Of course. When $\nu=1$, if you apply the operator $\partial_t-\partial^2_{xx}$ to the last integral you obtain precisely $f(t,x)$ so the equation is $$ f_t - f_{xx} = (\partial_t-\partial^2_{xx}) J_0^2 + f.$$

EDIT: you seem to know already the answer, so I stop here :) You edited your question when I was writing my answer...

Of course. When $\nu=1$, if you apply the operator $\partial_t-\partial^2_{xx}$ to the last integral you obtain precisely $f(t,x)$ so the equation is $$ f_t - f_{xx} = (\partial_t-\partial^2_{xx}) J_0^2 + f.$$

EDIT: you seem to know already the answer, so I stop here :) You edited your question when I was writing my answer...

By the way, if you want to solve the PDE just set $ f(t,x) = e^{t} g(t,x) $ and the equation in $g$ is a homogeneous heat equation. This sounds like some textbook exercise, I musr say