Solving non-homogeneous heat eq'n with fourier series

[wakko101]

Homework Statement

The heat eq'n is Ut -4Uxx = 2t - xsin(x)
Ux(0,t) = Ux(pi,t) = 0, U(x,0)=x^2+1

Homework Equations

Using separation of variables, in obtaining the eigenvalues/eigenfunctions of X''=-lambdaX, it would appear that you would need to use a cosine series basis and expand the equation. But it seems to me that the xsin(x) on the RHS of the equation is causing problems and I'm not sure how to proceed.

The Attempt at a Solution

If you expand the RHS as a cosine Fourier series, the coefficients would be given by int(0 to pi)(2t + xsin(x))cos(nx) dx. I'm stumped...any suggestions/hints would be appreciated.

Cheers. =)

 

[HallsofIvy]

Yes, that's right:
\int_0^\pi (2t+ x sin(x)cos(nx))dx[/itex]<br /> (except for the normalizing factor in front)<br /> <br /> What is the problem? Of course, t is a constant here so the first part is just <br /> 2t = 2t cos(0x). the second part, <br /> \int_0^\pi x sin(x)cos(nx)dx<br /> you should be able to do with integration by parts. Let u= x, dv= sin(x)cos(nx)dx.

 

[wakko101]

I had thought of that. I suppose I'm not sure how to go about integrating dv=sin(x)cos(nx).