The problem is to solve $$\Delta\phi=\frac \lambda {\varepsilon_0}\delta(x-x',y-y')\quad;\quad \phi(0,y)=\phi(a,y)=\phi(x,0)=\phi(x,b)=0.$$
My idea was to try and represent the RHS as a series of eigenfunctions of $\Delta$, which will be denoted by $\phi_{n,m}$.
My problem is that the given solution has one series, and my solution ends up with two. I don't see what I've done wrong.
My solution:
I found $\phi_{n,m}$ by separation of variables:
$$\Delta\phi=-k^2\phi\,\,\,;\,\,\,\phi(0,y)=\phi(a,y)=\phi(x,0)=\phi(x,b)=0\\\,\\ X''Y+XY''=-k^2XY\\\,\\ \therefore\,\,\dfrac{X''}{X}=-p\quad\dfrac{Y''}{Y}=-q\quad p^2+q^2=k^2$$
Solving these ODEs with the boundary condition yields \begin{gather*} X_n = \sin(p_n x) \quad Y_m = \sin(q_m y) \qquad p_n=\frac{n\pi}a \quad q_m=\frac{m\pi}b\\ \therefore\phi_{n,m}=\sin\left(\frac{n\pi}ax\right)\sin\left(\frac {m\pi}b y\right) \end{gather*} Due to completeness $$\phi=\sum_{n=1}^\infty\sum_{m=1}^\infty \beta_{n,m}\phi_{n,m} \quad \beta_{n,m}=\frac{\langle\phi,\phi_{n,m}\rangle}{\langle\phi_{n,m},\phi_{n,m}\rangle}$$ with the inner product $$\langle f, g\rangle \equiv \int_0^a\!\!\int_0^b\!\! fg\,dx\,dy$$
Under the assumption that eigenfunction expansion formally works for $\delta$ as well, and that term-by-term differentiation holds: \begin{gather*} \begin{aligned} \Delta\phi &= \sum_{n=1}^\infty\sum_{m=1}^\infty \beta_{n,m}\Delta\phi_{n,m} \\ &= -\sum_{n=1}^\infty\sum_{m=1}^\infty k_{n,m}^2 \beta_{n,m}\phi_{n,m} \\ &= \frac{4\lambda}{\varepsilon_0 a b} \sum_{n=1}^\infty\sum_{m=1}^\infty \sin\left(\frac{n\pi}a x'\right) \sin\left(\frac{m\pi}b y'\right)\sin\left(\frac{n\pi}a x\right) \sin\left(\frac{m\pi}a y\right) \end{aligned}\\ \therefore \beta_{n,m} = \frac{-4\lambda\sin\left(\frac{n\pi}a x'\right) \sin\left(\frac{m\pi}b y'\right)} {ab\varepsilon_0\left[ \left(\frac{n\pi}a\right)^2 \left(\frac{m\pi}b\right)^2\right]} \end{gather*}
The solution, however, was given as $$\phi(x,y)=\sum_{n=1}^\infty \frac{-4\lambda \sin\left(\frac{n\pi}a x'\right) \sin\left(\frac{n\pi}b y'\right)} {ab\varepsilon_0 \left[\left(\frac{n\pi} a\right)^2 \left(\frac{n\pi}b\right)^2\right]} \sin\left(\frac{n\pi}a x\right)\sin\left(\frac{n\pi}b y\right)$$
What have I done wrong? Thank you for your time.