t + t sin t t cos t sin t. t cos t sin t dt t 2 = exp 2 log t log(t cos t sin t) = Multiplying by this factor and then integrating, we conclude that

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1 ODEs, Homework #4 Soluions. Check ha y ( = is a soluion of he second-order ODE ( cos sin y + y sin y sin = 0 and hen use his fac o find all soluions of he ODE. When y =, we have y = and also y = 0, so ( cos sin y + y sin y sin = sin sin = 0 and y = is a soluion, indeed. Le us now use reducion of order o find a second soluion of he form y 2 = y v = v. Differeniaing, we ge y 2 = v + v, y 2 = 2v + v and so y 2 = v is also a soluion, provided ha 0 = ( cos sin y 2 + y 2 sin y 2 sin = ( cos sin v + (2 cos 2 sin + 2 sin v. This gives rise o he firs-order linear equaion ( 2 v + + sin cos sin v = 0. Noing ha ( cos sin = sin, we see ha an inegraing facor is ( 2 µ = exp + sin cos sin d ( 2 = exp 2 log log( cos sin = cos sin. Muliplying by his facor and hen inegraing, we conclude ha (µv = 0 = v = C µ = C ( ( cos sin C sin = 2 = v = C sin + C 2 = y 2 = C sin + C Show ha he zero soluion is he only bounded soluion of x ( = x + xy 2, y ( = y x 2 y. If x(, y( is a soluion of he given sysem, hen E( = x 2 + y 2 saisfies E ( = 2xx + 2yy = 2x 2 + 2y 2 = 2E( = E( = E(0e 2. Thus, E( can only remain bounded for all imes, if i is idenically zero.

2 3. For which of he following ODEs is he zero soluion sable? Asympoically sable? (a x = 2y, y = 2x (b x = x + 2y, y = x (c x = x 5y, y = 5x + y In he firs case, we have y = Ay, where he marix A and is eigenvalues are [ ] 0 2 A = = λ = 0 = λ = ± 2i. 2 0 Since hose have zero real par, he zero soluion is sable bu no asympoically. In he second case, he marix A and is eigenvalues are [ ] 2 A = = λ 2 λ 2 = 0 = λ =, 2. 0 Since one of he eigenvalues is posiive, he zero soluion is unsable. In he hird case, finally, he marix A and is eigenvalues are [ ] 5 A = = λ 2 2λ + 26 = 0 = λ = ± 5i. 5 Since hose have posiive real par, he zero soluion is unsable. 4. For which of he following ODEs is he zero soluion sable? Asympoically sable? (a x ( = x( 2 (b x ( = x( 3 (c x ( = x( cos In he firs case, separaion of variables gives d = x2 = x = + C = x( = + C. If we now impose he iniial condiion x(0 = x 0, hen we end up wih x 0 = /C = x( = = x 0 /x 0 x 0 and his implies blow up a ime = /x 0. Thus, he zero soluion is unsable. In he second case, separaion of variables gives d = x3 = 2 x 2 = + C = x( 2 = Imposing he iniial condiion x(0 = x 0, we hen easily ge x 2 0 = /C = x( 2 = 2 + /x 2 0 = x 2 0 2x C. In paricular, x( 0 as and he zero soluion is asympoically sable.

3 In he las case, separaion of variables gives d = x cos = log x = sin + C = x( = x 0e sin, so i easily follows ha he zero soluion is sable bu no asympoically. 5. Use he subsiuion z = log y( o solve he equaion y = y(log y. Since z = log y, we have z = y y and so z = y y This is a separable equaion which gives dz d = log y = z. = z = log(z = + C = z = Ce = y = e z = e Ce Check ha y ( = e is a soluion of he second-order ODE ( 2 + y ( 2 2y ( + 2y = 0 and hen use his fac o find all soluions of he ODE. When y = e, we have y = y = e, so ( 2 + y ( 2 2y ( + 2y = ( e = 0 and y = e is a soluion, indeed. Using reducion of order, we shall now find a second soluion of he form y 2 = y v = e v. Differeniaing, we ge y 2 = e v, y 2 = e v + e v, y 2 = e v + 2e v + e v and hus y 2 = e v is also a soluion, provided ha 0 = ( 2 + y 2 ( 2 2y 2 ( + 2y 2 = ( 2 + e v + ( e v. This gives rise o he firs-order linear ODE v and he corresponding inegraing facor is ( ( µ = exp d = exp 2 + v = d = 2 e +.

4 Muliplying by his facor and hen inegraing, we conclude ha (µv = 0 = v = C µ = C ( ( + e C e = 2 = v = C e + C 2 = y 2 = C + C 2 e. 7. Check wheher he zero soluion is a sable or unsable soluion of y = Ay when [ ] [ ] [ ] A =, A =, A = In he firs case, we have r A = 0 and de A =, so he eigenvalues are λ 2 (r Aλ + de A = 0 = λ 2 = 0 = λ =,. Since one of hose is posiive, he zero soluion is unsable. In he second case, we have r A = 6 and de A = 9, so he eigenvalues are λ 2 (r Aλ + de A = 0 = λ 2 + 6λ + 9 = 0 = λ = 3, 3. Since hose are boh negaive, he zero soluion is sable. In he hird case, we have r A = 2 and de A = 2, so he eigenvalues are λ 2 (r Aλ + de A = 0 = λ 2 + 2λ + 2 = 0 = λ = ± i. Since hose are complex wih negaive real par, he zero soluion is sable. 8. Use he subsiuion w = /y( o solve he equaion y + y = y 2 log. Seing y = w, we ge y = w 2 w, hence also w 2 w + w = w 2 log = w w = log This is a firs-order linear equaion wih inegraing facor ( d µ = exp = exp( log =. We now muliply by his facor and inegrae o ge ( w = log log = w = d. 2 2.

5 To compue he inegral, le u = log and dv = 2 d. Then v = and so log d = u dv = uv v du = log 2 d. 2 Combining he las wo equaions, we now ge w = log 2 d = log + + C, so we may finally conclude ha w = log + + C = y = w = log + + C. 9. Show ha he zero soluion is an unsable soluion of he sysem x ( = x + 2y + xy, y ( = y 2x x 2. Hin: find and solve he ODE saisfied by E( = x( 2 + y( 2. Following he hin, le E( = x 2 + y 2 and noe ha E ( = 2xx + 2yy = 2x 2 + 4xy + 2x 2 y + 2y 2 4xy 2x 2 y = 2E(. This gives a separable firs-order ODE which can be easily solved o ge E ( = 2E( = E( = Ce 2 = E( = E(0e 2. Since E( measures disance from he origin, his means ha soluions which sar ou near he origin do no remain near he origin, so he zero soluion is unsable. 0. Le a R and consider he second-order equaion y ( + 2ay ( + y( = 0. For which values of a is he zero soluion sable? Asympoically sable? In his case, he associaed characerisic equaion has roos λ 2 + 2aλ + = 0 = λ = a ± a 2. These are real and negaive when a, bu real and posiive when a. In he remaining case < a <, he roos are complex wih real par equal o a. Thus, he zero soluion is sable when a 0 and asympoically sable when a > 0.