For given conditions, $\frac{\partial u}{\partial t}=0$ at $x=0$ & $u=0$ at $t=0$. The solution of the differential equation $$\frac{\partial^2u}{\partial x\partial t}=e^{-t}\cos x$$ is given by
a) $(1-e^{-t})\sin x\qquad $ b) $(-e^{-t})\sin x+g(x)\qquad $ c) $(1+e^{-t})\cos x\qquad $ d) $(-e^{-t})\sin x\qquad $
my try: $$\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial t}\right)=e^{-t}\cos x$$ $$\int \partial \left(\frac{\partial u}{\partial t}\right)=\int e^{-t}\cos x\ \partial x$$ $$\frac{\partial u}{\partial t} =e^{-t}\sin x+f(t)$$ putting $x=0$ & $\frac{\partial u}{\partial t} =0$ i get $f(t)=0$ so $$\frac{\partial u}{\partial t} =e^{-t}\sin x$$ $$\int \partial u =\int e^{-t}\sin x\partial t$$ $$u=-e^{-t}\sin x+g(x)$$ now, putting $t=0$ & $u=0$ i get $g(x)=\sin x$, so the solution is $$u=-e^{-t}\sin x+\sin x$$$$=(1-e^{-t})\sin x$$ so the option is (a) but my book says the correct option is (d) i don't know why. please explain where i am wrong or what should be the right answer? thank you very much
I checked your proof and to me it is fine. $u(t,x) = (1- e^{-t})\sin x$ is the uniquely determined function with $$\frac{\partial^2u}{\partial x\partial t}=e^{-t}\cos x$$ that suffices the given start values. I mean if $u(t,x) = - e^{-t}\sin x$ then we get $u(0, x) = \sin x \not \equiv 0$ obviously. I suppose that the book is just wrong at this point. Good job noticing!