Consider the inhomogeneous elliptic equation $u_{xx} + u_{yy} = 1$ (this is often called a Poisson equation) in the disc $x^2 + y^2 < 1$, with the boundary condition $u = a$ on the boundary $x^2 + y^2 = 1$ of the disc. Find the solution $u$, noting that $u$ will be a function of $r = \sqrt{x^2 + y^2}$.
I know that there will be a homogenous and a particular solution.
I think that the homogenous solution directly follows Poisson's Formula
$$u(r,\theta ) = \frac{(a^2 - r^2)}{2\pi}\int_{0}^{2\pi} \frac{h(\theta )}{a^2 - 2 \arccos(\theta - \phi ) + r^2} \, \mathrm d \theta$$
where, in this case, $h(\theta ) = a$. I'm pretty sure that $a=1$ in this case but can someone confirm?
But then I have no idea how to solve for the particular solution.
Both the inhomogeneous function and the boundary condition are radially symmetric, so it's sufficient to find a solution $u(r)$ such that
\begin{cases} \frac{1}{r}(ru_r)_r = 1 \\ u(1) = a \end{cases}
Find the general solution by inverting each operation
\begin{align} (ru_r)_r &= r \\ ru_r &= \frac{r^2}{2} + c_1 \\ u_r &= \frac{r}{2} + \frac{c_1}{r} \\ u &= \frac{r^2}{4} + c_1\ln r + c_2 \end{align}
$u(0)$ needs to be finite, so $c_1 = 0$, and the boundary condition $u(1)=a$ gives $ c_2 = a - \frac14 $.
Hence, the final solution is
$$ u(r) = \frac{r^2-1}{4} + a $$