Sturm-Liouville problem involving $dy/dx$ $$ \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + \lambda y=0 \ , y(-1)=y(1) = 0 $$
I have 2 ideas
1- $ (\frac{dy}{dx} +2y )' + \lambda y  = 0  \\ \text{which means weighted function is } w=1 $
2- $ \frac{dy}{dx} (p \frac{dy}{dx}) + \lambda w(x)y = 0 \\ \text{again weighted function is w=1.} $
I ignored dy/dx term .. I saw an example which just did that.
Please guide me.
I need to construct Strum-Liouville system and find its orthogonality using weighted function.
Thank you
 A: You given ODE:
$$y''+2y'+\lambda y=0 ~~~(1) $$
Can be writte in S_L form as
$$e^{-2x}\left(\frac{d}{dx} e^{2x} \frac{dy}{dx}\right)+\lambda y=0.~~~~(2)$$
ODE (1) can be solved by letting $y=e^{mx}$, we get $$m^2+2m+\lambda=0 \implies m_{1,2}=-1\pm \sqrt{1-\lambda}.$$
In order to have periodic solutions, let $q^2=\lambda-1$
So $$Y=e^{-x}[C_1 e^{iqx} +C_2 e^{-iqx}]$$
Applying $y(-1)=0=y(1)$, we get
$$y=A e^{-x} \cos qx, x \in[-1,1], q=(n+1/2)\pi, n=0,2,3,..$$
So the eigenvalues are $\lambda_n=(n+1/2)^2\pi^2+1$, with eigenfunctions as
$$y_n(x)=A e^{-x} \cos [(n+1/2) \pi x].$$
The given ODE can be cast in a simple S-L problem as (2), then $$-\frac{1}{w(x)}\frac{d}{dx} \left(p(x)\frac{dy_n}{dx}\right) =\mu_n y_n, x \in[a,b]$$
implies that the two solutions $y_m, y_n$  are orthogonal as
$$\int_{a}^{b} w(x) y_m(x) y_n (x) dx=C \delta_{m,n}.$$
Here, in this case (1,2), $w(x)=p(x)=e^{2x}.$
We see the orthogonality here as
$$A^2\int_{-1}^{1} e^{2x} e^{-2x} \cos[(m+1/2)\pi x] \cos[(n+1/2)\pi x] dx= A^2 \int_{0}^{1}[\cos[(m+n)\pi x]+ \cos[(m-n)\pi x]] dx=A^2\frac{\sin (m-n) \pi}{(m-n) \pi}=A^2 \delta_{m,n} $$
A: Let $y=e^{-x}f$. Then
\begin{align}
         y' &= \;\;\;\;\;\;\;\;\;\;\;\;\;e^{-x}f'-e^{-x}f \\
         y'' &= e^{-x}f''-2e^{-x}f'+e^{-x}f\\
      y''+2y'+\lambda y &= e^{-x}f''+(\lambda -1)e^{-x}f.\\
\end{align}
The equation in $f$ becomes
$$
                           f''+(\lambda-1)f=0 \\
                      f(-1)=f(1)=0.
$$
The corresponding solutions $f$ are scalar multiples of the following for $n=0,1,2,3,\cdots$:
$$
            s_n(x)= \sin(n\pi x),\;\; \lambda_n=1+n^2\pi^2 \\
            c_n(x) = \cos((n+1/2)\pi x),\;\;\; \lambda_n=1+(n+1/2)^2\pi^2
$$
The original solutions $y_n$ are scalar multiples of $e^{-x}s_n(x)$ and $e^{-x}c_n(x)$.
