ODE is
$$ y'' + 3y' + (2+\lambda)y = 0, \quad y(0) = y(L) = 0$$
To show that it is a SLP, I simply did this
$$ (e^{3x}y')' + 2e^{3x}y + \lambda e^{3x} = 0 $$
The weight function then is $w(x) = e^{3x}$
Eigenvalues of the problem are
$$ \lambda = \frac{1}{4} + \frac{n^2 \pi^2}{L}$$
I need to prove orthogonality of eigen-functions. I am confused about the relation I need yo use for this purpose, is it $\int \sin{\lambda_m x} \ \sin{\lambda_n x} \ \ dx = 0$ or $\int e^{3x}\sin{\lambda_m x} \ \sin{\lambda_n x} \ \ dx = 0$ ? Which is it and why? Thanks in advance.
Best Answer
The characteristic equation for this ODE is
$$\eta^{2}+3\eta+\left(2+\lambda\right)=0$$
with roots
$$\eta=\dfrac{-3\pm\sqrt{1-4\lambda}}{2}=\dfrac{-3\pm i\sqrt{4\lambda-1}}{2}$$
Note that I wrote this with an imaginary part because your solution must be oscillatory if you want to satisfy the boundary conditions. From the boundary condition at $x=0$ you'll get the solution is
$$y_{\lambda}=Ce^{-\frac{3}{2}x}\sin\left(\dfrac{\sqrt{4\lambda-1}}{2}x\right)$$
The factor of $e^{-\frac{3}{2}x}$ in each of those eigenfunctions will cancel with the weight in the inner product, so the orthogonality relation boils down to the orthogonality of the sine function.