6 Second Order Linear Differential Equations

Transcription

1 6 Second Order Linear Differential Equations A differential equation for an unknown function y = f(x) that depends on a variable x is any equation that ties together functions of x with y and its derivatives. It is a second order differential equation if it involves d2 y dy, possibly together with, y and functions of x, but dx2 dx no higher order derivatives. For example d 2 y d = sin x, dx2 2 y dx 2 + xy2 = e x2, ( d 2 y dx 2 ) 2 + e x dy dx = ln(x2 + 1) are all second order differential equations in the unknown function y, but d 3 y dx 3 + dy dx = 0 is third order. A linear second order differential equation is one of the form p(x) d2 y dx 2 + q(x)dy + r(x)y = f(x) dx ( ) for functions p, q, r and f of the variable x. It does not feature products of y with its derivatives, or other such functions. For example d 2 y d = sin x, dx2 are both linear, but none of the following are linear: d 2 y dx 2 + xy2 = 0, 2 y dy + ex + (cos x)y = tanx dx2 dx d 2 y dx 2 cosy = ex, y d2 y ( dy ) 2 dx 2 + = 0. dx In general, linear equations are easier to solve than nonlinear equations. Indeed, often it is not known how to obtain an exact solution to a nonlinear equation, so instead one approximates the solution by studying related linear equations. The general linear second order differential equation ( ) is called homogeneous if f(x) is identically zero, i.e. f(x) = 0 for all x R, otherwise it is nonhomogeneous. This is the same terminology used earlier for matrix equations, since we have the following result analogous to the Theorem 3.4: Theorem 6.1. Consider the linear second order ODE p(x)y + q(x)y + r(x)y = f(x) (N) and its associated homogeneous equation p(x)y + q(x)y + r(x)y = 0. (H) (a) If y 1 and y 2 are solutions of (H) then so are y 1 + y 2, ky 1 and ky 2 for any number k. (b) If y 1 is a solution of (N) and y 0 a solution of (H), then y 1 + y 0 is a solution of (N). On the other hand, if y 2 is another solution of (N) then y 1 y 2 is a solution of (H). Remark. As a consequence of this, we see that to find the general solution of a nonhomogeneous linear ODE it suffices to find one solution, together with all solutions of the associated homogeneous equation. 109

2 Homogeneous linear equations with constant coefficients Proof. The proof is basically the same as that for the corresponding matrix algebra result (Theorem 3.4). For example if y 1 and y 2 both satisfy (N) then So d dx (y 1 y 2 ) = y 1 y 2 and d 2 dx 2 (y 1 y 2 ) = y 1 y 2. p(x) d2 dx 2 (y 1 y 2 ) + q(x) d dx (y 1 y 2 ) + r(x)(y 1 y 2 ) = p(x) ( y 1 y 2) ( + q(x) y 1 y 2) + r(x)(y1 y 2 ) = [ p(x)y 1 + q(x)y 1 ] [ + r(x)y 1 p(x)y 2 + q(x)y 2 ] + r(x)y 2 = f(x) f(x) = 0, hence the difference y 1 y 2 solves (H). The other parts can be shown by similarly. Heuristically speaking, since a second order ODE should in some sense require two integrations to find a solution, two items of information ought to be required to specify a particular solution. For example: d 2 y dx 2 = 0 dy = A y = Ax + B, dx for constants A and B. To find A and B we could pose an initial value problem: y(x 0 ) = a, y (x 0 ) = b or a boundary value problem: y(x 1 ) = c, y(x 2 ) = d, for x 1 x 2. For the equation above either case will lead to unique values of A and B, since the functions y = x and y = 1 are linearly independent neither is a constant multiple of the other. 6.1 Homogeneous linear equations with constant coefficients The simplest linear second order ODE is ay + by + cy = 0, (H) where a, b, c R are constants, a 0. Since the function y = e rx has derivatives y = re rx and y = r 2 e rx, i.e. multiples of the original function, it is a potential candidate for a solution to (H). Substituting these functions into (H) gives ar 2 e rx + bre rx + ce rx = (ar 2 + br + c)e rx = 0. Now e rx 0 for all x R, so the only way this equation can be satisfied is if ar 2 + br + c = 0, (A) that is, if r is a solution to the auxiliary equation (A), i.e. if r = b ± b 2 4ac. 2a Depending on the nature of the solutions r above, we have different types of solution: 110

3 b 2 4ac > 0 In this case there are two distinct real solutions r 1 and r 2 to (A). Thus e r1x and e r2x are both solutions. Since r 1 r 2 these functions are linearly independent, hence the general solution is y = A 1 e r1x + A 2 e r2x for constants A 1 and A 2. b 2 4ac = 0 In this case the only solution to (A) is r = b/2a, but we want two linearly independent solutions. Another is given by taking y = xe rx for this r since then y = e rx + rxe rx, y = 2re rx + r 2 xe rx and so ay + by + cy = (ar 2 + br + c)xe rx + (2ar + b)e rx = 0 because ar 2 + br + c = 0 by choice of r, and 2ar + b = 2a ( b/2a) + b = 0 as well. Hence the general solution is y = (A 1 + A 2 x)e rx for constants A 1 and A 2. Note that xe rx is a nonconstant multiple of e rx, so these are linearly independent functions. b 2 4ac < 0 Since the discriminant of (A) is negative, we will get two distinct complex roots, but since a, b, c R we find that the roots are conjugate to each other: r = p ± iq for p = b 4ac b 2a, and q = 2. 2a Thus the general solution to (H) in this case is y = A 1 e px+iqx + A 2 e px iqx = A 1 e px( cosqx + i sinqx ) + A 2 e px( cos( qx) + i sin( qx) ) = e px( (A 1 + A 2 )cosqx + i(a 1 A 2 )sin qx ) = e px( B 1 cosqx + B 2 sin qx ) for constants B 1 and B 2. Although we have made use of complex exponentials, it is always possible to choose A 1 and A 2 so that the numbers B 1 and B 2 are real, if that is an issue. Indeed, inverting the equations that define B 1 and B 2 in terms of A 1 and A 2 we get A 1 = 1 2 (B 1 ib 2 ), A 2 = 1 2 (B 1 + ib 2 ). Exercise 6.2. Solve the initial value problem y 5y + 6y = 0, y(0) = 1, y (0) =

4 Homogeneous linear equations with constant coefficients Exercise 6.3. Solve the boundary value problem y + 2y + y = 0, y(0) = 1, y(1) = 1. Exercise 6.4. Solve the boundary value problem d2 x dt 2 +2dx dt +10x = 0, x(0) = 2, x( ) π 6 = 1. Example 6.5 (S05 7(a i)). Find the general solution to the ordinary differential equation d 2 y dx 2 + 8dy + 41y = 0. dx 112

5 The auxiliary equation is r 2 + 8r + 41 = 0 r = 8 ± = 8 ± = 4 ± 5i. Hence the general solution is y = e 4x (Acos5x + B sin 5x), where A and B are constants. Example 6.6. Find the general solution of the ordinary differential equation 16 d2 z dt 2 +24dz dt + 9z = 0. The auxiliary equation is 16r r + 9 = (4r + 3) 2 = 0, which has repeated root 3 4. Thus the general solution is z = (A + Bt)e 3t/ Nonhomogeneous linear equations with constant coefficients From Theorem 6.1 we know that to find the general solution to ay + by + cy = f(x) (N) it is enough to guess one particular solution y P, and then add to this the general solution y C of the associated homogeneous equation (known as the complementary equation) got by setting f(x) 0. What makes an appropriate guess is to some extent found by trial and error, but there are a number of obvious choices depending on what appears on the right hand side of (N), as given below: f(x) = Trial solution y P px n a n x n + + a 1 x + a 0 p cosqx (or p sinqx) k 1 cosqx + k 2 sin qx pe qx px n cosqx (or px n sin qx) px n e qx pe qx cosrx (or pe qx sinrx) For example if f(x) = x then trying y P = a 1 x + a 0 we have and so need ke qx (a n x n + + a 1 x + a 0 )(k 1 cosqx + k 2 sinqx) (a n x n + + a 1 x + a 0 )e qx e qx (k 1 cosrx + k 2 sin rx) y P = a 1, y P = 0, a 0 + b a 1 + c(a 1 x + a 0 ) = ca 1 x + (ca 0 + ba 1 ) = x. Equating coefficients gives the following linear system for a 0 and a 1 : ca 1 = 1, ca 0 + ba 1 = 0 a 1 = 1 c, a 0 = b c 2 provided c 0. Note that if c = 0 then we cannot solve the system above for a 1 and a 0 the linear system is inconsistent. However, this is not actually a surprise: if c = 0 then the complementary equation is ay +by = 0, which has auxiliary equation r(ar+b) = 0, with roots of 0 and b/a. As a result, assuming b 0, the solution to the complementary equation is y C = A+Be bx/a, so that the constant term a 0 in our trial solution y P is already a solution to the complementary equation. That is, this part of the trial solution disappears when we calculate ay P + by P. In general, if the usual trial solution y P turns out to be a solution of the complementary equation then the next step is to try the function xy P as the trial solution instead of y P. So above we should use y P = x(a 1 x + a 0 ) = a 1 x 2 + a 0 x y P = 2a 1x + a 0, y P = 2a 1, 113

6 Nonhomogeneous linear equations with constant coefficients and so we need to solve a(2a 1 ) + b(2a 1 x + a 0 ) = x a 1 = 1 2b, a 0 = a b 2, where we now have to assume that b 0. If b = 0 as well, then 0 is a repeated root of the auxiliary equation, and we should try an extra power of x, i.e. y P = x 2 (a 1 x + a 0 ), and then find that we need to take a 1 = 1/6a, a 0 = 0 to get ay P = x. Exercise 6.7 (A04 7(a)). Find the general solution of y + y 6y = 7e 2x. Exercise 6.8. Find the general solution of d2 r dθ 2 + 2dr + r = sin 2θ. dθ Exercise 6.9 (A04 7(c)). Solve the boundary value problem y + 4y + 8y = x, y(0) = 1, y ( π 4 ) =

7 Exercise 6.10 (S03 5(b)). Find a particular solution to d2 x dt 2 x = 3et. 115

8 Nonhomogeneous linear equations with constant coefficients Exercise Solve the initial value problem y 6y + 9y = 8e 3x, y(0) = 1, y (0) = 1. Example 6.12 (S05 7(aii)). Find the general solution to d2 y dx 2 4dy dx + 5y = e x. The auxiliary equation is r 2 4r + 5 = 0 r = 4 ± = 4 ± 4 2 = 2 ± i. Thus the solution to the complementary equation is y C = e 2x (Acosx + B sin x). For a particular solution to the nonhomogeneous equation try y P = ke x, then y P = ke x, y P = ke x y P 4y P + 5y P = ke x + 4ke x + 5ke x = 10ke x, 116

9 so we want 10k = 1. Thus the general solution is where A and B are constants. y = y P + y C = 1 10 e x + e 2x (Acos x + B sinx), Example 6.13 (S05 7(b)). Solve the initial value problem The auxiliary equation is d 2 y dx 2 dy dx = 5 cos3x, y(0) = 3, y (0) = 1. r 2 r = r(r 1) = 0 r = 0, 1. Thus the solution to the complementary equation is y C = A + Be x, where A and B are constants. For a particular solution try y P = C cos3x + D sin 3x, then and so y P = 3C sin 3x + 3D cos3x, y P = 9C cos3x 9D sin 3x y P y P = 9C cos3x 9D sin 3x ( 3C sin3x + 3D cos3x) = ( 9C 3D)cos3x + (3C 9D)sin3x. For y P to be a solution of the nonhomogeneous equation we need 9C 3D = 5, and 3C 9D = 0 C = 3D, hence 27D 3D = 30D = 5 So the general solution is But now D = 1 6, C = 1 2. y = y C + y P = A + Be x 1 2 cos3x 1 sin 3x 6 y(0) = 3 = A + B 1 2, and y = Be x sin 3x 1 2 cos3x, so y (0) = 1 = B 1 2 Thus B = 3 2 and A = 3 1 = 2, and so the solution of the initial value problem is y = ex 1 2 cos3x 1 sin 3x. 6 Example Find the general solution to d2 y dx 2 4dy 21y = sin2x. dx The auxiliary equation of the complementary equation is r 2 4r 21 = 0, i.e. (r 7)(r + 3) = 0, which has roots 3, 7. Thus the general solution of the complementary equation is y C = Ae 3x + Be 7x. For a particular solution to the nonhomogeneous equation try y P = C sin2x + D cos2x y P = 2C cos2x 2D sin2x, y P = 4C sin 2x 4D cos2x. 117

10 Nonconstant coefficients So we need y P 4y P 21y P = ( 4C + 8D 21C)sin 2x + ( 4D 8C 21D)cos2x = sin 2x, thus we must solve 25C + 8D = 1 and 8C 25D = 0. From the second equation we get D = C; substituting this into the first gives 25C 25 C = C = 1, hence C = 689 and D = Thus the general solution is y = Ae 3x + B 7x 25 8 sin 2x cos2x Example Solve the initial value problem 2 d2 y dx dy dx + 25y = 75, y(0) = 1, y (0) = 4. The complementary equation has auxiliary equation 2m 2 +10m+25 = 0, which has roots 10± = 5 2 ± 5 2 i. Thus the complementary function is y C = e (Acos 5x/2 5x 2 ) + B sin 5x 2. For a particular solution try y P = C, a constant. Then y P = y P = 0, and so we need Thus the general solution is 2y P + 10y P + 25y P = 25C = 75 C = 3. y = 3 + e 5x/2( Acos 5x 2 + B sin 5x 2 y = 5 2 e 5x/2( Acos 5x 2 + B sin 5x ) e 5x/2( Asin 5x 2 + B cos 5x ) 2 So y(0) = 3+A = 1 A = 4, and y (0) = B = 4 so that B = 2 5 ( 4+10) = ( ) Hence the solution is y = 3 + e 5x/2 4 cos 5x sin 5x Nonconstant coefficients There are a few circumstances when it is not too difficult to solve the equation ) p(x)y + q(x)y + r(x)y = f(x). (N) The first is when r(x) 0, since if we set u = y, i.e. let u be the first derivative of y, then u = y and so the equation becomes p(x)u + q(x)u = f(x) u + q(x) p(x) u = f(x) p(x). That is, we can obtain a first order linear ODE for the function u, solve this using standard techniques, and then recover y since y = y dx = u dx. Exercise 6.16 (A04 7(d)). Solve x 2 y + 2xy =

11 Exercise Solve t d2 x dt 2 + dx dt = 1. Example By writing u = dy, find the general solution of the second order ordinary dx differential equation 2 d2 y dx 2 + ex = 3 dy dx. Let u = y, then u = y, so if 2y + e x = 3y then 2u + e x = 3u, and thus u 3 2 u = ex. The integrating factor for this first order equation for u is e and so 3 2 dx = e 3x/2 u 3 2 u = ex e 3x/2 u 3 2 e 3x/2 u = d dx (e 3x/2 u) = e 3x/2 e x = e x/2. 119

12 Nonconstant coefficients Integrating this last equation we get e 3x/2 u = 2e x/2 +A, A a constant. So u = 2e x +Ae 3x/2. But then u = y = dy dx = 2ex + Ae 3x/2, and integrating again gives where B and C are constants (with B = 2 3 A). y = 2e x + Be 3x/2 + C, Note. Above we had constant coefficients, so could have used our earlier methods. Another straightforward situation is if we have found one solution y 1 to the homogeneous equation p(x)y + q(x)y + r(x)y = 0. (H) To find the general solution, try y = v(x)y 1, where we want to find v(x). Since y = v y 1 + vy 1, y = v y 1 + 2v y 1 + vy 1, substituting these into the equation (H) we get, after some rearrangement, p [ v y 1 + 2v y 1 + vy 1] + q [ v y 1 + vy 1] + rvy1 = v [ py 1 + qy 1 + ry 1] + py1 v + [ 2py 1 + qy 1] v = py 1 v + [ 2py 1 + qy 1] v = 0 which is a second order ODE for v, in which the function v does not explicitly appear, only its derivatives. So we can now set u = v and solve the resulting first order ODE for u, integrate this to get v, and multiply by y 1 to get more solutions of (H). Exercise Verify that y 1 = x is a solution of x 2 y x(x + 2)y + (x + 2)y = 0. Find the general solution by trying y 2 = xv(x). 120

13 Exercise Verify that y 1 = x 2 is a solution of d2 y dx 2 3 dy x dx + 4 y = 0, and hence find x2 the general solution. Finally, suppose we have found two solutions y 1 and y 2 to the homogeneous equation (H) and wish to solve (N). We try y = v 1 y 1 + v 2 y 2 for functions v 1 and v 2. Then y = v 1 y 1 + v 2 y 2 + v 1 y 1 + v 2y 2, y = (v 1 y 1 + v 2 y 2) + v 1 y 1 + v 1y 1 + v 2 y 2 + v 2y 2 121

14 Nonconstant coefficients Substituting these into (N) yields py + qy [ + ry = v 1 py 1 + qy 1 + ry ] [ 1 + v2 py 2 + qy 2 + ry ] 2 + p(v 1y 1 + v 2y 2) + p(v 1y 1 + v 2y 2 ) + q(v 1y 1 + v 2y 2 ) Thus if we can choose v 1 and v 2 so that = p(v 1 y 1 + v 2 y 2 ) + p(v 1 y 1 + v 2 y 2) + q(v 1 y 1 + v 2 y 2) = f(x) v 1y 1 + v 2y 2 = 0 and p(v 1y 1 + v 2y 2) = f then y = v 1 y 1 + v 2 y 2 will be a solution of the nonhomogeneous equation (N). Exercise Verify that y 1 = x is a solution of x 2 y +xy y = 0. Find another solution y 2, and hence find the general solution to x 2 y + xy y = xlnx. 122

15 Exercise Given that secxdx = ln tan x+secx, find the general solution of y +y = tan x. 123

16 Exercises 6.4 Exercises 1. Find the general solution of the following differential equations: (i) y y 6y = 0 (ii) y 9y + 20y = 0 (iii) y + 10y + 29y = 0 (iv) y + y + y = 0 (v) y 10y + 25y = 0 (vi) y 4y + y = 0 2. Solve the following initial value problems: (i) y + 2y + 4y = 0, y(0) = 1, y (0) = 0 (ii) y + 3y = 0, y(0) = 3, y (0) = 6 (iii) y 2y 5y = 0, y(0) = 0, y (0) = 3 (iv) y + 4y = 0, y(0) = 1, y (0) = 8 3. Solve the following boundary value problems: (i) y 16y = 0, y(0) = 5, y( 1 4 ) = 5e (ii) y 2y = 0, y(0) = 1, y( 1 2 ) = e 2 (iii) y 2y + 2y = 0, y(0) = 3, y( π 2 ) = 0 (iv) y + 4y + 5y = 0, y( π 2 ) = 14e π, y( 3π 2 ) = 14e 3π 4. Find the general solution of the following differential equations: (i) y y 2y = 2x (iii) y 4y + 13y = 3e 2x 5e 3x (ii) y 6y + 8y = 3e x (iv) y + 2y = 4 cos 2 3x 5. Solve the following initial value problems: (i) y + 4y = 34 cosx + 8, y(0) = 3, y (0) = 2 (ii) y + y = 5 sin2x 4, y(0) = 0, y (0) = 3 (iii) y 3y = 2e 2x sin x, y(0) = 1, y (0) = 2 (iv) y 6y + 9y = 4e 3x, y(0) = 1, y (0) = 2 6. Find the general solution of the following differential equations: (i) xy = 2 + y (ii) 3y 2y = 8x + 2 (iii) y + x(y ) 2 = 0 124

17 7. In each of the following, verify that the function y 1 is a solution of the given differential equation. Use the substitution y 2 = u(x)y 1 to find the general solution. (i) y 1 x y 8 x 2 y = 0, y 1(x) = x 4 (ii) y + 2x 1 x 2 y 2 1 x 2 y = 0, y 1(x) = x ( (iii) y ) ( 1 y x x 2 ) x 2 y = 0, y 1 (x) = x 8. In each of the following show that the given functions y 1 and y 2 solve the associated homogeneous problem, and hence find the general solution of the nonhomogeneous equation. (i) x 2 y 2y = 3x 2 1, y 1 (x) = x 2, y 2 (x) = x 1. (ii) xy (1 + x)y + y = x 2 e 2x, y 1 (x) = 1 + x, y 2 (x) = e x. (iii) x 2 y 3xy + 4y = x 2 lnx, y 1 (x) = x 2, y 2 (x) = x 2 lnx. 125