Use separation of variables to solve the PDE $$u_{tt}=u_{xx}+\sin t$$ for $$t \ge0, \ \ \ \ \ -\infty <x < +\infty$$ such that $$u(x, 0)=0 $$ and $$u_x(0,t)=0$$.
I substituted in $u(x,t)=X(x)T(t)$ and got $$X(x)T''(t)=X''(x)T(t)+\sin t$$Then do I divide by $X(x)T(t)$? And what do I do after that and so on?
So when your PDE has a non-homogeneous part such as $\sin(t)$, you want to solve for the steady state solution fist, and then do a change of variables to turn it into a problem that's solvable by separation of variables. In other words, if we can get rid of the $\sin(t)$ term, then the PDE simplifies into something that you probably already know how to solve.
By inspection, let $v(x,t) = u(x,t) + \sin(t)$. Then, \begin{align} v_{tt} &= u_{tt} - \sin(t)\\ v_{xx} &= u_{xx}\\ \end{align} Which transforms the PDE into \begin{align} (v_{tt} + \sin(t)) &= v_{xx} + \sin(t)\\ \Rightarrow v_{tt} &= v_{xx} \end{align} For the boundary conditions, \begin{align} v(x,0) &= u(x,0) + \sin(0) = u(x,0) = 0\\ v_x(0,t) &= u_x(0,t) + \frac{\partial}{\partial x}\sin(t) = 0 + 0 = 0 \end{align} In summary, we have that \begin{align} v_{tt} &= v_{xx}\\ v(x,0) &= 0 \\ v_x(0,t) &= 0 \end{align} Which you can solve by separation of variables.