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Homework Statement
Consider the PDE
which has the solution
The Attempt at a Solution
So what I am having trouble is solving it using this method.
I am going to say that my $$u(x,t) = \sum_{n=1}^{\infty} u_n(t) \sin(nx)$$ and $$x \sin(t) = \sum_{n=1}^{\infty}h_n(t)\sin(nx)$$
The reason I chose sine for my inhomogeneous term is because my book recommends it. But I think it is because if I use cosine, I would get a $$\frac{a_0}{2}$$ term and it would be difficult.
To solve for the coefficients of $$h_n(t)$$, I get $$h_n(t) = \frac{2}{\pi}\int_{0}^{\pi} x\sin(t) \sin(nx) dx = \frac{2\sin(t)(-1)^n}{n}$$
Substituting everything into $$u_{tt} = u_{xx} + x\sin(t)$$ gives me
$$ \sum_{n=1}^{\infty}u''_n(t) \sin(nx) + \sum_{n=1}^{\infty}u_n(t)n^2\sin(nx) = \sum_{n=1}^{\infty}\frac{2\sin(t)(-1)^n}{n}\sin(nx)$$
Dividing out that sine, I'll get
$$ \sum_{n=1}^{\infty}u''_n(t) + \sum_{n=1}^{\infty}u_n(t)n^2 = \sum_{n=1}^{\infty}\frac{2\sin(t)(-1)^n}{n}$$
Here is where I am stuck, can someone tell me what value of n to use?
Because I could get an ODE with many ns
Did I overlooked something?
Thank you very much