ODE Homework Solutions of Linear Homogeneous Equations; the Wronskian

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1 ODE Homework Solutions of Linear Homogeneous Equations; the Wronskian 1. Verify that the functions y 1 (t = e t and y (t = te t are solutions of the differential equation y y + y = 0 Do they constitute a fundamental set of solutions? [ 3. #] Sol. Observe that y 1 y 1 + y 1 = e t e t + e t = 0 y y + y = (e t + te t (e t + te t + te t = 0 which shows that y 1, y are solutions of the equation. Furthermore W (e t, te t = et te t e t e t + te t = et + te t te t = e t 0 Hence y 1, y forms a fundamental set of solutions.. Verify that the functions y 1 (x = x and y (x = sin x are solutions of the differential equation (1 x cot xy xy + y = 0, 0 < x < π Do they constitute a fundamental set of solutions? [ 3. #7] Sol. Observe that (1 x cot xy 1 xy 1 + y 1 = 0 x + x = 0 (1 x cot xy xy + y = (1 x cot x sin x x cos x + sin x = sin x + x cos x x cos x + sin x = 0 which shows that y 1, y are solutions of the equation. Furthermore W (x, sin x = x sin x 1 cos x = x cos x sin x Since π cos π sin π = 1 0, so W (x, sin x 0.Hence y 1, y forms a fundamental set of solutions.

2 3. Find the Wronskian of two solutions of the differential equation t y t(t + y + (t + y = 0 with out solving the equation [ 3. #9] Sol. Rewrite the equation as the form y t(t + y + t + y = 0 t t Then by Abel s formula, the Wronskian of two solutions is ( t(t + W (t = c exp dt = c exp ( t + ln t = c t e t for some constant c. t. Find the Wronskian of two solutions of the differential equation x y + xy + (x + ν = 0 with out solving the equation [ 3. #31] Sol. Rewrite the equation as the form Bessel s equation y + 1 x y + x + ν y = 0 x Then by Abel s formula, the Wronskian of two solutions is ( 1 W (x = c exp x dx = c exp ( ln x = c x for some constant c.. If y 1 and y are a fundamental set of solutions of t y y + (3 + ty = 0 and if W (y 1, y ( = 3, find the value of W (y 1, y (6. [ 3. #3] Sol. Rewrite the equation as the form y t y t y = 0 t Then by Abel s formula, the Wronskian of two solutions is ( ( W (y 1, y (t = c exp t dt = c exp = ce t t for some constant c. Since W (y 1, y ( = 3, we have that ce 1 = 3 c = 3e. That is, the Wronskian of y 1 and y is W (y 1, y (t = 3e t t

3 Thus W (y 1, y (6 = 3e 6 6 = 3e Exact Equations. The equation P (xy +Q(xy +R(xy = 0 is said to be exact if it can be written in the form [P (xy ] + [f(xy] = 0, where f(x is to be determined in terms of P (x, Q(x, and R(x. The latter equation can be integrated once immediately, resulting in a first order linear equation for y that can be solved. By equating the coefficients of the preceding equations and then eliminating f(x, the necessary condition for exactness is P (x Q (x + R(x = 0. It can be shown that this is also a sufficient condition. Determine whether the equation xy (cos xy + (sin xy = 0, x > 0 is exact. If so,, solve the equation [ 3. #] Sol. By letting P (x = x, Q(x = cos x, R(x = sin x, then P Q + R = 0 sin x + sin x = 0 which shows that the equation is exact. Assume that the equation can be rewrite as the form (xy + [f(xy] = 0, then xy (cos xy + (sin xy = xy + y + f(xy + f (xy which implies that f(x = cos x 1. Thus the equation can be reformulate as the form (xy ((cos x + 1y = 0 Integrating on both side, we get xy (cos x + 1y = c 1, that is, y cos x+1 y = c 1 x x, which is a 1st order linear equation with integrating factor ( x cos s + 1 µ(x = exp ds s The general solution for the equation is y(x = c x 1 µ(s µ(x s ds + c µ(x cos τ+1dτ ( x τ cos s + 1 x exp ( x s = c 1 s for some constants c 1, c. ds + c exp s ds 7. The Adjoint Equation. If a second order linear homogeneous equation is not exact, it can be made exact by multiplying by an appropriate integrating factor µ(x. Thus we require that µ(x be such that µ(xp (xy + µ(xq(xy + µ(xr(xy = 0 can be

4 written in the form [µ(xp (xy ] + [f(xy] = 0. By equating coefficients in these two equations and eliminating f(x, the function µ must satisfy P µ + (P Qµ + (P Q + Rµ = 0. This equation is known as the adjoint of the original equation and is important in the advanced theory of differential equations. In general, the problem of solving the adjoint differential equation is as difficult as that of solving the original equation, so only occasionally is it possible to find an integrating factor for a second order equation. Find the adjoint of the differential equation (1 x y xy + α(α + 1y = 0. Legendre s equation [ 3. #8] Sol. By letting P (x = 1 x, Q(x = x, R(x = α(α + 1, then P Q + R = + + α(α + 1 = α(α + 1 which shows that the equation is not exact if α 0, 1. Furthermore, P Q = x + x = x, then the adjoint of the equation is (1 x µ xµ + α(α + 1µ = 0 which is the given differential equation itself Complex Roots of the Characteristic Equation 8. Find the general solution of the differential equation 9y + 9y y = 0 [ 3.3 #1] Sol. The characteristic equation is 9r + 9r = 0 Thus we get r =, 1. Therefore the general solution is of the 3 3 form y(t = c 1 e t 3 + c e t 3

5 9. Find the solution of the initial value problem. y + y + y = 0, y(0 = 1, y (0 = 0 Sketch the graph of the solution and describe its behavior for increasing t. [ 3.3 #18] Sol. The characteristic equation is r + r + = 0 Thus we get r = ± i. Therefore the general solution is of the form y(t = c 1 e t cos t + c e t sin t Thus y (t = (c c 1 e t cos t (c 1 + c e t sin t. y(0 = 1, y (0 = 0, so c 1 = 1 c c 1 = 0 Since and we have that c 1 = 1, c =. Hence the solution of the initial value problem is y(t = e t cos t + e t sin t = e t sin(t + θ where θ = sin 1 1. Note that both e t 0 as t. Hence y(t = 0. The graph of solution is as follows lim t 10. Consider the initial value problem y + y + 6y = 0, y(0 =, y (0 = α 0.

6 (a Find the solution y(t of this problem. (b Find α so that y = 0 when t = 1. (c Find, as a function of α, the smallest positive value of t for which y = 0. (d Determine the limit of the expression found in part (c as α. [ 3.3 #] Sol. (a The characteristic equation is r + r + 6 = 0 Thus we get r = 1 ± i. Therefore the general solution is of the form y(t = c 1 e t cos t + c e t sin t Thus y (t = ( c c 1 e t cos t ( c 1 + c e t sin t. Since y(0 =, y (0 = α, so c 1 = c c 1 = α and we have that c 1 =, c = α+. Hence the solution of the initial value problem is y(t = e t cos t + α + e t sin t (b If y(1 = 0, then e 1 cos + α+ e 1 sin = 0. That is, cos = α + sin α = cot (c If y(t = 0, then cos t + α+ sin t = 0, that is, tan t = α + Note that α 0, then α > and we have that < 0. α+ Thus, the time t = t(α for which y = 0, is + kπ for some k N. Hence the smallest positive value of t for t = 1 tan 1 ( α + which y = 0 is t(α = 1 ( tan 1 + π = π 1 ( tan 1 α + α +

7 (d ( π lim t(α = lim 1 ( tan 1 α α α + = π 1 ( lim tan 1 = π α α Consider the initial value problem y + ay + (a + 1y = 0, y(0 = 1, y (0 = 0. (a Find the solution y(t of this problem. (b For a = 1 find the smallest T such that y(t < 0.1 for t > T. (c Repeat part (b for a = 1, 1, and. (d Using the results of parts (b and (c, plot T versus a and describe the relation between T and a. [ 3.3 #6] Sol. (a The characteristic equation is r + ar + (a + 1 = 0 Thus we get r = a ± i. Therefore the general solution is of the form y(t = c 1 e at cos t + c e at sin t Thus y (t = (c ac 1 e t cos t (c 1 + ac e t sin t. Since y(0 = 1, y (0 = 0, so c 1 = 1 c ac 1 = 0 and we have that c 1 = 1, c = a. Hence the solution of the initial value problem is y(t = e at cos t + ae at sin t = 1 + a e at sin(t + θ where θ = sin a. (b Note that y(t 1 + a e at, the inequality 1 + a e at 0.1 implies that t 1 a ln( a

8 Therefore T 1 ln( a a. For a = 1, the solution y(t becomes y(t = ( e t sin t + π By using Newton s method, to solve y(t = 0.1, we have that T (c For a = 1, 1 and, the corresponding T is T , T 1 7.8, T respectively. (d The plots of T versus a is as follows Note that the estimate T a approach the graph 1 a ln( a as a gets large. 1. Euler Equations. An equation of the form t d y dt + αtdy + βy = 0, t > 0, (ii dt where α and β are real constants, is called an Euler equation. (a Let x = ln t and calculate dy and d y in terms of dy and dt dt dx d y. dx (b Use the results of part (a to transform Eq. (ii into d y + (α 1dy + βy = 0. dx dx (iii Observe that Eq. (iii has constant coefficients. If y 1 (x and y (x form a fundamental set of solutions of Eq. (iii,

9 then y 1 (ln t and y (ln t form a fundamental set of solutions of Eq. (ii. [ 3.3 #3] Sol. (a Let x = ln t, then dx dt = 1 t. Thus dy dt = dy dx dx dt = 1 dy t dx d y dt = d ( dy = d ( 1 dy dt dt dt t dx = 1 dy t dx + 1 d ( dy t dx dx (b By (a, Eq. (ii becomes ( t 1 dy t dx + 1 d y t dx 13. Solve the equation d y + (α 1dy dx dx + βy = 0 t y + 7ty + 10y = 0 = 1 t dy dx + 1 t d dt = 1 t dy dx + 1 t d y dx ( dy dx + αt 1 dy t dx + βy = 0 [ 3.3 #1] Sol. By letting x ln t, the above problem shows that the equation t y + 7ty + 10y = 0 can be transformed to the equation ÿ + 6ẏ + 10y = 0 where = d. The characteristic equation for the new differential dx equation is r + 6r + 10 = 0 Thus we get r = 3 ± i. Therefore the general solution is of the form y(x = c 1 e 3x cos x + c e 3x sin x Since x = ln t, hence the general solution for the original equation is y(t = c 1 t 3 cos ( ln t + c t 3 sin ( ln t 3.. Repeated Roots; Reduction of Order

10 1. Find the general solution of the differential equation y 6y + 9y = 0 [ 3. #6] Sol. The characteristic equation is r 6r + 9 = (r 3 = 0 Thus we get r = 3 with multiplicity. Therefore the general solution is of the form y(t = c 1 e 3t + c te 3t 1. Consider the initial value problem y + 1y + 9y = 0, y(0 = 1, y (0 =. (a Solve the initial value problem and plot its solution for 0 t. (b Determine where the solution has the value zero. (c Determine the coordinates (t 0, y 0 of the minimum point. (d Change the second initial condition to y (0 = b and find the solution as a function of b. Then find the critical value of b that separates solutions that always remain positive from those that eventually become negative. [ 3. #1] Sol. (a The characteristic equation is r + 1r + 9 = (r + 3 = 0 Thus we get r = 3 with multiplicity. Therefore the general solution is of the form y(t = c 1 e 3t + c te 3t e 3t Thus y (t = ( c 3c 1 1, y (0 =, so c 1 = 1 c 3 c 1 = 3c te 3t. Since y(0 = and we have that c 1 = 1, c =. Hence the solution of the initial value problem is y(t = e 3t t 3t e = t e 3t

11 The graph of solution for 0 t is as follows (b Let y(t = 0, then t 3t e = 0 t = 0 t =. (c Observe that y (t = 3t e 6 1t e 3t = 1t 16 e 3t. Then y (t = 0 if and only if t = t 0 = 16. Note that 1 y (t = 78 t e 3t and y ( = 1 e 8 > 0. Hence y(t has minimum y 0 = y ( 16 1 = 3 e 8 at t = t 0. That is, the minimum point for the graph of y(t is (t 0, y 0 = ( 16, 1 3 e 8. (d If the initial data becomes y(0 = 1, y (0 = b, then the solution for the initial value problem becomes y(t = (3 + bt + e 3t = e 3t b te 3t Then we can see that the solution y(t remains positive for all t if and only if b+3 0, therefore, the critical value is b = Consider the initial value problem 16y + y + 9y = 0, y(0 = a > 0, y (0 = 1. (a Solve the initial value problem. (b Find the critical value of a that separates solutions that become negative from those that are always positive. [ 3. #18] Sol. (a The characteristic equation is 16r + r + 9 = (r + 3 = 0

12 Thus we get r = 3 with multiplicity. Therefore the general solution is of the form y(t = c 1 e 3t + c te 3t e 3t Thus y (t = ( c 3c 1 a, y (0 = 1, so c 1 = a c 3 c 1 = 1 3c te 3t. Since y(0 = and we have that c 1 = a, c = 3a. Hence the solution of the initial value problem is y(t = ae 3t 3a + te 3t (3a t + a = e 3t (b Since y(t = ae 3t + 3a te 3t. The solution remains positive for all t if and only if 3a 0, so the critical value is a = (a Consider the equation y + ay + a y = 0. Show that the roots of the characteristic equation are r 1 = r = a, so that one solution of the equation is e at. (b Use Abel s formula to show that the Wronskian of any two solutions of the given equation is W (t = y 1 (ty (t y 1(ty (t = c 1 e at, where c 1 is a constant. (c Let y 1 (t = e at and use the result of part (b to obtain a differential equation satisfied by a second solution y (t. By solving this equation, show that y (t = te at. [ 3. #0] Sol. (a The characteristic equation is r + ar + a = (r + a = 0 Thus the root of characteristic equation are r 1 = r = a, and thus e at is a solution of the equation. (b By Abel s formula, the Wronskian is ( W (t = c 1 exp adt = c 1 e at for some constant c 1. Without loss of generality, we may take c 1 = 1.

13 (c Let y 1 (t = e at, then y 1(t = ae at. By the definition of the Wronskian and the result of (b, the second solution y (t must satisfy the differential equation That is, e at y (t + ae at y (t = W (t = e at y (t + ay (t = e at The integrating factor µ(t = exp adt = e at, we obtain that y (t = e at dt = te at + c e at for some constant c, by taking c = 0, we get y (t = te at. 18. Use the method of reduction of order to find a second solution of the differential equation t y ty + 6y = 0, t > 0;, y 1 (t = t [ 3. #] Sol. Let y (t = t v(t. Substituting into the equation, we get t (v + tv + t v t(tv + t v + 6t v = 0 That is, v = 0, which implies that v(t = at + b for some constants a, b. Thus, y (t = at 3 +bt. By setting a = 1, b = 0, we get y (t = t Use the method of reduction of order to find a second solution of the differential equation xy y + x 3 y = 0, x > 0;, y 1 (x = sin x [ 3. #7] Sol. Let y (x = v(x sin x = vy 1. Substituting into the equation, we get x(v y 1 + v y 1 + vy 1 (v y 1 + vy 1 + x 3 vy 1 = 0 That is, xy 1 v +xy 1v y 1 v +(xy 1 y 1+x 3 y 1 v = xy 1 v +(xy 1 y 1 v = 0 Let w = v, then we have that w = ( 1 y 1 x y 1 w, which is a separable equation. Thus dw w = ( 1 x y 1 y 1 dx ln w = ln x ln y 1 + c 0

14 which implies that w(x = v (x = c 1x y 1 By integrating w(x, we get xdx v(x = c 1 sin x = c 1 = c 1 cos x sin x + c = c 1x sin x, where c 1 = e c 0 csc udu = c 1 cot u + c for some constants c 1, c. Thus, y (t = c 1 cos x + c sin x. By setting c 1 =, c = 0, we get y (t = cos x. 0. Use the method of reduction of order to find a second solution of the differential equation (x 1y xy + y = 0, x > 1;, y 1 (x = e x [ 3. #9] Sol. Let y (x = v(xe x = vy 1. Substituting into the equation, we get (x 1(v y 1 + v y 1 + vy 1 x(v y 1 + vy 1 + vy 1 = 0 That is, (x 1y 1 v + ((x 1y 1 xy 1 v = 0 Let w = v, then we have that w = ( x y 1 x 1 y 1 w, which is a separable equation. Thus dw w = ( which implies that x x 1 y 1 dx ln w = x+ln(x 1 ln y 1 +c 0 y 1 w(x = v (x = c 1(x 1e x y 1 = c 1 (x 1e x, where c 1 = e c 0 By integrating w(x, we get v(x = c 1 (x 1e x dx = c 1 xe x + c for some constants c 1, c. Thus, y (t = c 1 x+c e x. By setting c 1 = 1, c = 0, we get y (t = x.

15 1. The differential equation y + δ(xy + y = 0 arises in the study of the turbulent flow of a uniform stream past a circular cylinder. Verify that y 1 (x = exp ( δx is one solution and then find the general solution in the form of an integral. [ 3. #3] Sol. Note that So y y 1(x = δxe δx, y 1(x = δe δx + δ x e δx 1 +δ(xy 1+y 1 = δe δx +δ x e δx δ x e δx +δe δx = 0 which shows that y 1 (x is indeed a solution of the equation. Let y (x = y 1 (xv(x, substituting into the equation, we get That is, v y 1 + v y 1 + vy 1 + δ ( x(v y 1 + vy 1 + vy 1 = 0 y 1 v + (y 1 + δxy 1 v = 0 Let w = v, then we have that w = ( y 1 y 1 + δxw, which is a separable equation. Thus dw (y w = 1 + δx dx ln w = ln y 1 δ x + c 0 y 1 for some constant c 0, which implies that w(v = v (x = c 1 y 1 e δx Integrating the above, we obtain and thus = c 1 e δx, where c1 = e c 0 v(x = c 1 x e δs ds y (x = e δx c1 x e δs ds = c1 x e δ(s x ds Therefore, the general solution for the equation is x y(x = c 1 e δ(s x ds + c e δx

16 . The method in above problem can be extended to second order equations with variable coefficients. If y 1 is a known nonvanishing solution of y +p(ty +q(ty = 0, show that a second solution y satisfies ( y y 1 = W (y 1,y, where W (y y1 1, y is the Wronskian of y 1 and y. Then use Abel s formula to determine y. [ 3. #33] Sol. Suppose the y (t is the second solution for the equation which linearly independent of y 1, then by definition of the Wronskian, ( y y 1 y = y 1y = W (y 1, y y 1 y1 y1 According to Abel s formula, we have that Hence ( t W (y 1, y (t = exp p(sds ( y W (y 1, y t = y 1 y 1 (t = exp ( t p(sds y 1 (t Therefore, we can obtain y (t by t exp ( s p(τdτ y (t = y 1 (t ds y 1 (s 3. Find a second independent solution of the equation (x 1y xy + y = 0, x > 1;, y 1 (x = e x [ 3. #3] Sol. Rewrite the equation as the form y x x 1 y + 1 y = 0. x 1 According to above problem, the second independent solution y (x is x exp ( s τ τ 1 y (x = y 1 (x dτ x exp ( s τ 1+1 ds = e dτ x τ 1 ds y 1 (s e s x exp ( s ( = e x τ 1 dτ x ds = e x exp ( s + ln(s 1 + c 0 ds e s e s x = e x c 1 (s 1e s x ds = c e s 1 e x (s 1e s ds, where c 1 = e c 0 = c 1 e x( xe x + c = c1 x + c 1 c e x By taking c 1 = 1, c = 0, we get y (x = x.