Solve Particular Solution Differential Equation: {\theta}''(t)-{\theta}(t)=tsint

[Precursor]

Homework Statement
Find a particular solution to the differential equation: {\theta}''(t)-{\theta}(t)=tsint

The attempt at a solution
So I started by using the particular solution {\theta}_{p}=(At+B)(Csint+Dcost)
Before I continue with the rest of the solution, is this correct so far?

 

[Lavabug]

Looks right, you should have 4 coefficients to determine for a particular solution of that form.

 

[Precursor]

So I continue by finding the first and second derivatives of that particular solution:
{\theta}'_{p}=(At+B)(Ccost-Dsint)+A(Csint+Dcost)
{\theta}''_{p}=(At+B)(-Csint-Dcost)+A(Ccost-Dsint)+A(Ccost-Dsint)
Now I substitute back into the differential equation and get
-2(At+B)(Csint+Dcost)+2A(Ccost-Dsint)=tsint
From here I'm having a bit of difficulty. How do I solve for the coefficients?

 

[Lavabug]

Well I would've personally multiplied everything out in your trial solution before differentiating, so you'd have Atcost + Btsint + Ccost + Dsint. Remember your coefficients arbitrary, but in your case you must try 4.

Then differentiate for Y'p and Y''p, plug into your homogenous equation, regroup/factor everything in terms of sint, cost, tsint, tcost, then set equal to tsint and solve the system of equations.

These problems stress one's neatness and eyesight, be careful haha.

 

[Lavabug]

Did it myself(tempted as my final is coming up lol), and this is the correct solution:
http://www.wolframalpha.com/input/?i=y''+-y+=+xsinx

(y and x are your theta and t, wolfram is a lifesaver!)

Also always be sure to check that your particular solutions are linearly independent with the solutions to the homogeneous part. Clear sails on this one as you've got real-valued exponentials for your homogeneous part.

 

[SteamKing]

The coefficients are ultimately determined using the initial conditions, where y(t) or y'(t) are given for some value t. If no initial conditions are given, the coefficients remain unknown.

 

[Lavabug]

You mean the coefficients of the homogeneous part of the solution yes?

 

[Precursor]

So I've worked it through and I get

AC = -1/2
AD = 0
AD = -BC
AC = BD

This doesn't seem right though.

 

[Lavabug]

I have no idea how you reached that. Here's how I did it:
http://img685.imageshack.us/img685/1977/p1000999sx.jpg

 

[Precursor]

So this particular solution is not correct: {\theta}_{p}=(At+B)(Csint+Dcost)

Because when I multiply through I get: ACtsint+ADtcost+BCsint+BDcost


Oh, do I replace the coefficients AC, AD, BC, BD to A, B, C, D?

 

[Lavabug]

Yep, that's what I meant by "your coefficients are arbitrary".

 

[Precursor]

Alright I've finally got the correct solution. But you managed to do it showing a lot less work. Did you skip some simplification steps?

But thanks for the help!

 

[Lavabug]

You're welcome. I don't expand anything out unless I have to, avoids a possible source of mistakes in my experience. When I plug my particular and its derivatives back in I just factor the x terms out of everything(identifying them term by term) so I can see my coefficients clearly and what each equation is equal to: (0,1,0,0) in this case. Very "linear", some people do the same thing but with columns.

 

[viko]

If you've already learned complex calculus, I strongly recommend you to use complex solutions, it makes it much easier

 

[Lavabug]

Do you mean trying a trial solution with complex exponentials, so you'd only have 2 coefficients to solve? Never thought of that, but it sounds like a clever idea.