Solution of Sturm-Liouville problem Here I have been asked one question in my homework if someone can give a hint, I would be grateful for it. If a function $f(x)$ has an expansion  
$$f(x)=\alpha_0 x+\sum_{n=1}^{\infty} \alpha_{n} \sin( \mu_{n}x)$$ for $0<x<1$ where $\tan(\mu_{n})=\mu_n$, obtain the solution to problem 
$$y^{''}+\lambda y=f(x),$$ where
$0<x<1$, $y(0)=0$ and $y(1)=y^{'}(1)$.
In the form $$y(x)=\sum_{n=1}^{\infty} \frac{\alpha_n}{\sigma-\lambda_n} \sin(\mu_n x)$$
I tried to find eigenvalue and eigenfunctions of the homogeneous part and didn't find a satisfactory answer.
Please mention the eigenvalues and eigenfunction (if possible) and suggest the way how to handle it.
Thanking you in anticipation.
 A: *

*Step 1: Find eigenfunctions of the differential operator.


We're looking for solutions of  $y'' +\lambda y = \lambda_n y$. The general solution if $\lambda_n \ne \lambda$ is $y = A\cos(\sqrt{\lambda-\lambda_n}x) + B\sin(\sqrt{\lambda-\lambda_n} x)$. $y(0) = 0$ implies $A = 0$, while $y(1) = y'(1)$ implies
$$
B\sin(\sqrt{\lambda-\lambda_n} ) = B\sqrt{\lambda-\lambda_n}\cos(\sqrt{\lambda-\lambda_n})\Longrightarrow \sqrt{\lambda - \lambda_n} = \tan(\sqrt{\lambda-\lambda_n}) \Longrightarrow \lambda_n = \lambda - \mu_n^2
$$
So the eigenfunctions are $\left|n\right> = \sin(\mu_n x)$ with eigenvalues $\lambda_n = \lambda - \mu_n^2$. However, this gives a degenerate eigenfunction for $n=0$, as $\mu_0 = 0$. This is the case $\lambda = \lambda_n$ that we ignored before, and by inspection we see $\left| 0 \right> = x$. Note that these functions are all orthogonal.


*

*Step 2: Take inner products with the differential equation


I'm going to use bra-ket notation here because it's easier for working with eigenfunctions. Letting $D$ be the differential operator $D = d^2/dx^2 + \lambda$, we have for our equation $D\left|y\right> = \left|f\right>$. Noting that $f(x) = \sum_{m = 0}^\infty\alpha_m\left|m\right>$, we take the inner product of the equation with an eigenfunction $\left<n\right|$ to get
$$
\left<n\right|D\left|y\right> = \lambda_n\left<n\right|\left. y\right> = \left< n\right|\left. f\right> = \sum_{m = 0}^\infty \alpha_n\left<n\right|\left.m\right> = \alpha_n \left<n \right|\left. m\right> \Longrightarrow \frac{\left<n\right|\left. y\right>}{\left<n \right|\left. n\right>} = \frac{\alpha_n}{\lambda_n},
$$
where we used the orthogonality of the eigenfunctions to eliminate the sum.


*

*Step 3: Use eigenfunction expansion to find the solution


Resolving the identity operator as  $I =\sum_{n=0}^\infty\left|n\right>\left<n\right|/\left<n\right|\left. n\right>$, we have
$$
\left|y\right> = \left[\sum_{n=0}^\infty\frac{\left|n\right>\left<n\right|} {\left<n \right|\left. n\right>}\right]\left|y\right> = \sum_{n=0}^\infty \frac{\left<n\right|\left. y\right>}{\left<n\right|\left. n\right>}\left| n \right> = \sum_{n=0}^\infty \frac{\alpha_n}{\lambda_n}\left| n\right> = \frac{\alpha_0}{\lambda} x + \sum_{n= 1}^\infty \frac{\alpha_n}{\lambda - \mu_n^2}\sin(\mu_n x)
$$
which is the desired form for your solution $y(x)$.
A: Try a solution of the form
$$
         y = \beta_0 x + \sum_{n=1}^{\infty}\beta_n\sin(\mu_n x)
$$
Substituting into your equation,
\begin{align}
    f&= \alpha_0x + \sum_{n=1}^{\infty}\alpha_n\sin(\mu_n x) \\
  &=
    y''+\lambda y \\
   &= \lambda\beta_0 x + \sum_{n=1}^{\infty}\beta_n(-\mu_n^2+\lambda)\sin(\mu_n x)
\end{align}
So $\beta_0=\alpha_0/\lambda$, $\beta_n = \alpha_n/(\lambda-\mu_n^2)$. This will work provided $\lambda\ne 0$ and $\lambda\ne\mu_n^2$ for $n=1,2,3\cdots$. Assuming these restrictions, the solution $y$ is
$$
     y =\frac{\alpha_0}{\lambda}+\sum_{n=1}^{\infty}\frac{\alpha_n}{\lambda-\mu_n
^2}\sin(\mu_n x).
$$
