given that $y''=-\lambda y$, $y(0)=0$, $y(1)=0$ and $\phi_{n}(x)=\sqrt{2}\sin(n\pi x)$, $\lambda=n^{2}\pi^{2}$.
solve: $y''=\sin(2\pi x)-\sin(3\pi x)$, $y(0)=0$, $y(1)=0$.
The main idea of this question is using the data to solve it, and not other super-techniques.
what I thought is taking $$ \phi_{2}=\sqrt{2}\sin(2\pi x)\\\phi_{3}=\sqrt{2}\sin(3\pi x)\\\frac{\frac{\phi_{2}}{4}-\frac{\phi_{3}}{9}}{\pi^{2}\sqrt{(2)}} $$ which should also be a solution, as it is a linear combination of solutions therefore according to Sturm–Liouville theory, must be also a solution.
Am I correct?
Sturm-Liouville theory states that for each $\lambda_{n}$ there is an eigenfunction $\phi_{n}(x)$ that is unique up to an arbitrary constant. Then $\phi_{n}''(x)=-\lambda_{n}\phi_{n}(x)$ where $\lambda_{n}=n^{2}\pi^{2}$ for $n\in\mathbb N.$
For the case $n=2$ we have $\phi_{2}''(x)=-4\pi^{2}\phi_{2}(x)$ and for the case $n=3$ we have $\phi_{3}''(x)=-9\pi^{2}\phi_{3}(x)$. We want to find a solution $y(x)=A\phi_{2}(x)+B\phi_{3}(x)$ to the given problem so substituting we have $y''(x)=A\phi_{2}''(x)+B\phi_{3}''(x)=-4A\pi^{2}\phi_{2}(x)-9B\pi^{2}\phi_{3}(x)=\frac{1}{\sqrt{2}}(\phi_{2}(x)-\phi_{3}(x)).$
Then comparing sides gives $A=-\frac{1}{4\pi^{2}\sqrt{2}}$ and $B=\frac{1}{9\pi^{2}\sqrt{2}}$, so the required solution is $y(x)=-\frac{\sin(2\pi x)}{4\pi^{2}}+\frac{\sin(3\pi x)}{9\pi^{2}}.$