Problem. Consider boundary value problem: \begin{cases} \Delta u(x,y)+2u(x,y)=x-\alpha, & \text{in $\Omega$,} \\ u(x,y)=0, & \text{on $\partial\Omega$,} \\ \end{cases} where $\alpha$ is real, $\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ and $\Omega=(0,\pi)\times(0,\pi)$. The question is for which $\alpha$ we have at least one weak solution for this PDE.
Function $u$ is said to be weak solution to this problem iff $u\in H_0^1(\Omega)$ and equation is satisfied in sence of distributions.
Solution. To solve this problem we should use Fredholm alternitive. For our case it takes the form (formally):
$Lu=f$ has solution for all permissable $f$ that satisfies $(f,v)_{L^2(\Omega)}=\int_\Omega fv\,dx=0$, where v is solution for adjoint homogeneous equation $L^*v=0$.
Adjoint homogeneous equation has next form:
\begin{cases} \Delta v(x,y)+2v(x,y)=0, & \text{in $\Omega$,} \\ v(x,y)=0, & \text{on $\partial\Omega$,} \\ \end{cases}
To solve it you can use method called separation of variables, on rectangle variables are separated very easy. You get $v(x,y)=C\,sinx\,siny$, where $C$ is arbitrary constant. We fix $C=1$.
Now we can easily find for which $f$ our main equation has at least one solution.
$$(f,v)_{L^2(\Omega)}=\int_\Omega fv\,dx=\int_\Omega(x-\alpha)sinx\,siny\,dx=2\pi-4\alpha=0 \text{ when } \alpha=\frac{Pi}{2}$$
Your solution is essentially correct; just one small mistake (with the integral over $\Omega$). Hint: