Given the orthonormal basis $\{\phi_n\}$ in $L^2$ that solves Sturm-Liouville eigenvalue problem
$y''=-\lambda y$ which also satisfies $y(0)=y(1)=0$ (1),
find the series expansion in this basis of the solution to:
$y''=x-x^{2}$.(2)
my idea: I shall take the orthonormal basis that solves Sturm-Liouville in $L^2$. This will be $\{\phi_n\}$, and I find a series that satisfies (2).
Meaning $y=\sum_{n\in\mathbb{Z}}\phi_{n}a_{n}$ represents a function that satisfies (2).
first I shall use the inner product of $L^2$: $$ b_{n}=\langle x-x^{2},\sqrt{2}\sin(n\pi x)\rangle =\int_{0}^{1}(x-x^{2})\sqrt{2}\sin(n\pi x)dx\\ =-\frac{\sqrt{2}(-2+2\cos(n\pi)+n\pi \sin(n\pi))}{n^{3}\pi^{3}}$$ thus $$\Rightarrow b_{n}=\begin{cases} \begin{array}{c} n\in\mathbb{Z_{\textrm{even}}}\\ n\in\mathbb{Z_{\textrm{odd}}} \end{array} & \begin{array}{c} 0\\ \frac{4\sqrt{2}}{n^{3}\pi^{3}} \end{array}\end{cases}$$ now this solution also satisfies (1) as it's a linear combination of solution to (1). Hence we've found a joint solution for both.
Am I correct?
Edit:
Hmm for a solution to be solution of 1 it must also have only be a square of integer. yet it might be still correct?
Edit: $n^{2}\pi^{2}a_{n}=b_{n}\\\Rightarrow a_{n}=\begin{cases} \begin{array}{c} n\in\mathbb{Z_{\textrm{even}}}\\ n\in\mathbb{Z_{\textrm{od}}} \end{array} & \begin{array}{c} 0\\ \frac{4\sqrt{2}}{n^{5}\pi^{5}} \end{array}\end{cases}$
Expanding both sides of $y''=r(x)=x-x^2$ gives the equation $$ -\sum\lambda_na_n\phi_n=\sum b_n\phi_n \\ \implies a_n=-\frac{b_n}{\lambda_n} $$ where $$ b_n=\langle r(x),\phi_n(x)\rangle $$ as you computed. This gives that the non-trivial coefficients are those with odd index $n$ and $$ a_n=-\frac{4\sqrt2}{n^5\pi^5} $$
Test: Direct integration and application of the boundary conditions $y(0)=y(1)=0$ gives as solution $$ y(x)=-\frac{x}{12}(x^3-2x^2+1) =-\frac{x(x-1)}{12}(x^2-x-1) =-\frac{x(1-x)}{12}(1+x(1-x)) $$ Now computing the expansion coefficients gives $$ a_n=⟨y(x),ϕ_n(x)⟩=-\frac{2\sqrt2(1-\cos(n\pi))}{n^5\pi^5} $$ which confirms the solution.