I wish to solve this PDE in 3 dimensions
$$(\Delta-\lambda^2) u(\vec r)=-c$$
where $\lambda$ and $c$ are real and $c$ is a constant. The function $u$ is defined in a closed volume and boundaries conditions are given for $u$.
Wikipedia solve the problem assuming that $u$ and its derivatives vanish at infinite, which is wrong in my case. But it could be a good way to start maybe. So the solution with this assumption is
$$ u(\vec r) = \int d^3r' \frac{e^{-\lambda |\vec r - \vec r'| }}{4 \pi |\vec r - \vec r'|} $$
Then I fail to integrate this. Either cartesian or spherical coordinates, I am stuck immediately.
I also tried to Fourier transform the equation using $u(\vec r)=\int d^3 q~~ \tilde{u}(\vec q) e^{-i \vec q \vec r}$ which leads us to
$$ (q^2-\lambda^2)\tilde{u}(\vec q)=-c ~\delta(\vec q) \\ \\ \tilde{u}(\vec q)=-c ~ \delta (\vec q) \frac{1}{q^2 - \lambda ^2} \\ u(\vec r)=-c \int d^3q~~\delta (\vec q) \frac{1}{q^2 - \lambda ^2} e^{-i \vec q \vec r} \\ \\ u(\vec r)=\frac{c}{\lambda^2} $$
Which isn't correct obviously.