How to solve $\frac{\partial^2u}{\partial x\partial t}=-\frac{\partial u}{\partial t}+\sin(x)e^{-x}$ How to solve $$\frac{\partial^2u}{\partial x\partial t}=-\frac{\partial u}{\partial t}+\sin(x)e^{-x}$$ I'm looking for one solution only. I was reading about separable variables, but I'm not sure if it will work in this case, as it has an extra term and a cross derivative?
Could anyone suggest me a way to solve it? Is $u(t,x)=X(x)T(t)$ still a good choice?

Since the free term is the product of an exponential and a sine, it is reasonable to guess a solution of the form
$$e^{-x}(a\cos x+b\sin x).$$
Plugging this in the differential equation should give you the desired solution:
$$\frac{1}{5}e^{-x}(2\cos x +\sin x).$$
