2nd-Order Linear Equations

Transcription

1 4 2nd-Order Linear Equations 4.1 Linear Independence of Functions In linear algebra the notion of linear independence arises frequently in the context of vector spaces. If V is a vector space over the field R, then the vectors v 1, v 2,..., v n V are linearly independent if c 1 v 1 + c 2 v c n v n = 0 implies that c k = 0 for all 1 k n. The vectors in a vector space can be functions. If I R is an interval, then we denote by V I the vector space over R consisting of functions f : I R. Given f V I and c R, we define scalar multiplication of c with f as yielding a new function cf V I given by (cf)(x) = cf(x) for all x I. If f, g V I, we define addition of f with g as yielding a new function f + g V I given by (f + g)(x) = f(x) + g(x) for all x I. These operations are consonant with conventions established in elementary algebra. If an interval I is not specified at the outset of an analysis involving real-valued functions f 1, f 2,..., f n of a single real variable, then we take n I = Dom(f i ) = Dom(f 1 ) Dom(f 2 ) Dom(f n ) i=1 provided that this results in an interval, and we carry out the analysis in the vector space V I. A linear combination of f 1, f 2,..., f n V I is an expression of the form c 1 f 1 + c 2 f c n f n for some choice of constants c 1, c 2,..., c n R, which of course is itself a function in V I given by (c 1 f 1 + c 2 f c n f n )(x) = c 1 f 1 (x) + c 2 f 2 (x) + + c n f n (x) for all x I. To write c 1 f 1 + c 2 f c n f n = 0 means in this context that (c 1 f 1 + c 2 f c n f n )(x) = 0

2 2 for all x I; that is, c 1 f 1 + c 2 f c n f n is the zero function on I, 0 : I {0}. Definition 4.1. If f 1, f 2,..., f n are functions with common domain I R, then f 1, f 2,..., f n are linearly independent on I if implies that c k = 0 for all 1 k n. c 1 f 1 + c 2 f c n f n = 0 Functions that are not linearly independent on I are said to be linearly dependent on I. Thus, f 1, f 2,..., f n are linearly dependent on I if there can be found constants c 1, c 2,..., c n R, not all zero, such that (c 1 f 1 + c 2 f c n f n )(x) = 0 for all x I.

3 3 4.2 Homogeneous Equations with Constant Coefficients A differential equation of the form a n y (n) + a n 1 y (n 1) + + a 2 y + a 1 y + a 0 y = f(t), (1) where each a i is a constant, a n 0, and y is taken to be a function of t, is called an nth-order nonhomogeneous linear differential equation with constant coefficients. If f(t) = 0 on some open interval I R, then we obtain a n y (n) + a n 1 y (n 1) + + a 2 y + a 1 y + a 0 y = 0, (2) which is an nth-order homogeneous 1 differential equation with constant coefficients. (The constant coefficients imply the equation is linear, so that adjective can at least be omitted.) Now, if we set a i = 0 for all i > 2, then (2) becomes a 2 y + a 1 y + a 0 y = 0, (3) a second-order homogeneous equation with constant coefficients. This type of equation, where it is understood that a 2 0, will be the primary focus of attention in this section and the next. Note that the left side of (3) is a linear combination of the functions y, y, and y. Any solution to (3) must be a function ϕ defined on some open interval I such that a 2 ϕ (x) + a 1 ϕ (x) + a 0 ϕ(x) = 0 for all x I, and since we are assuming that a 2 0 it follows that ϕ, ϕ, and ϕ must be linearly dependent on I. Observe that, regardless of the choice for constants a 0, a 1, and a 2, the zero function y 0 is always a solution to (3). Proposition 4.2. Let I R be an open interval. If y 1, y 2 : I R are solutions to a 2 y + a 1 y + a 0 y = 0 on I, then any linear combination of y 1 and y 2 is also a solution on I. Proof. Suppose that y 1, y 2 : I R are solutions to a 2 y + a 1 y + a 0 y = 0. Then and a 2 y 1(x) + a 1 y 1(x) + a 0 y 1 (x) = 0 a 2 y 2(x) + a 1 y 2(x) + a 0 y 2 (x) = 0 for all x I, which is to say that a 2 y 1 + a 1 y 1 + a 0 y 1 = 0 and a 2 y 2 + a 1 y 2 + a 0 y 2 = 0 on I. Now, for any c 1, c 2 R, let c 1 y 1 + c 2 y 2 be a linear combination of y 1 and y 2. What we must show is that y = c 1 y 1 + c 2 y 2 is a solution to (3), and so we substitute c 1 y 1 + c 2 y 2 for y in the equation: a 2 y + a 1 y + a 0 y = a 2 (c 1 y 1 + c 2 y 2 ) + a 1 (c 1 y 1 + c 2 y 2 ) + a 0 (c 1 y 1 + c 2 y 2 ) = a 2 (c 1 y 1 + c 2 y 2) + a 1 (c 1 y 1 + c 2 y 2) + a 0 (c 1 y 1 + c 2 y 2 ) = c 1 (a 2 y 1 + a 1 y 1 + a 0 y 1 ) + c 2 (a 2 y 2 + a 1 y 2 + a 0 y 2 ) 1 This use of the word homogeneous is in no way related to how the word was used in Section 2.6.

4 4 = c c 2 0 = 0. The result is zero, which shows that c 1 y 1 + c 2 y 2 does indeed satisfy (3) for all x I, and therefore it is a solution to the equation. The following existence-uniqueness theorem will be used to prove practical results in this section, and while the existence part of its proof is straightforward (see Theorem 4.7 below), the uniqueness part is not and is deferred to a much later chapter. Theorem 4.3. Let a 0, a 1, a 2, b 0, b 1, t 0 R. If a 2 0, then there exists a unique solution y(t) to the initial value problem that is valid on (, ). a 2 y + a 1 y + a 0 y = 0, y(t 0 ) = b 0, y (t 0 ) = b 1 (4) Remark. In actual fact the statement of Theorem 4.3 can be applied to any open interval I. That is, if I (, ) is open and t 0 I, then there exists a unique solution y(t) to the IVP (4) that is valid on I. The proof of Theorem 4.3 would need to be little altered to show this. Corollary. If ϕ : I R is a solution to a 2 y + a 1 y + a 0 y = 0 on an open interval I, then there exists a unique function ψ(t) that is a solution to a 2 y + a 1 y + a 0 y = 0 on (, ) such that ψ = ϕ on I. Proof. Suppose that ϕ(t) is a solution to a 2 y + a 1 y + a 0 y = 0 on an interval I. Let t 0 I, and let b 0 = ϕ(t 0 ) and b 1 = ϕ (t 0 ). We thus have the IVP a 2 y + a 1 y + a 0 y = 0, y(t 0 ) = ϕ(t 0 ), y (t 0 ) = ϕ (t 0 ), (5) and by construction ϕ(t) is a solution to this IVP. Now, by Theorem 4.3 there exists a unique solution ψ(t) to the IVP (5) that is valid on (, ), and thus is must also be valid on I (, ). By the remark following Theorem 4.3 we know there can only be one solution to (5) that is valid on I, and therefore we must have ψ = ϕ on I. Because of the corollary above, any function y(t) that is given to be a solution to (3) can be assumed to be valid on (, ). Indeed, Theorem 4.7 below makes it clear that (3) has precisely two linearly independent solutions on (, ), and gives their explicit forms. To construct the machinery that proves the theorem we start with the following lemma. Lemma 4.4. Let y 1 and y 2 be two solutions to a 2 y + a 1 y + a 0 y = 0 on (, ). If y 1 and y 2 are linearly independent on (, ), then for all t (, ). y 1 (t)y 2(t) y 1(t)y 2 (t) 0 (6)

5 Proof. The contrapositive of the conditional statement in the lemma reads as follows: If there exists some τ (, ) such that y 1 (τ)y 2(τ) y 1(τ)y 2 (τ) = 0, then y 1 and y 2 are linearly dependent on (, ). It is this equivalent statement that we shall set out to prove. Suppose that there exists some real number τ such that y 1 (τ)y 2(τ) y 1(τ)y 2 (τ) = 0. (7) There are three cases that we may consider. Case 1: y 1 (τ) 0. Let r = y 2 (τ)/y 1 (τ), and consider the initial value problem a 2 y + a 1 y + a 0 y = 0, y(τ) = y 2 (τ), y (τ) = y 2(τ). (8) Clearly y 2 (t) is a solution to this IVP, valid for all t (, ). Now, by Proposition 4.2, since y 1 (t) is a solution to the ODE (3) on (, ), so too is ϕ(t) = ry 1 (t). Indeed and by (7) ϕ(τ) = ry 1 (τ) = y 2(τ) y 1 (τ) y 1(τ) = y 2 (τ), ϕ (τ) = ry 1(τ) = y 2(τ) y 1 (τ) y 1(τ) = y 2(τ) y 1 (τ) y1(τ)y 2(τ) = y y 2 (τ) 2(τ), so that ϕ(t) is seen to be a solution to the IVP (8) on (, ). Since by Theorem 4.3 the solution to an IVP must be unique, we conclude that ϕ = y 2 on (, ). That is, (ry 1 y 2 )(t) = 0 for all < t <, which shows that y 1 and y 2 are linearly dependent on (, ). Case 2: y 1 (τ) = 0 and y 1(τ) 0. From (7) it follows that y 2 (τ) = 0. Let r = y 2(τ)/y 1(τ), and again consider the initial value problem (8). As before, y 2 (t) is a solution to the IVP. Once more, Proposition 4.2 implies that ϕ(t) = ry 1 (t) is a solution to (3), and since and ϕ(τ) = ry 1 (τ) = y 2(τ) y 1(τ) y 1(τ) = 0 = y 2 (τ) ϕ (τ) = ry 1(τ) = y 2(τ) y 1(τ) y 1(τ) = y 2(τ), it s seen that ϕ(t) is also a solution to the IVP (8). Since by Theorem 4.3 the solution to an IVP must be unique, we conclude that ϕ = y 2. That is, ry 1 y 2 = 0, which shows that y 1 and y 2 are linearly dependent on (, ). Case 3: y 1 (τ) = 0 and y 1(τ) = 0. The zero function y 0 satisfies the initial value problem a 2 y + a 1 y + a 0 y = 0, y(τ) = 0, y (τ) = 0, and since y 1 (t) also satisfies this IVP, by Theorem 4.3 it must be that y 1 is the zero function. Thus, since y 1 0, we have c 1 y 1 + c 2 y 2 = 0 on (, ) if we choose c 1 = 1 and c 2 = 0, which shows that y 1 and y 2 are, once again, linearly dependent on (, ). Proposition 4.5. If y 1 and y 2 are two linearly independent solutions to a 2 y +a 1 y +a 0 y = 0 on (, ), then for any t 0, b 0, b 1 R there exist unique constants c 1, c 2 R such that c 1 y 1 +c 2 y 2 satisfies the IVP (4) on (, ). 5

6 6 The proof of this proposition is not complicated and will be included here eventually. Proposition 4.6. The general solution to a 2 y + a 1 y + a 0 y = 0 is a two-parameter family of solutions c 1 y 1 + c 2 y 2, where y 1 and y 2 are linearly independent functions on (, ). Proof. Let y 1 (t) and y 2 (t) be two linearly independent functions on (, ) that are particular solutions to a 2 y + a 1 y + a 0 y = 0. Now, suppose that ϕ(t) is any arbitrary solution to a 2 y +a 1 y +a 0 y = 0 on (, ), which certainly implies that ϕ(t) is everywhere differentiable. Fix < t 0 <, and let b 0 = ϕ(t 0 ) and b 1 = ϕ (t 0 ). Since t 0, b 0, b 1 R, by Proposition 4.5 there exist constants c 1, c 2 R such that c 1 y 1 + c 2 y 2 satisfies the initial value problem a 2 y + a 1 y + a 0 y = 0, y(t 0 ) = b 0, y (t 0 ) = b 1. However ϕ(t) also satisfies the initial value problem, and since by Theorem 4.3 the solution to the IVP must be unique, it follows that ϕ(t) = c 1 y 1 (t)+c 2 y 2 (t) on (, ). So any solution to the ODE a 2 y + a 1 y + a 0 y = 0 is expressible as a linear combination of y 1 and y 2, and therefore the general solution to a 2 y + a 1 y + a 0 y = 0 is a two-parameter family of solutions of the form c 1 y 1 + c 2 y 2. The auxiliary equation corresponding to a 2 y + a 1 y + a 0 y = 0 is the quadratic equation a 2 r 2 + a 1 r + a 0 = 0. If a 2, a 1, a 0 R with a 2 0, then the auxiliary equation s solution set may consist of two real numbers (called distinct real roots), one real number (called a repeated root), or a complex conjugate pair. Theorem 4.7. If the auxiliary equation associated with (3) has distinct real roots r 1 and r 2, then y 1 (t) = e r 1t and y 2 (t) = e r 2t are solutions to (3), and hence y(t) = c 1 e r 1t + c 2 e r 2t is the general solution. If the auxiliary equation has a repeated root r, then y 1 (t) = e rt and y 2 (t) = te rt are solutions to (3), and hence y(t) = c 1 e rt + c 2 te rt is the general solution. In actual fact if the auxiliary equation has distinct roots r 1 = α + βi and r 2 = α βi that constitute a complex conjugate pair, then it is still the case that y 1 (t) = e r 1t and y 2 (t) = e r 2t are solutions to (3), and hence y(t) = c 1 e r 1t + c 2 e r 2t is the general solution. However, there is a nicer way to present the general solution that does not feature the imaginary unit i. We will investigate how this is done in the next section. Example 4.8. Solve the initial value problem y 4y + 3y = 0, y(0) = 1, y (0) = 1/3.

7 Solution. The auxiliary equation associated with y 4y +3y = 0 is r 2 4r +3 = 0. Factoring the left side gives (r 3)(r 1) = 0, and so we obtain two distinct real roots: r 1 = 1 and r 2 = 3. By Theorem 4.7 it follows that y 1 (t) = e t and y 2 (t) = e 3t are particular solutions to the ODE, and so the general solution is y(t) = c 1 e t + c 2 e 3t. Now, using the initial condition y(0) = 1 we obtain 1 = c 1 e 0 + c 2 e 3(0), or c 1 + c 2 = 1. Also, since y (t) = c 1 e t + 3c 2 e 3t, the initial condition y (0) = 1/3 gives 1/3 = c 1 e 0 + 3c 2 e 3(0), or c 1 + 3c 2 = 1/3. We thus have the system of equations { c1 + c 2 = 1 7 c 1 + 3c 2 = 1 3 Solving this system, we find that c 1 = 4/3 and c 2 = 1/3, and so y(t) = 4 3 et 1 3 e3t is the solution to the initial value problem. Example 4.9. Solve the initial value problem y 6y + 9y = 0, y(0) = 2, y (0) = 25/3. Solution. The auxiliary equation associated with y 6y +9y = 0 is r 2 6r +9 = 0. Factoring the left side gives (r 3)(r 3) = 0, and so we obtain a repeated real root: r = 3. By Theorem 4.7 it follows that y 1 (t) = e 3t and y 2 (t) = te 3t are particular solutions to the ODE, and so the general solution is y(t) = c 1 e 3t + c 2 te 3t. Now, using the initial condition y(0) = 2 we obtain 2 = c 1 e 0 + c 2 (0)e 3(0), or c 1 = 2. Also, since y (t) = 3c 1 e 3t + c 2 e 3t + 3c 2 te 3t, the initial condition y (0) = 25/3 gives 25/3 = 3c 1 e 0 + c 2 e 3(0) + 3c 2 (0)e 3(0), or 3c 1 + c 2 = 25/3. We thus have the system of equations { c1 = 2 Solving for c 2 results in c 2 = 7/3, and so 3c 1 + c 2 = 25 3 y(t) = 2e 3t te3t is the solution to the initial value problem. Example Solve the initial value problem y 2y y + 2y = 0, y(0) = 2, y (0) = 3, y (0) = 5. Solution. As with the second-order equations considered in previous examples we look for solutions of the form e rt, and so substitute e rt for y in the ODE to obtain ( e rt ) 2 ( e rt ) ( e rt ) + 2 ( e rt ) = 0, which becomes r 3 e rt 2r 2 e rt re rt + 2e rt = 0.

8 8 Dividing out e rt (which never equals zero) then yields the auxiliary equation Now, r 3 2r 2 r + 2 = 0. r 3 2r 2 r + 2 = 0 r 2 (r 2) (r 2) = 0 (r 2)(r 2 1) = 0, which finally becomes (r 2)(r 1)(r + 1) = 0, and so the roots of the auxiliary equation are 1, 1, and 2. From these roots come three linearly independent solutions to the ODE: y 1 (x) = e t, y 2 (t) = e t, y 3 (t) = e 2t. We conclude that the three-parameter family of functions constitutes the general solution to the ODE. Now, from (9) we obtain and y(t) = c 1 e t + c 2 e t + c 3 e 2t (9) y (t) = c 1 e t + c 2 e t + 2c 3 e 2t y (t) = c 1 e t + c 2 e t + 4c 3 e 2t, and these equations, together with the given initial conditions, form the system of equations { c1 + c 2 + c 3 = 2 c 1 + c 2 + 2c 3 = 3 c 1 + c 2 + 4c 3 = 5 The solution to the system is (c 1, c 2, c 3 ) = (0, 1, 1), and therefore from (9) we find that is the solution to the initial value problem. y(t) = e t + e 2t

9 9 4.3 Auxiliary Equations With Complex Roots In 7.2 of the Calculus Notes the number e is defined to be the unique real number for which ln(e) = 1, 2 and the inverse of the natural logarithm function is defined to be the base-e exponential function exp : R (0, ). Thus, for all x R and y (0, ), exp(x) = y if and only if ln(y) = x. We further define e x = exp(x) for all x R. In 10.3 of Calculus Notes it was further established that x n exp(x) = n! for all x R. We have e = e 1 = exp(1) = as an immediate result. Recall the set of complex numbers, n=0 n=0 1 n n! = 1 n! C = {x + iy : x, y R}, where of course i is the number for which i 2 = 1. We may extend the domain of the exponential function to obtain a function exp : C C by defining n=0 exp(x + iy) = e x (cos y + i sin y), which in 3.3 of the Complex Analysis Notes [COM] is proven to be given by z n exp(z) = n! n=0 for all z C. By definition we set e z = exp(z), so that e x+iy = exp(x + iy) = e x (cos y + i sin y) = for all x, y R. From 3.3 of [COM] we have (x + iy) n n=0 n! (10) (11) e z+w = exp(z + w) = exp(z) exp(w) = e z e w (12) for any z, w C, and so a familiar law of exponents is seen to hold in our strange new world. The historically significant equation e iθ = cos θ + i sin θ, is known as Euler s formula, which derives from (11) by setting x = 0 and y = θ. This formula figures prominently in the developments to come. Henceforth we will frequently write complex numbers as α + iβ, where α, β R are to be taken as constants. Any complex-valued function f : D R C of a single real variable 2 In 7.6 of the Calculus Notes it is proven that e = lim n (1 + 1/n) n, which some authors present as the definition of e.

10 t can be cast in the form f(t) = u(t) + iv(t), where u and v are real-valued functions. The derivative of the function f(t) = u(t) + iv(t) is defined to be f f(t + h) f(t) (t) = lim, h 0 h and f is said to be differentiable at t if f (t) exists in C. It is straightforward to show that f (t) = u (t) + iv (t), and so f is differentiable at t if and only if u and v are differentiable at t in the sense defined in basic calculus. With this established, we can determine the derivative of e (α+iβ)t : d [ ] e (α+iβ)t = d ( e αt cos βt + ie αt sin βt ) dt dt = d ( e αt cos βt ) + i d ( e αt sin βt ) dt dt = ( αe αt cos βt βe αt sin βt ) + i ( αe αt sin βt + βe αt cos βt ) = e αt (α cos βt + iα sin βt + iβ cos βt β sin βt) = e αt (α + iβ)(cos βt + i sin βt) = (α + iβ) e αt (cos βt + i sin βt) = (α + iβ)e (α+iβ)t, where the last equality follows from (11). More concisely, for any z C we have 10 (e zt ) = ze zt. (13) We are now ready to return to the study of differential equations of the form where as usual a 2, a 1, a 0 R and a 2 0. The auxiliary equation a 2 y + a 1 y + a 0 y = 0, (14) a 2 r 2 + a 1 r + a 0 = 0 has roots r 1 = a 1 + a 2 1 4a 2 a 0 and r 2 = a 1 a 2 1 4a 2 a 0 ; 2a 2 2a 2 that is, we have r 1 = α + iβ and r 2 = α iβ, where α = a 1 a 2 and β = 1 4a 2 a 0 2a 2 2a 2 are real numbers. It is easy to verify that y 1 (t) = e (α+iβ)t and y 2 (t) = e (α iβ)t are solutions to (14), and therefore y(t) = c 1 e (α+iβ)t + c 2 e (α iβ)t (15) is the general solution to (14). However, since differential equations are used to model physical systems, it would seem reasonable that solutions should be expressible entirely in terms of realvalued functions. This is not the case with (15). Is there another way to express the general solution to (14) without resorting to complex values? Indeed there is.

11 Lemma Let ϕ(t) = u(t) + iv(t) be a solution to a 2 y + a 1 y + a 0 y = 0, where u(t) and v(t) are real-valued functions. Then u(t) and v(t) are also solutions. Proof. Since ϕ(t) is a solution to (14), we have and hence Rearranging terms then gives a 2 ϕ (t) + a 1 ϕ (t) + a 0 ϕ(t) = 0, a 2 [u (t) + iv (t)] + a 1 [u (t) + iv (t)] + a 0 [u(t) + iv(t)] = 0. [a 2 u (t) + a 1 u (t) + a 0 u(t)] + i[a 2 v (t) + a 1 v (t) + a 0 v(t)] = 0. In order for a complex number u + iv to be zero, it must be that u = 0 and v = 0. Thus we obtain a 2 u (t) + a 1 u (t) + a 0 u(t) = 0 and a 2 v (t) + a 1 v (t) + a 0 v(t) = 0, which shows that u(t) and v(t) are both solutions to (14). The crowning result of this section now follows. Theorem If a 2 y + a 1 y + a 0 y = 0 has auxiliary equation with complex conjugate roots α ± iβ, then two linearly independent solutions are y 1 (t) = e αt cos βt and y 2 (t) = e αt sin βt, and therefore y(t) = c 1 e αt cos βt + c 2 e αt sin βt is the general solution. Proof. Suppose that a 2 r 2 + a 1 r + a 0 = 0 has complex conjugate roots α ± iβ, so that in particular a 2 (α + iβ) 2 + a 1 (α + iβ) + a 0 = 0. (16) Then ϕ(t) = e (α+iβ)t is a solution to a 2 y + a 1 y + a 0 y = 0, which we now verify. By (13) we have ϕ (t) = (α + iβ)e (α+iβ)t and ϕ (t) = (α + iβ) 2 e (α+iβ)t. Substituting these expressions into (14) gives a 2 (α + iβ) 2 e (α+iβ)t + a 1 (α + iβ)e (α+iβ)t + a 0 e (α+iβ)t = 0, and thus e (α+iβ)t [ a 2 (α + iβ) 2 + a 1 (α + iβ) + a 0 ] = 0, which by (16) is equivalent to 0 = 0. This confirms that ϕ(t) satisfies (14). Now, by (11) we may write ϕ(t) = e αt+iβt = e αt cos βt + ie αt sin βt By Lemma 4.11 both e αt cos βt and e αt sin βt are solutions to (14) that are easily verified to be linearly independent. Therefore by Proposition 4.6 the general solution to (14) may be expressed as y(t) = c 1 e αt cos βt + c 2 e αt sin βt, 11 where c 1 and c 2 are arbitrary constants.

12 With the above machinery in place, we are now in a position to build some cities on the Moon. Example Find the general solution to 2y 2y + 13y = 0. Solution. The auxiliary equation is 2r 2 2r + 13 = 0, which solves to give r = ( 2) ± ( 2) 2 4(2)(13) 2(2) = 2 ± That is, r = α ± iβ with α = 1/2 and β = 5/2. By Theorem 4.12 = 2 ± 10i 4 y(t) = c 1 e t/2 cos(5t/2) + c 2 e t/2 sin(5t/2) = 1 2 ± 5 2 i. 12 is the general solution to the ODE. Example Solve the initial value problem y + 9y = 0, y(0) = 1, y (0) = 1. Solution. The auxiliary equation is r = 0, which solves to give r = ± 9 = ±3i. That is, r = α ± iβ with α = 0 and β = 3. By Theorem 4.12 y(t) = c 1 cos 3t + c 2 sin 3t is the general solution to the ODE. From the initial condition y(0) = 1 we then obtain 1 = c 1 cos 0 + c 2 sin 0, and thus c 1 = 1. Next, given that y (t) = 3c 1 sin 3t + 3c 2 cos 3t and y (0) = 1, we obtain and thus c 2 = 1/3. Therefore we conclude that is the solution to the IVP. 1 = 3c 1 sin 0 + 3c 2 cos 0 y(t) = cos 3t + 1 sin 3t 3 The following example shows how the idea of Theorems 4.7 and 4.12 can be extended to handle higher-order linear homogeneous equations with constant coefficients. The full theory is developed formally in Chapter 6, but for now we take an informal approach. Example Find the general solution to y (4) + 13y + 36y = 0. Solution. As ever we search for solutions of the form e rt. Since the ODE is a fourth-order equation it is a fact that the general solution will be a four-parameter family of linearly independent functions. To find four such functions, we substitute e rt into the ODE to obtain the auxiliary equation r r = 0.

13 Factoring yields (r 2 + 4)(r 2 + 9) = 0, which solves to give two complex conjugate pairs as solutions: r = ±2i, ±3i. This in fact means that the general solution to the ODE is y(t) = c 1 e 2i + c 2 e 2i + c 3 e 3i + c 4 e 3i by a natural extension of Theorem 4.7; but as with the second-order equations we ve been considering thus far there is a way to cast the general solution in a form that does not feature the imaginary unit i. By (11) we may write e 2i as cos 2t + i sin 2t and e 3i as cos 3t + i sin 3t. By the same argument as given in the proof of Lemma 4.11, we find that cos 2t, sin 2t, cos 3t, and sin 3t are also solutions to the ODE, and since they are linearly independent it follows that the general solution to the ODE may be expressed as y(t) = c 1 cos 2t + c 2 sin 2t + c 3 cos 3t + c 4 sin 3t. Remember: this is not a different general solution than the one found before, merely a different expression for the general solution. The two families of functions are the same. 13

14 4.4 Nonhomogeneous Equations with Constant Coefficients We now consider second-order nonhomogeneous linear differential equation with constant coefficients, which is to say equations of the form a 2 y + a 1 y + a 0 y = f(t) (17) for a 2 0 and f(t) 0. The term f(t) is sometimes referred to as the nonhomogeneity. To solve such equations we will use the Method of Undetermined Coefficients as outlined in the following theorem. Theorem 4.16 (Method of Undetermined Coefficients). Let P m (t) be a nonzero polynomial of degree m, and let y p denote a particular solution to a 2 y + a 1 y + a 0 y = f(t). 1. If f(t) = P m (t)e αt, then where y p (t) = t s e αt m (a) s = 0 if α is not a root of a 2 r 2 + a 1 r + a 0 = 0 (b) s = 1 if α is a single root of a 2 r 2 + a 1 r + a 0 = 0 (c) s = 2 if α is a double root of a 2 r 2 + a 1 r + a 0 = 0 14 A k t k, (18) 2. If f(t) = P m (t)e αt cos βt or f(t) = P m (t)e αt sin βt for β 0, then m m y p (t) = t s e αt cos βt A k t k + t s e αt sin βt B k t k, (19) where (a) s = 0 if α + βi is not a root of a 2 r 2 + a 1 r + a 0 = 0 (b) s = 1 if α + βi is a root of a 2 r 2 + a 1 r + a 0 = 0 In the first case above the coefficients that must be determined are A 0, A 1,..., A m, and in the second case they are A 0,..., A m and B 0,..., B m. It is understood that a 0, a 1, a 2, α, β, s, and m are all fixed constants wherever they appear. There are no arbitrary constants present in the statement of the theorem since the theorem is presenting a particular solution to (17), not a general solution. The task of constructing a general solution to (17), as well as dealing with equations where f(t) is some linear combination of the functions addressed in the theorem, will be undertaken in the next section using the Superposition Principle. Proofs of some parts of Theorem 4.16 will appear at the end of this section, while proofs of other parts will be left as exercises. Remark. The Method of Undetermined Coefficients as outline in Theorem 4.16 in fact applies to higher-order equations of the form (1), as well as first-order equations a 1 y + a 0 y = f(t); but the focus here will be second-order equations. See the exercises at the end of the section for applications of the method that involve equations of order three and four.

15 Example Find a form for a particular solution y p (t) to 2y + 7y 15y = f(t) for each expression for f(t). (a) f(t) = 3te 8t (b) f(t) = 9t 3 e 5t (c) f(t) = (7t 2 t)e 4t (d) f(t) = t cos 6t Solution. (a) The function 3te 8t fits the form P m (t)e αt in part (1) of the Method of Undetermined Coefficients, with P m (t) = 3t being a first-degree polynomial so that m = 1, and e αt = e 8t so that α = 8. Hence by Theorem 4.16(1) we have 1 y p (t) = t s e 8t A k t k = t s e 8t (A 0 + A 1 t). To determine s, observe that the auxiliary equation 2r 2 + 7r 15 = 0 has distinct real roots 5 and 3/2, and so α = 8 is not a root of the auxiliary equation, and we conclude that s = 0 by Theorem 4.16(1a). Therefore 15 is the desired form. y p (t) = (A 1 t + A 0 )e 8t (b) The function 9t 3 e 5t fits the form P m (t)e αt, with P m (t) = 9t 3 being a third-degree polynomial so that m = 3, and e αt = e 5t so that α = 5. Hence by Theorem 4.16(1) we have 3 y p (t) = t s e 5t A k t k = t s e 5t (A 0 + A 1 t + A 2 t 2 + A 3 t 3 ). The auxiliary equation 2r 2 + 7r 15 = 0 has real roots 5 and 3/2, so α = 5 is a single root of the auxiliary equation and so we conclude that s = 1 by Theorem 4.16(1b). Therefore y p (t) = t(a 3 t 3 + A 2 t 2 + A 1 t + A 0 )e 5t is the desired form. (c) The function (7t 2 t)e 4t fits the form P m (t)e αt, with P m (t) = 7t 2 t being a seconddegree polynomial so that m = 2, and e αt = e 4t so that α = 4. Hence by Theorem 4.16(1) we have 2 y p (t) = t s e 4t A k t k = t s e 5t (A 0 + A 1 t + A 2 t 2 ). The auxiliary equation has real roots 5 and 3/2, and since α = 4 is not a root we conclude that s = 0 by Theorem 4.16(1a). Therefore y p (t) = (A 2 t 2 + A 1 t + A 0 )e 4t

16 16 is the desired form. (d) The function t cos 6t fits the form P m (t)e αt cos βt in part (2) of the Method of Undetermined Coefficients, with P m (t) = t being a first-degree polynomial so that m = 1, e αt = 1 so that α = 0, and cos βt = cos 6t so that β = 6. Hence by Theorem 4.16(2) we have 1 1 y p (t) = t s e 0t cos 6t A k t k + t s e 0t sin 6t B k t k = t s (A 0 + A 1 t) cos 6t + t s (B 0 + B 1 t) sin 6t. The auxiliary equation has roots 5 and 3/2, and since α + iβ = 6i is not a root we conclude that s = 0 by Theorem 4.16(2a). Therefore is the desired form. y p (t) = (A 1 t + A 0 ) cos 6t + (B 0 + B 1 t) sin 6t Example Find a particular solution to 2y + 7y 15y = 3te 8t. Solution. From Example 4.17(a) it was found that a particular solution would be of the form From this we obtain and y p (t) = (A 1 t + A 0 )e 8t. y p(t) = 8(A 1 t + A 0 )e 8t + A 1 e 8t y p(t) = 64(A 1 t + A 0 )e 8t + 16A 1 e 8t. Substituting these expressions into the ODE gives 3te 8t = 2y p + 7y p 15y p and hence = 2 [ 64(A 1 t + A 0 )e 8t + 16A 1 e 8t] + 7 [ 8(A 1 t + A 0 )e 8t + A 1 e 8t] 15 [ (A 1 t + A 0 )e 8t] = (169A 1 t + 169A A 1 )e 8t, 169A 1 t + (169A A 1 ) = 3t. (20) The objective here is not to solve for t, but to find A 1 and A 0 such that (20) is satisfied for all < t <. This requires the coefficients of t to be the same on both sides, so that 169A 1 = 3; and also the constant terms must match, so 169A A 1 = 0. (There is no constant term on the right, which is to say it is zero.) This gives us a simple system of equations, { 169A 1 = 3 169A A 1 = 0 Solving this system yields A 1 = 3/169 and A 0 = 9/2197.

17 17 Therefore is a particular solution to the ODE. ( 3 y p (t) = 169 t ) e 8t Example Find a particular solution to y y = t sin t. Solution. Here t sin t fits the form P m (t)e αt sin βt in Theorem 4.16(2), with P m (t) = t being a first-degree polynomial so that m = 1, e αt = 1 so that α = 0, and sin βt = sin t so that β = 1. Hence, by the Method of Undetermined Coefficients a particular solution to the ODE will have the form 1 1 y p (t) = t s e 0t cos t A k t k + t s e 0t sin t B k t k = t s (A 1 t + A 0 ) cos t + t s (B 1 t + B 0 ) sin t. The auxiliary equation is r 2 1 = 0, which has roots r = ±1. So, since α + iβ = i is not a root of the auxiliary equation, by Theorem 4.16(2a) we conclude that s = 0 and hence y p (t) takes the form y p (t) = (A 1 t + A 0 ) cos t + (B 1 t + B 0 ) sin t. From this we get and y p(t) = (A 1 + B 1 t + B 0 ) cos t + ( A 1 t A 0 + B 1 ) sin t y p(t) = ( A 1 t A 0 + 2B 1 ) cos t + ( 2A 1 B 1 t B 0 ) sin t. Substituting these expressions into the ODE gives t sin t = y p(t) y p (t) = ( A 1 t A 0 + 2B 1 ) cos t + ( 2A 1 B 1 t B 0 ) sin t [(A 1 t + A 0 ) cos t + (B 1 t + B 0 ) sin t] = ( 2B 1 )t sin t + ( 2A 1 )t cos t + ( 2A 0 + 2B 1 ) cos t + ( 2A 1 2B 0 ) sin t. Equating the coefficients of the linearly independent functions t sin t, t cos t, cos t, and sin t on each side, we obtain the system 2B 1 = 1 2A 1 = 0 2A 0 + 2B 1 = 0 2A 1 B 0 = 0 Solving this system gives A 0 = 1/2, A 1 = 0, B 0 = 0, and B 1 = 1/2. Therefore y p (t) = 1 2 cos t 1 2 t sin t is a particular solution to the ODE.

18 The Superposition Principle The key to finding the general solution to a 2 y + a 1 y + a 0 y = f(t), as opposed to particular solutions, is the following theorem. Theorem 4.20 (Superposition Principle). For each 1 k n let y k (t) be a solution to a 2 y + a 1 y + a 0 y = f k (t) on an interval I k. Then for any constants b 1,..., b n the function n k=1 b ky k (t) is a solution to on the interval n k=1 I k. a 2 y + a 1 y + a 0 y = n b k f k (t) (21) k=1 Proof. The proof will be by induction. The statement of the theorem is trivially true in the case when n = 1. Let n 1 be an arbitrary fixed positive integer, and assume the theorem is true for this fixed n value. Suppose that y n+1 (t) is a solution to a 2 y + a 1 y + a 0 y = f n+1 (t) on an interval I n+1, so that a 2 y n+1(t) + a 1 y n+1(t) + a 0 y n+1 (t) = f n+1 (t) (22) for all t I n+1. Now, for any constants b 1,..., b n+1 R, our inductive hypothesis implies that n k=1 b ky k (t) is a solution to (21) on I = n k=1 I k, which is to say ( n ) ( n ) n n a 2 b k y k (t) + a 1 b k y k (t) + a 0 b k y k (t) = b k f k (t) k=1 k=1 k=1 k=1 for all t I, or equivalently a 2 n k=1 b k y k(t) + a 1 n b k y k(t) + a 0 k=1 n b k y k (t) = k=1 Meanwhile, multiplying both sides of (22) by b n+1 gives n b k f k (t). (23) a 2 b n+1 y n+1(t) + a 1 b n+1 y n+1(t) + a 0 b n+1 y n+1 (t) = b n+1 f n+1 (t) (24) for all t I n+1. Let t I I n+1. Then t I n+1 and t I are both true, so that equations (23) and (24) hold simultaneously and may be added to obtain a 2 n k=1 b k y k(t) + a 1 n b k y k(t) + a 0 k=1 n k=1 b k y k (t) + a 2 b n+1 y n+1(t) + a 1 b n+1 y n+1(t) + a 0 b n+1 y n+1 (t) = From this we conclude that n+1 n+1 n+1 n+1 a 2 b k y k(t) + a 1 b k y k(t) + a 0 b k y k (t) = b k f k (t) k=1 k=1 k=1 k=1 n b k f k (t) + b n+1 f n+1 (t). k=1 k=1

19 for all t I I n+1 = n+1 k=1 I k. Therefore the statement of the theorem is verified to be true when n + 1 is substituted for n, which completes the proof. Next comes yet another existence-uniqueness theorem, this time for initial value problems involving the nonhomogeneous differential equation (17), where f(t) 0. Theorem Let a 0, a 1, a 2, b 0, b 1, t 0 R. If y p is a particular solution to a 2 y + a 1 y + a 0 y = f(t) on an open interval I containing t 0, and y 1 and y 2 are linearly independent solutions to a 2 y + a 1 y + a 0 y = 0, then there exists a unique solution to the IVP on I of the form y p + c 1 y 1 + c 2 y 2. a 2 y + a 1 y + a 0 y = f(t), y(t 0 ) = b 0, y (t 0 ) = b 1 (25) Proof. Suppose that y p is a particular solution to (17) on an open interval I containing t 0, and y 1 and y 2 are linearly independent solutions to (3) on (, ). Proposition 4.6 implies that c 1 y 1 + c 2 y 2 is a solution to (3) for any choice of constants c 1 and c 2, and so, noting that f(t) = 0 + f(t), the Superposition Principle implies that y p + c 1 y 1 + c 2 y 2 is a solution to (17) on I. In order to have a solution to the IVP (25) we must find c 1 and c 2 so that { yp (t 0 ) + c 1 y 1 (t 0 ) + c 2 y 2 (t 0 ) = b 0 y p(t 0 ) + c 1 y 1(t 0 ) + c 2 y 2(t 0 ) = b 1 A little algebra will show that the only way to satisfy the system is to have and c 1 = [b 0 y p (t 0 )]y 2(t 0 ) [b 1 y p(t 0 )]y 2 (t 0 ) y 1 (t 0 )y 2(t 0 ) y 1(t 0 )y 2 (t 0 ) c 2 = [b 1 y p(t 0 )]y 1 (t 0 ) [b 0 y p (t 0 )]y 1(t 0 ), y 1 (t 0 )y 2(t 0 ) y 1(t 0 )y 2 (t 0 ) which is valid so long as the denominators are not zero. But Lemma 4.4 assures us that the denominators are not zero, and so y p + c 1 y 1 + c 2 y 2 with c 1 and c 2 as given above constitutes a solution to the IVP that is valid on I. Is the solution unique? If ϕ is any solution to (25) on I, then the function ˆϕ : I R defined by ˆϕ(t) = y p (t) + c 1 y 1 (t) + c 2 y 2 (t) ϕ(t) for all t I is a solution to the IVP a 2 y + a 1 y + a 0 y = 0, y(t 0 ) = 0, y (t 0 ) = 0 (26) that is valid on I. Now, by the corollary to Theorem 4.3 we can extend ˆϕ to a function ψ that satisfies (3) on (, ), so that ψ(t) = ˆϕ(t) for all t I. Now, ψ is a solution to the IVP (26) that is valid on (, ), and so too is y 0. Since any solution to (26) must be unique by Theorem 4.3, we conclude that ψ(t) = 0 for all < t <. Thus for t I we have ˆϕ(t) = 0, and so ϕ(t) = y p (t) + c 1 y 1 (t) + c 2 y 2 (t) 19

20 on I. That is, any proposed solution to (25) must be none other than y p + c 1 y 1 + c 2 y 2 itself, and so the solution is unique. Proposition If y p is a particular solution to a 2 y +a 1 y +a 0 y = f(t) on an open interval I, and y 1 and y 2 are any two linearly independent solutions to a 2 y + a 1 y + a 0 y = 0, then the general solution to a 2 y + a 1 y + a 0 y = f(t) on I is y p + c 1 y 1 + c 2 y 2. Proof. Let y p be a particular solution to (17) on an open interval I, and let y 1 and y 2 be any two linearly independent solutions to (3). Suppose ϕ is any solution to (17) on I. Fix t 0 I, and set b 0 = ϕ(t 0 ) and b 1 = ϕ (t 0 ). Then ϕ satisfies the IVP (25) on I. By Theorem 4.21 there exist constants c 1 and c 2 such that y p + c 1 y 1 + c 2 y 2 also satisfies the IVP (25) on I. Because any solution to the IVP must be unique, it follows that ϕ(t) = y p (t) + c 1 y 1 (t) + c 2 y 2 (t) on I. Therefore any solution to (17) must be of the form y p + c 1 y 1 + c 2 y 2 for some choice of constants c 1 and c 2. Example Determine the form of a particular solution to Solution. First consider the equation y + 5y + 6y = sin t cos 2t. (27) y + 5y + 6y = sin t. Consulting Theorem 4.16 we have f 1 (t) = P m (t)e αt sin βt with P m (t) = 1, e αt = 1, and sin βt = sin t, so that m = 0, α = 0, and β = 1. Now, the auxiliary equation r 2 + 5r + 6 = 0 has roots 3 and 2, and since α + iβ = i is not a root, Theorem 4.16(2a) implies that s = 0 and the particular solution takes the form 0 0 y p (t) = t 0 e 0t cos t A k t k + t 0 e 0t sin t B k t k = A 0 cos t + B 0 sin t. Next we consider the equation y + 5y + 6y = cos 2t. We have f 2 (t) = P m (t)e αt cos βt with P m (t) = 1, e αt = 1, and cos βt = cos 2t, so that m = 0, α = 0, and β = 2. Since α + iβ = 2i is not a root of the auxiliary equation, Theorem 4.16(2a) implies that s = 0 and the particular solution takes the form 0 0 y p (t) = t 0 e 0t cos 2t C k t k + t 0 e 0t sin 2t D k t k = C 0 cos 2t + D 0 sin 2t. Now, by the Superposition Principle a particular solution to (27), where sin t cos 2t = f 1 (t) + f 2 (t), has the form A 0 cos t + B 0 sin t + C 0 cos 2t + D 0 sin 2t, 20

21 21 which is a linear combination of four linearly independent functions. Example Find a general solution to and then find the solution to the initial value problem Solution. First we consider the equation y + 2y + y = t e t, (28) y + 2y + y = t e t, y(0) = 0, y (0) = 2. y + 2y + y = t 2 + 1, which has nonhomogeneity f 1 (t) = t that is of the form P m (t)e αt with P m (t) = t and e αt = 1, so that m = 2 and α = 0. The associated auxiliary equation r 2 + 2r + 1 = 0 has repeated real root r = 1, which make clear that α = 0 is not a root and so, by Theorem 4.16(1) we conclude that 2 y p1 (t) = t 0 e 0t A k t k = A 2 t 2 + A 1 t + A 0 (29) is the form of a particular solution. Substituting this, along with y p 1 (t) = 2A 2 t + A 1 and y p 1 (t) = 2A 2, into y + 2y + y = t gives which we may rewrite as 2A 2 + 2(2A 2 t + A 1 ) + (A 2 t 2 + A 1 t + A 0 ) = t 2 + 1, A 2 t 2 + (4A 2 + A 1 )t + (2A 2 + 2A 1 + A 0 ) = t 2 + 0t + 1. Equating coefficients gives rise to the system { A 2 = 1 A 1 + 4A 2 = 0 A 0 + 2A 1 + 2A 2 = 1 Solving this system yields A 0 = 7, A 1 = 4, and of course A 2 = 1. Putting these values into (29) delivers y p1 (t) = t 2 4t + 7. Next we turn to the equation y + 2y + y = e t, which has nonhomogeneity f 2 (t) = e t that is of the form P m (t)e αt with P m (t) = 1 and e αt = e t, so that m = 0 and α = 1. Since α = 1 is not a root of the auxiliary equation, by Theorem 4.16(1) we conclude that 0 y p2 (t) = t 0 e t A k t k = A 0 e t (30)

22 22 is the form of a particular solution. Substituting this into y + 2y + y = e t gives 4A 0 e t = e t, and so we must have 4A 0 = 1, or A 0 = 1/4. Putting this result into (30) results in y p2 (t) = e t /4. The Superposition Principle now implies that y p (t) = y p1 (t) + y p2 (t) = t 2 4t + 7 e t /4 is a particular solution to (28). Since the auxiliary equation has the repeated real root 1, by Theorem 4.7 the corresponding homogeneous equation y + 2y + y = 0 has general solution y h (t) = c 1 e t + c 2 te t, and so by Proposition 4.22 the general solution to (28) is y = y p + y h, or y(t) = t 2 4t et + c 1 e t + c 2 te t. (31) Finally we turn to the matter of solving the initial value problem. From (31) and the initial condition y(0) = 0 we obtain 0 = 7 1/4 + c 1, or c 1 = 27/4. Now, differentiating (31) gives y (t) = 2t et e t + c 2 e t c 2 te t, which together with the initial condition y (0) = 2 implies that c 2 = 1/2. Therefore is the solution to the IVP. Example Find a general solution to Solution. Start with y(t) = t 2 4t et 27 4 e t 1 2 te t y (4) 5y + 4y = 10 cos t 20 sin t. (32) y (4) 5y + 4y = 10 cos t. Referring to Theorem 4.16 we have f 1 (t) = P m (t)e αt cos βt with P m (t) = 10, e αt = 1, and cos βt = cos t, so that m = 0, α = 0, and β = 1. The auxiliary equation r 4 5r = 0 factors first as (r 2 4)(r 2 1) = 0, and then as (r 2)(r + 2)(r 1)(r + 1) = 0, so that it can be seen to have roots r = ±2, ±1. Since α + iβ = i is not a root of the auxiliary equation, we set s = 0 in (19) to obtain 0 0 y p1 (t) = t 0 e 0t cos t A k t k + t 0 e 0t sin t B k t k = A 0 cos t + B 0 sin t as the form of a particular solution to the ODE.

23 Next we consider y (4) 5y + 4y = 20 sin t. Again referring to Theorem 4.16, we have f 2 (t) = P m (t)e αt sin βt with P m (t) = 20, e αt = 1, and cos βt = cos t, so that m = 0, α = 0, and β = 1. Since α + iβ = i is not a root of the auxiliary equation r 4 5r = 0 we set s = 0 in (19) to obtain 0 0 y p2 (t) = t 0 e 0t cos t C k t k + t 0 e 0t sin t D k t k = C 0 cos t + D 0 sin t as the form of a particular solution to the ODE. (Note: the symbols A 0 and B 0 are already in use in this problem, so C 0 and D 0 are used here instead.) Now, by the Superposition Principle the form for a particular solution to (32) is given by y p = y p1 + y p2 ; that is, y p (t) = (A 0 cos t + B 0 sin t) + (C 0 cos t + D 0 sin t). However, the four terms on the right-hand side of the equation do not represent four linearly independent functions. If we let A = A 0 + C 0 and B = B 0 + D 0, we may write simply Substituting this into (32) gives y p (t) = A cos t + B sin t. 10 cos t 20 sin t = y (4) p (t) y p(t) + 4y p (t) = (A cos t + B sin t) 5( A cos t B sin t) + 4(A cos t + B sin t) = 10A cos t + 10B sin t, Matching coefficients gives 10A = 10 and 10B = 20, so that A = 1 and B = 2, and the particular solution becomes y p (t) = cos t 2 sin t. A general solution for the corresponding homogeneous equation y (4) 5y + 4y = 0 is y h (t) = c 1 e 2t + c 2 e 2t + c 3 e t + c 4 e t, since the roots of the auxiliary equation are ±2, ±1. Therefore by Proposition 4.22 (or rather a natural extension of it) we conclude that y(t) = cos t 2 sin t + c 1 e 2t + c 2 e 2t + c 3 e t + c 4 e t 23 is the general solution to (32).

24 Method of Variation of Parameters We start by reviewing what we know so far. Let y 1, y 2 be linearly independent solutions to a 2 y + a 1 y + a 0 y = 0. (33) on (, ). By Proposition 4.6 the general solution to this equation is the two-parameter family of functions {c 1 y 1 + c 2 y 2 : c 1, c 2 R}. By Proposition 4.22 a general solution to will be a two-parameter family of functions a 2 y + a 1 y + a 0 y = f(t) (34) {y p + c 1 y 1 + c 2 y 2 : c 1, c 2 R}, where y p is some particular solution to (34). The only real difficulty encountered in practice is finding a particular solution. The Method of Undetermined Coefficients may be employed in cases where f(t) is of the form P m (t)e αt cos βt or P m (t)e αt sin βt (or some linear combination of these), but what if it isn t? In such eventualities it would be essential to have some other method of solving nonhomogeneous equations of the form (34). One way forward is the Method of Variation of Parameters, which obtains a particular solution of the form y p (t) = v 1 (t)y 1 (t) + v 2 (t)y 2 (t) for all t in some open interval. That is, the parameters c 1, c 2 in the general solution c 1 y 1 + c 2 y 2 to (33) are replaced by functions v 1, v 2 of t to get a particular solution to (34) of the form v 1 y 1 + v 2 y 2. Define the functions w 1 = y 2 f a 2 (y 1 y 2 y 1y 2 ) and w 2 = y 1 f a 2 (y 1 y 2 y 1y 2 ). Since y 1 and y 2 are linearly independent solutions to (33) on (, ) by hypothesis, by Lemma 4.4 we have y 1 (t)y 2(t) y 1(t)y 2 (t) 0 for all < t <. This implies that w 1 and w 2 are both continuous on any open interval I where f is continuous, and so by Proposition 2.8 there exist functions v 1, v 2 such that v 1(t) = w 1 (t) and v 2(t) = w 2 for all t I; that is, we have differentiable functions v 1, v 2 : I R such that v 1 y 2 f = and v y 1 f a 2 (y 1 y 2 y 1y 2 = 2 ) a 2 (y 1 y 2 y 1y 2 ). Rearranging these two equations gives us the system { y1 v 1 + y 2 v 2 = 0 y 1v 1 + y 2v 2 = f/a 2 (35)

25 25 and so, defining y p : I R by y p = v 1 y 1 + v 2 y 2, we have y p = (v 1y 1 + v 1 y 1) + (v 2y 2 + v 2 y 2) = (v 1y 1 + v 2y 2 ) + (v 1 y 1 + v 2 y 2) = v 1 y 1 + v 2 y 2 using the first equation in the system (33), and thus y p = v 1y 1 + v 1 y 1 + v 2y 2 + v 2 y 2. Now, recalling that y 1 and y 2 are solutions to (33), from the second equation in (35) we obtain f = a 2 (v 1y 1 + v 2y 2) = (a 2 y 1 + a 1 y 1 + a 0 y 1 ) v 1 + (a 2 y 2 + a 1 y 2 + a 0 y 2 ) v 2 + (a 2 v }{{}}{{} 1y 1 + a 2 v 2y 2) 0 0 = a 2 (v 1y 1 + v 1 y 1 + v 2y 2 + v 2 y 2) + a 1 (v 1 y 1 + v 2 y 2) + a 0 (v 1 y 1 + v 2 y 2 ) = a 2 y p + a 1 y p + a 0 y p That is, the function y p = v 1 y 1 + v 2 y 2 with v 1 (t) = 1 y 2 (t)f(t) a 2 y 1 (t)y 2(t) y 1(t)y 2 (t) dt and v 2(t) = 1 a 2 y 1 (t)f(t) y 1 (t)y 2(t) y 1(t)y 2 (t) dt satisfies (34) on I, and therefore is a particular solution to (34). This formally justifies the Method of Variation of Parameters. Example Find a general solution to on the interval ( π/4, π/4). y + 4y = tan 2t (36) Solution. The auxiliary equation r = 0 has roots ±2i, so the general solution to the corresponding homogeneous equation y + 4y = 0 is y h (t) = c 1 cos 2t + c 2 sin 2t. We see that y 1 (t) = cos 2t and y 2 (t) = sin 2t are two linearly independent solutions to y +4y = 0. By the Method of Variation of Parameters a proposed particular solution to (36) is of the form y p (t) = v 1 (t) cos 2t + v 2 (t) sin 2t, where v 1 and v 2 have derivatives v 1 and v 2 that satisfy the system (35): { v 1 cos 2t + v 2 sin 2t = 0 2v 1 sin 2t + 2v 2 cos 2t = tan 2t From the first equation in the system we obtain v 1(t) = tan 2tv 2, which when substituted into the second equation yields 2v 2(t) sin 2t tan 2t + 2v 2(t) cos 2t = tan 2t,

26 and thus v 2(t) tan 2t = 2 sin 2t tan 2t + 2 cos 2t. Multiplying the fraction by cos 2t/ cos 2t, we obtain v 2(t) = sin 2t 2 sin 2 2t + 2 cos 2 2t = sin 2t 2. Now, v 1(t) = v 2(t) sin 2t tan 2t = tan 2t = sin2 2t 2 2 cos 2t. Note that both v 1 and v 2 are continuous on the interval ( π/4, π/4). Next we determine v 1 and v 2 (up to an arbitrary constant c) by integration: sin 2 2t cos 2 v 1 (t) = 2 cos 2t dt = 2t 1 dt = 1 (cos 2t sec 2t) dt 2 cos 2t 2 sin 2t = ln sec 2t + tan 2t + c, and v 2 (t) = 1 cos 2t sin 2t dt = + c. 2 4 Finally we determine y p, setting the arbitrary constants in the expressions for v 1 (t) and v 2 (t) equal to zero (the simplest choice, though any choice will do): y p (t) = v 1 (t) cos 2t + v 2 (t) sin 2t [ sin 2t = ln ] sec 2t + tan 2t cos 2t sin 2t cos 2t 4 = 1 4 (cos 2t) ln sec 2t + tan 2t. 26 The general solution to (36) is y(t) = y h (t) + y p (t), or cos 2t y(t) = c 1 cos 2t + c 2 sin 2t ln sec 2t + tan 2t 4 for all π/4 < t < π/4. In the example above we could have found a general solution to (36) on any open interval where tan(2t) is continuous, such as (π/4, 3π/4). The resultant expression for y(t) would have been the same.

27 Free Mechanical Vibrations Newton s Second Law of Motion states that the force acting on an object equals the mass of the object times the object s acceleration. In vector form we write this as F = ma, or simply F = ma if motion is along a line, in which case working with scalar quantities is sufficient. If y(t) is the position of an object on a line at time t, then y (t) gives the velocity along the line, and y (t) gives the acceleration. That is, a(t) = y (t), and so by Newton s Second Law we obtain the differential equation F = my. If an object O is set upon a horizontal surface and attached to a spring that is anchored to a wall, we create a so-called mass-spring system. Setting O in motion, we can expect a few forces to act on it. First there is the force exerted by the spring (the tension), which by Hooke s Law is proportional to the extent to which the unanchored end of the spring that is attached to O is displaced from the position where the tension is zero. The constant of proportionality is typically denoted by k, where k > 0 is called the spring constant or stiffness. Let y(t) be the position of O at time t, with y = 0 at the point where the spring is tension-free (the equilibrium position), y > 0 where the spring is stretched, and y < 0 where the spring is compressed. Then Hooke s Law states that F spring = ky, where the negative sign reflects that F spring is directed in the positive direction when y < 0, and in the negative direction when y > 0. Next, there is the force of friction exerted by the surface upon which O is moving. This force is generally considered to be proportional to the velocity y of O, with the constant of proportionality being denoted by b, where b > 0 is called the damping constant. Thus we have F friction = by, where the negative sign reflects that F friction is directed in the direction opposite the direction that O is moving. Any other forces acting on O are collectively called external forces and denoted by F ext. The net force on O is thus F friction + F spring + F ext, and if m is the mass of O, it follows from Newton s Second Law that which can be written as my = F friction + F spring + F ext = by ky + F ext, my + by + ky = F ext, a second-order linear differential equation with constant coefficients. If we consider no other forces to be acting on O other than spring tension and friction, then we set F ext = 0 and obtain simply my + by + ky = 0, (37) which is the model for a physical state known as free mechanical vibration, the kind of model that we ll be analyzing in this section.

28 Recall that a nonconstant function y(t) is periodic if there exists some constant p > 0 such that y(t) = y(t + p) for all t Dom(y), in which case p is called the period of the function. Assuming y(t) is a continuous periodic function with period p and t 0 Dom(y), let y max = max {y(t) : t [t 0, t 0 + p)} and y min = min {y(t) : t [t 0, t 0 + p)}. We define the amplitude A of the function y(t) to be 28 and the natural frequency ν to be A = y max y min, (38) 2 ν = 1 p. (39) A period value p is generally regarded as having the unit seconds/cycle, so that a natural frequency value ν has unit cycles/second. A mass-spring system for which the damping constant b is zero, which is to say there is no friction acting on the object, is called an undamped system. In such a system the object O will exhibit an oscillatory motion modeled by a periodic position function y(t). Example A 2 kg object O is attached to a spring with stiffness k = 50 N/m. Initially O is displaced 1/4 m to the left of the equilibrium point and given a velocity of 1 m/s to the left. Neglecting damping, find the equation of motion of O, the times O is at the equilibrium position, and also the period, natural frequency, and amplitude of its motion. Solution. To find the equation of motion means to find an expression for y(t) which gives the position of O at time t. Referring to (37), we obtain an initial value problem 2y + 50y = 0, y(0) = 1/4, y (0) = 1, where y(0) = 1/4 indicates that O is initially 1/4 m to the left of the equilibrium point, and y (0) = 1 indicates that O initially has a velocity of 1 m/s in the leftward direction. The auxiliary equation 2r = 0 has roots α ± iβ = ±5i, so that α = 0 and β = 5, and by Theorem 4.12 the general solution to the ODE is y(t) = e αt (c 1 cos βt + c 2 sin βt) = c 1 cos 5t + c 2 sin 5t. y(t) t 1 t 2 t t Figure 1. An undamped system.