Does anyone know how to solve the following pde.
\begin{align}
&\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2} +K(xt)u \quad t>0,-\infty<x<\infty \\
&u(0,x)=1
\end{align}
where K(x) is a known function. I would be satisfied with K(x)=cos x. I tried a few things, but I am still not able to solve in closed form.
Change variables $\tau=xt$, then the pde becomes
\begin{equation}\label{Eq1}
x\frac{\partial u}{\partial \tau}=\frac{\partial^2 u}{\partial x^2} +K(\tau)u
\end{equation}
Denote the Fourier transform of $u(\tau,x)$
\begin{equation}
\hat{u}=\hat{u}(\tau,\lambda)=\int^\infty_{-\infty}e^{i\lambda x}u(\tau,x)dx
\end{equation}
(i) Taking the Fourier transform of both sides
\begin{equation}
-i\frac{\partial^2 \hat{u}}{\partial \tau\partial\lambda}=-\lambda^2 \hat{u}+K(\tau)\hat{u}=\hat{u}(K(\tau)-\lambda^2)
\end{equation}
(ii) Remove the mixed partial
\begin{align}
\xi &=\tau+a\lambda\\
\eta &=\tau+b\lambda
\end{align}
The equation becomes
\begin{equation}
-ia(\frac{\partial^2 \hat{u}}{\partial \xi^2} -\frac{\partial^2 \hat{u}}{\partial \eta^2})=\hat{u}(K(\xi+\eta)-\frac{(\eta-\xi)^2}{4a^2})
\end{equation}
Also, one may change variables $y=xt$, in the original pde. Then it becomes
\begin{equation} \frac{\partial u}{\partial t} =t^2\frac{\partial^2 u}{\partial y^2}+K(y)u \end{equation}
Now one can take the Laplace transform to get \begin{equation} \lambda \hat{u} =\frac{\partial^2\partial^2 u}{\partial y^2\partial \lambda^2}+k(y)\hat{u} \end{equation}
It the original equation looks similiar to a Schrodinger equation.
