I am considering the following PDE: $$ \begin{cases} -\Delta u = 4 &\text{in}\;\Omega = \{(x,y) : x^{2} + y^{2} < 1\}\\ u = 0 &\text{on}\;\partial\Omega. \end{cases} $$
The exact solution is radial and given by $u(x,y) = 1-x^{2}-y^{2} = 1-r^{2}$. I'm working with the equation in polar coordinates (assuming that the solution is radial): $$ \begin{cases} -\dfrac{1}{r}\dfrac{d}{dr}\left( r\dfrac{du}{dr} \right) = 4 &r<1\\ u(r=1) = 0. \end{cases} $$
Treating this as a one-dimensional ODE, I obtain a general solution like $u(r) = -r^{2}-C_{1}/(2r^{2}) + C_{2}$. However, I see two issues with this approach. The first is that I cannot solve for both $C_{1}$ and $C_{2}$ with the given boundary condition. The second is that this solution is not well-defined at $r=0$ unless I take $C_{1}=0$ (and indeed, doing so and solving for $C_{2}$ gives the correct solution). I'm just curious what the proper way to deal with the boundary conditions is, especially if I didn't know the solution in advance already.
You might have made a simple calculation mistake. If you solve the ODE \begin{align} -\frac{1}{r}(r u')'= 4 \end{align} you should get \begin{align} u(r) = \underbrace{c_1\log r + c_2}_{\text{homogeneous solution}}-r^2. \end{align}
Now to finish the problem, you will need a clear definition of what you mean by a solution to your original problem. If you are looking for classical solution defined on the disk $x^2+y^2<1$ which is continuous up to the boundary, i.e. $u$ has to be at least twice continuously differentiable, then you can safely discard the the log term since $\log(x^2+y^2)$ is not continuous at the origin. Hence you have \begin{align} u(r) = c -r^2. \end{align} Plug in the boundary condition $r=1$ you will get $u(r) =1-r^2$.
However, if you are considering more general solutions, say distributional solutions, then we see that \begin{align} -\Delta u = -\Delta(c_1\log(x^2+y^2)+c_2 -r^2) = -2\pi c_1\delta_0+4 =4 \end{align} which means $c_1$ is again zero.