solve this ODE?

Question

hey. what is the solution to this ode? I don't need the work or anything, just the answer. Just to confirm my own answer.

Anyways, um

dy/dt = -cy + asin(wt)

c, a, w are just constants.

Thanks

Thanks scarlet. I realized after that I could just plug it in and check, but oh well. I got the same thing btw. You'll get best answer whenever I can choose (if I remember, sorry if I dont).

Answer 1

If you have an answer, just differentiate it and plug it back into the equation. That will confirm whether it is correct.

I'll give you the working anyway, so that if our answers differ you can see where the difference comes from.

dy/dt + cy = a sin ωt.
=> e^(ct) dy/dt + e^(ct) (cy) = e^(ct) a sin ωt
=> d/dt (e^(ct) y) = e^(ct) a sin ωt
=> e^(ct) y = ∫e^(ct) a sin ωt dt.

Now for this sort of integral, since e^(ct) differentiates to a multiple of itself, and sin goes to cos and back to sin, the trick is to integrate by parts twice and get back a known function plus a multiple of the original integral. Then you can determine the integral algebraically.

Let K = ∫e^(ct) a sin ωt dt
= e^(ct) (1/c) a sin ωt - ∫e^(ct)(1/c) (aω) cos ωt dt
= e^(ct)(1/c) a sin ωt - [e^(ct)(1/c^2)(aω) cos ωt - ∫e^(ct)(1/c^2)(-aω^2) sin ωt dt]
= e^(ct)(1/c) a sin ωt - e^(ct)(1/c^2)(aω) cos ωt + (-ω^2/c^2) K
So (1 + ω^2/c^2) K = e^(ct)(1/c) a sin ωt - e^(ct)(1/c^2)(aω) cos ωt
and thus K = [e^(ct)(1/c) a sin ωt - e^(ct)(1/c^2)(aω) cos ωt] / (1 + omega;^2 / c^2)

So y = e^(-ct) K
= [(1/c) a sin ωt - (1/c^2)(aω) cos ωt] / (1 + omega;^2 / c^2)
= (ac sin ωt - aω cos ωt) / (c^2 + ω^2).

To confirm that this is the correct answer, plug it back in:
dy/dt = (acω cos ωt + aω^2 sin ωt) / (c^2 + ω^2)
so dy/dt + cy = (acω cos ωt + aω^2 sin ωt + ac^2 sin ωt - acω cos ωt) / (c^2 + ω^2)
= a sin ωt (ω^2 + c^2) / (c^2 + ω^2)
= a sin ωt.
Hence dy/dt = -cy + a sin ωt as required.