Solve the following Goursat problem for $u(x,t)$: $$u_{tt} = u_{xx} + 4cos(2t)$$ $(t >0,−t <x <t)$
$$u(t,t) = t^2, \ \ \ \ \ \ \ u(−t,t) = \text{arctan} \ t $$
My thoughts are as follows convert in the form of $$u_{\zeta \eta}$$ and then we can solve. I have an example which I was just following which is attached but I'm not too sure how to go about solving with the extra $ 4cos(2t)$.
I've attempted to do the same but not too sure how to continue / know if this is the right method. $$u_{\zeta \eta} + 4cos(2(\eta - \zeta)) = 0 $$
If this is correct how do we go to $u(x,t)$ from this? (Perhaps, integrate w.r.t $\eta$ followed by $\zeta$ ?
I was hoping someone could help me with this, it would be really appreciated (and if you do, could you try explain this method - I want to learn this method please)
Thank you!
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Edit: If someone could explain the boundary conditions also that would be helpful because it isn't in the regular form of $$u(x,0)= \phi_1(x)$$ - it's really different, which I'm a bit confused about :/
$\textbf{EDIT 2}:$ I've just attempted to do the following to change the variables from $u \rightarrow v$. Define $$u = u_p + v$$, where $u_p = -cos(2t)$ this yields $$v_{tt} = v_{xx}$$. THen using method from above get $$v_{\zeta \eta} = 0$$. Now, I'm confused again.