Question

Particular solution to ODE
y'' -2y' +5y = e^2t * (1-tsin(3t))

Answer 1

I already solved the part after distributing.
e^2t - e^2t *t*sin(3t)

so now I need help solving -e^2t*t*sin(3t)

Answer 2

Can you show me what you've done so far in trying to solve it?

Answer 3

So far, I used superposition to split up the particular term into e^2t and -e^2t*t*sin(3t).

I have already solved the homogeneous case (setting right hand side to =0) and solving for y''-2y'+5y=e^2t, I get Xp1 = e^2t * 1/5

I have never encountered 3 terms and that is why I am slightly stumbled.

Answer 4

Oh ok, well what methods do you know about, I want to help you do it with a method you're comfortable with since I know of several ways you can attack this thing.

Answer 5

Undetermined coefficients I'm assuming is what you're doing?

Answer 6

Yes that would be the most ideal method. I am just having trouble with this since its 3 terms and I never had an example of this before.

Answer 7

Hi Silen. I can help you with this, but it would take some time.
How quickly do you need a solution?

Answer 8

4 hours would be fine. If you could just tell me how to get started on the initial undetermined coefficient that would be a great start.

Answer 9

You have a homogeneous part and a nonhomogeneous part to this ODE. Have you arrived at the solution to the homogeneous part?

Answer 10

Oh my mistake, you posted the solution to the nonhomogeneous part where you only consider \(e^{2t}\)... Sorry!

Answer 11

For the remaining part, since \(e^{2t}t\sin 3t\) isn't linearly dependent relative to either the homogeneous solution pair (\(y_h=e^t(\cos 2t+\sin2t)\)) as well as the first particular solution (\(y_p=\dfrac{1}{5}e^{2t}\)), you can set the remaining guess solution to something like \(y_p=e^{2t}(At+B)\sin3t\).

Answer 12

Actually, a slight adjustment: Since the degree of this ODE is 2 and a first order derivative is included, you may need to consider a cosine term as well.

Answer 13

@sithsandgiggles Ok so basically e^2t*(At+B)sin(3t)+ e^2t(Ct+D)cos(3t) ?

Answer 14

@sithsandgiggles Ok so basically e^2t*(At+B)sin(3t)+ e^2t(Ct+D)cos(3t) ?

Answer 15

That's what I would try, yep

Answer 16

Taking those derivatives might be a nightmare but hopefully it works.

Answer 17

Siths is right on the mark. I'm going through the 1st and 2nd derivative of that large expression but it's proving very tedious. e^2t*(At+B)sin(3t)+ e^2t(Ct+D)cos(3t)
still working on it

Answer 18

I've had no access to the site for quite some time due to some server issue.

Answer 19

It was a nightmare!