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I recently asked this question about converting the DE

$$y'' + 2y' + (\lambda + 1)y = 0$$

to Sturm-Liouville form:

$$\frac{d}{dx}\left( e^{2x} y' \right) + (\lambda e^{2x} + e^{2x})y = 0$$

I am now trying to get the solutions satisfying the boundary conditions

$$y(0) = y(\pi) = 0$$

And I then want to use integration to check that the eigenfunctions corresponding to different eigenvalues are orthogonal.

I'm presuming that the latter is done by just taking the inner product of the eigenfunctions, which means integrating their product with the weight function $r(x) = e^{2x}$ from $0$ to $1$?

But how do I get the solutions satisfying the BCs in the first place?

I would greatly appreciate it if people could please take the time to demonstrate this.

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    $\begingroup$ It's a second order, linear DE, so make a guess of the form $y = e^{\gamma x}$. This yields the solutions $$y = e^{-x} ( c_{1} e^{\sqrt{-\lambda} x} + c_{2} e^{ - \sqrt{-\lambda} x})$$ Now check the cases when $\lambda < 0, \ \lambda = 0, \ \lambda > 0$ and see if there are any non-trivial eigenfunctions satisfying the boundary conditions. $\endgroup$ – Mattos Aug 16 '18 at 8:49
  • $\begingroup$ @Mattos Got it. Thanks! $\endgroup$ – The Pointer Aug 16 '18 at 17:26
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If $\lambda>0$ solution of DE is $$y=C_1\, e^{-x} \sin{\left( x\, \sqrt{\lambda }\right) }+C_2\, e^{-x} \cos{\left( x\, \sqrt{\lambda }\right) }$$ Then eigenvalues and eigenfunctions are $$\lambda_k=k^2,$$ $$y_k=e^{-x}\sin(kx),$$ $k=1,2,\ldots$

If $\lambda\leqslant 0$ solution of DE satisfying the boundary conditions $$y(0) = y(\pi) = 0$$ is only $$y=0$$ Eigenfunctions are orthogonal with the weight function $r(x)=e^{2x}$: $$\langle y_k, y_m\rangle=\int_0^\pi e^{2x}y_ky_mdx=\int_0^\pi\sin(kx)\sin(mx)dx=0,$$ if $k\neq m$.

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There is an alternative way to convert $$ -y''-2y'-y = \lambda y $$

to standard form--namely, by setting $y(x)=e^{-x} f(x)$. Then $$ -(e^{-x}f)''-2(e^{-x}f)'-e^{-x}f=\lambda e^{-x}f \\ -(e^{-x}f''-2e^{-x}f'+e^{-x}f)-2(e^{-x}f'-e^{-x}f)-e^{-x}f=\lambda e^{-x}f \\ -f''=\lambda f $$

The conditions are $f(0)=f(\pi)=0$. The solutions $y$ are of the form $y=e^{-x}f(x)$.

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