the derivation is not entirely enlightening, so I will just give the steps to follow to solve y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = Q(x). (The method requires being able to solve the homogeneous equation, so we will typically apply this method to the constant-coecient equation y (n) + a n 1 y (n 1) + + a 1 y + a 0 y = 0.) Step 1: Solve the corresponding homogeneous equation y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = 0 and nd n (linearly independent) solutions y 1,, y n. Step 2: Look for functions v 1,, v n making y p = v 1 y 1 + v 2 y v n y n a solution to the original equation: do this by requiring v 1, v 2,, v n to satisfy the system of equations v 1 y 1 + v 2 y v n y n = 0 v 1 y 1 + v 2 y v n y n = 0... v 1 y (n 2) 1 + v 2 y (n 2) v n y n (n 2) = 0 v 1 y (n 1) 1 + v 2 y (n 1) v n y (n 1) n = Q(x) Solve the relations for v 1, v 2,, v n using Cramer's Rule (or any other method). This yields v i = W i(x) W (x), where W is the Wronskian of the functions y 1, y 2,, y n and W i is the same Wronskian determinant except with the ith column replaced with the column vector 0. 0 Q(x) Step 3: Integrate each of these relations to nd v 1,, v n. (Ignore constants of integration.) Step 4: Write down the particular solution to the nonhomogeneous equation, y p = v 1 y 1 + v 2 y v n y n..

9 Step 5: If asked, add the particular solution to the general solution to the homogeneous equation, to nd all solutions of the nonhomogeneous equation. This will yield y = y p + C 1 y C n y n. Plug in any extra conditions given to solve for coecients. Example: Find all functions y for which y + y = sec(x). Step 1: The homogeneous equation is y + y = 0 which has two independent solutions of y 1 = cos(x) and y 2 = sin(x). Step 2: We have Q(x) = sec(x). We have W = cos(x) sin(x) sin(x) cos(x) = 1. Also, W 1 = 0 sin(x) sec(x) cos(x) = sin(x) sec(x). Finally, W 2 = cos(x) 0 sin(x) sec(x) = cos(x) sec(x) = 1. Thus plugging in to the formulas gives v 1 = sin(x) sec(x) = tan(x) and v 2 = cos(x) sec(x) = 1. Step 3: Integrating yields v 1 = ln(cos(x)) and v 2 = x. Step 4: We obtain the particular solution of y p = ln(cos(x)) cos(x) + x sin(x). Step 5: The general solution is, therefore, given by y = [ln(cos(x)) cos(x) + x sin(x)] + C 1 sin(x) + C 2 cos(x). Example: Find all functions y for which y y + y y = e x. We could use undetermined coecients to solve this, but let's use variation of parameters. Step 1: The homogeneous equation is y y + y y = 0 which has characteristic polynomial r 3 r 2 + r 1 = (r 1)(r 2 + 1). So three independent solutions are y 1 = cos(x), y 2 = sin(x), and y 3 = e x. Step 2: We have Q(x) = e x. cos(x) sin(x) e x We have W = sin(x) cos(x) e x R 3+R 1 cos(x) sin(x) e x = sin(x) cos(x) e x cos(x) sin(x) e x 0 0 2e x = 2ex. 0 sin(x) e x Next, W 1 = 0 cos(x) e x e x sin(x) e x = e2x (sin(x) cos(x)). cos(x) 0 e x Also, W 2 = sin(x) 0 e x cos(x) e x e x = e2x (cos(x) + sin(x)). cos(x) sin(x) 0 Finally, W 3 = sin(x) cos(x) 0 cos(x) sin(x) e x = ex. Thus plugging in to the formulas gives v 1 = 1 2 ex (sin(x) cos(x)), v 2 = 1 2 ex (cos(x) + sin(x)), and v 3 = 1 2. Step 3: Integrating yields v 1 = 1 2 ex cos(x), v 2 = 1 2 ex sin(x), and v 3 = 1 2 x. Step 4: The particular solution is y p = 1 2 ex cos 2 (x) 1 2 ex sin 2 (x) xex = 1 2 ex xex. Step 5: The general solution is, therefore, given by y = 1 2 xex + C 1 cos(x) + C 2 sin(x) + C 3 e x. (Note that we absorbed the 1 2 ex term from the particular solution into C 3 e x.) Theorem (Abel's Identity): If y 1, y 2,, y n are any solutions to the homogeneous linear dierential equation y (n) +P n (x) y (n 1) + +P 2 (x) y +P 1 (x) y = 0, then the Wronskian W (y 1,, y n ) is equal to C e P n(x) dx, for some constant C.

10 Abel's Identity allows one to compute the Wronskian (up to a constant factor) without actually nding the solutions y 1, y 2,, y n. Another result of Abel's Identity is that the Wronskian is either zero everywhere (if C = 0) or nonzero everywhere (if C 0, since exponentials are never zero). Abel's Identity provides some relationships between the solution functions, which can sometimes be used to deduce information about them. For example, for a second-order linear dierential equation with two solutions y 1 and y 2, the Wronskian is W (x) = y 2y 1 y 1y 2. Therefore, if one solution y 1 is known, Abel's identity gives the value of W (x) (up to a value for C, which we can take to be 1 by rescaling), and therefore yields a rst-order linear dierential equation for the other solution y 2, which can then be solved for y 2. This procedure is known in general as reduction of order. The (very clever) proof of Abel's Identity involves showing that the Wronskian W satises the dierential equation W = P n (x) W, by dierentiating the determinant that denes W, and then using row operations to deduce that W = P n (x) W. 4.5 Second-Order Equations: Applications to Newtonian Mechanics One of the applications we somewhat care about is the use of second-order dierential equations to solve certain physics problems. Most of the examples involve springs, because springs are easy to talk about. Note: Second-order linear equations also arise often in basic circuit problems in physics and electrical engineering. All of the discussion of the behaviors of the solutions to these second-order equations also carries over to that setup Spring Problems and Damping Basic setup: An object is attached to one end of a spring whose other end is xed. The mass is displaced some amount from the equilibrium position, and the problem is to nd the object's position as a function of time. Various modications to this basic setup include any or all of (i) the object slides across a surface thus adding a force (friction) depending on the object's velocity or position, (ii) the object hangs vertically thus adding a constant gravitational force, (iii) a motor or other device imparts some additional nonconstant force (varying with time) to the object. In order to solve problems like this one, follow these steps: Step 1: Draw a diagram and label the quantity or quantities of interest (typically, it is the position of a moving object) and identify and label all forces acting on those quantities. Step 2: Find the values of the forces involved, and then use Newton's Second Law (F = ma) to write down a dierential equation modeling the problem. Also use any information given to write down initial conditions. In the above, F is the net force on the object i.e., the sum of each of the individual forces acting on the mass (with the proper sign) while m is the mass of the object, and a is the object's acceleration. Remember that acceleration is the second derivative of position with respect to time thus, if y(t) is the object's position, acceleration is y (t). You may need to do additional work to solve for unknown constants e.g., for a spring constant, if it is not explicitly given to you before you can fully set up the problem. Step 3: Solve the dierential equation and nd its general solution. Step 4: Plug in any initial conditions to nd the specic solution. Step 5: Check that the answer obtained makes sense in the physical context of the problem.

11 In other words, if you have an object attached to a xed spring sliding on a frictionless surface, you should expect the position to be sinusoidal, something like C 1 sin(ωt) + C 2 cos(ωt) + D for some constants C 1, C 2, ω, D. If you have an object on a spring sliding on a surface imparting friction, you should expect the position to tend to some equilibrium value as t grows to, since the object should be 'slowing down' as time goes on. Basic Example: An object, mass m, is attached to a spring of spring constant k whose other end is xed. The object is displaced a distance d from the equilibrium position of the spring, and is let go with velocity v 0 at time t = 0. If the object is restricted to sliding horizontally on a frictionless surface, nd the position of the object as a function of time. Step 1: Take y(t) to be the displacement of the object from the equilibrium position. The only force acting on the object is from the spring, F spring. Step 2: We know that F spring = k y from Hooke's Law (aka, the only thing we know about springs). Therefore we have the dierential equation k y = m y. We are also given the initial conditions y(0) = d and y (0) = v 0. Step 3: We can rewrite the dierential equation as m y + k y = 0, or as y + k m y = 0. The characteristic equation is then r 2 + k k m = 0 with roots r = ± m i. Hence the general solution is k y = C 1 cos(ωt) + C 2 sin(ωt), where ω = m. Step 4: The initial conditions give d = y(0) = C 1 and v 0 = y (0) = ωc 2 hence C 1 = d and C 2 = v 0 /ω. Hence the solution we want is y = d cos(ωt) + v 0 ω sin(ωt). Step 5: The solution we have obtained makes sense in the context of this problem, since on a frictionless surface we should expect that the object's motion would be purely oscillatory it should just bounce back and forth along the spring forever since there is nothing to slow its motion. We can even see that the k form of the solution agrees with our intuition: the fact that the frequency ω = increases with bigger m spring constant but decreases with bigger mass makes sense a stronger spring with larger k should pull back harder on the object and cause it to oscillate more quickly, while a heavier object should resist the spring's force and oscillate more slowly. Most General Example: An object, mass m, is attached to a spring of spring constant k whose other end is xed. The object is displaced a distance d from the equilibrium position of the spring, and is let go with velocity v 0 at time t = 0. A motor attached to the object imparts a force along its direction of motion given by R(t). If the object is restricted to sliding horizontally on a surface which imparts a frictional force of µ times the velocity of the object (opposite to the object's motion), set up a dierential equation modeling the problem. Here is the diagram:. As before we take y(t) to be the displacement of the object from the equilibrium position. The forces acting on the object are from the spring, F spring, from friction, F friction, and from the motor, F motor.

12 We know that F spring = k y from Hooke's Law (aka, the only thing we know about springs). We are also given that F fric = µ y, since the force acts opposite to the direction of motion and velocity is given by y. And we are just given F motor = R(t). Plugging in gives us the dierential equation k y µ y + R(t) = m y, which in standard form is m y + µ y + k y = R(t). We are also given the initial conditions y(0) = d and y (0) = v 0. Some Terminology: If we were to solve the dierential equation m y + µ y + k y = 0 (here we assume that there is no outside force acting on the object, other than the spring and friction), we would observe a few dierent kinds of behavior depending on the parameters m, µ, and k. Overdamped Case: If µ 2 4mk > 0 and R(t) = 0, we would end up with general solutions of the form C 1 e r1t +C 2 e r2t, which when graphed is just a sum of two exponentially-decaying functions. Physically, as we can see from the condition µ 2 4mk > 0, this means we have 'too much' friction, since we can just see from the form of the solution function that the position of the object will just slide back towards its equilibrium at y = 0 without oscillating at all. This is the overdamped case. [Overdamped because there is 'too much' damping.] Critically Damped Case: If µ 2 4mk = 0 and R(t) = 0, we would end up with general solutions of the form (C 1 + C 2 t)e rt, which when graphed is a slightly-slower-decaying exponential function that still does not oscillate, but could possibly cross the position y = 0 once, depending on the values of C 1 and C 2. This is the critically damped case. [Critically because we give the name 'critical' to values where some kind of behavior transitions from one thing to another.] Underdamped Case: If µ 2 4mk < 0 and R(t) = 0, we end up with general solutions of the form e αt [C 1 cos(ωt) + C 2 sin(ωt)], where α = µ 2m and 4mk ω2 µ2 = 4m 2. When graphed this is a sine curve times an exponentially-decaying function. Physically, this means that there is some friction (the exponential), but 'not enough' friction to eliminate the oscillations entirely the position of the object will still tend toward y = 0, but the sine and cosine terms will ensure that it continues oscillating. This is the underdamped case. [Underdamped because there's not enough damping.] Undamped Case: If there is no friction (i.e., µ = 0), we saw earlier that the solutions are of the form y = C 1 cos(ωt) + C 2 sin(ωt) where ω 2 = k/m. Since there is no friction, it is not a surprise that this is referred to as the undamped case Resonance and Forcing Suppose an object of mass m (sliding on a frictionless surface) is oscillating on a spring with frequency ω. Examine what happens to the object's motion if an external force F (t) = A cos(ωt) is applied which oscillates at the same frequency ω. From the solution to the Basic Example above, we know that ω = k/m so we must have k = m ω 2. Then if y(t) is the position of the object once we add in this new force R(t) = A cos(ωt + θ), Newton's Second Law now gives k y + R(t) = m y, or m y + k y = R(t). If we divide through by m and put in k = m ω 2 we get y + ω 2 y = A m cos(ωt). Now we use the method of undetermined coecients to nd a solution to this dierential equation. We would like to try something of the form y = D 1 cos(ωt) + D 2 sin(ωt), but this will not work because functions of that form are already solutions to the homogeneous equation y + ω 2 y = 0. Instead the method instructs that the appropriate solution will be of the form y = D 1 t cos(ωt) + D 2 t sin(ωt). We can use a trigonometric formula (the sum-to-product formula) to rewrite this as y = D t cos(ωt + φ), where φ is a phase shift. (We can solve for the coecients in terms of A, m, ω but it will not be so useful.) We can see from this formula that as t grows, so does the amplitude D t: in other words, as time goes on, the object will continue oscillating with frequency ω around its equilibrium point, but the swings back and forth will get larger and larger.

13 You can observe this phenomenon for yourself if you sit in a rocking chair, or swing an object back and forth you will quickly nd that the most eective way to rock the chair or swing the object is to push back and forth at the same frequency that the object is already moving at. We may work out the same computation with an external force F (t) = A cos(ω 1 t) oscillating at a frequency ω 1 ω. In this case (using the same argument as above) we have y + ω 2 y = A m cos(ω 1t). The trial solution (again by undetermined coecients) is y(t) = B cos(ω 1 t), where B = A/m ω 2 ω1 2. Thus the overall solution is B cos(ω 1 t), plus a solution to the homogeneous system. Now as we can see, if ω 1 and ω are far apart (i.e., the driving force is oscillating at a very dierent frequency from the frequency of the original system) then B will be small, and so the overall change B cos(ω 1 t) that the driving force adds will be relatively small. However, if ω 1 and ω are very close to one another (i.e., the driving force is oscillating at a frequency close to that of the original system) then B will be large, and so the driving force will cause the system to oscillate with a much bigger amplitude. As ω 1 approaches ω, the amplitude B will go to, which agrees with the behavior seen in the previous example (where we took ω 1 = ω). Important Remark: Understanding how resonance arises (and how to minimize it!) is a very, very important application of dierential equations to structural engineering. A poor understanding of resonance is something which, at several times in the not-too-distant past, has caused bridges to fall down, airplanes to crash, and buildings to fall over. We can see from the two examples that resonance arises when an external force acts on a system at (or very close to) the same frequency that the system is already oscillating at. Of course, resonance is not always bad. The general principle, of applying an external driving force at (one of) a system's natural resonance frequencies, is the underlying physical idea behind the construction of many types of musical instruments. Well, you're at the end of my handout. Hope it was helpful. Copyright notice: This material is copyright Evan Dummit, You may not reproduce or distribute this material without my express permission.

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