I need to solve the following Boundary Value problem
$$\frac{\partial^2w }{\partial x^2}+\frac{\partial^2w }{\partial y^2}=c$$
Boundary conditions are
$$w(x,h)=0$$
$$w\left(\frac{\pm h}{\sqrt3},y\right)=0$$
I am trying to solve it by Homotopy Perturbation Method, but unable to succeed to get desired answer which should be
$$w=\frac{c(y-h)(3x^2-y^2)}{4h}$$
You might want to verify your given solution. I don't think it satisfies the equation.
First, let
$$ w(x,y) = \frac{c}{6}(3x^2-h^2) + v(x,y) $$
then $v(x,y)$ is harmonic with boundary conditions
\begin{cases} v\left(\pm \frac{h}{\sqrt3},y\right) = 0 \\ v(x,h) = -\frac{c}{6}(3x^2-h^2) \end{cases}
The solution to this problem isn't unique. If we assume $v(x,y\to\infty)$ is bounded, then separation of variables gives
$$ v(x,y) = \sum_{n=1}^\infty c_n \exp\left(-\frac{n\pi\sqrt 3}{2h}(y-h)\right) \sin\left(\frac{n\pi}{2h}\big(\sqrt3 x+h\big)\right) $$
where
$$c_n = \frac{\sqrt 3}{h}\int_{-h/\sqrt3}^{h/\sqrt3} v(x,h) \sin\left(\frac{n\pi}{2h}\big(\sqrt3 x+h\big)\right) \ dx$$