How to solve $\frac{\partial^2u}{\partial x\partial t}=-\frac{\partial u}{\partial t}+\sin(x)e^{-x}$

Question

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?

If you denote $z=\frac{du}{dt}$, the equation becomes $\frac{dz}{dx}=-z+\sin x e^{-x},\Rightarrow,z(x)=a(x)e^{-x}$. Find $a(x),\Rightarrow, u(x,t)=z(x)t+b$ — Svyatoslav, Dec 03 '22 at 17:44

score 1 · Accepted Answer · answered Dec 03 '22 at 22:53

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: $$\frac{1}{5}e^{-x}(2\cos x +\sin x).$$

How to solve $\frac{\partial^2u}{\partial x\partial t}=-\frac{\partial u}{\partial t}+\sin(x)e^{-x}$

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