# Solving the BVP $y''+2y'+(\lambda+1)y=0$ for $y(0)=y(\pi)=0$

Convert the differential equation $$y''+2y'+(\lambda+1)y=0$$ to Sturm-Liouville form, and obtain the solutions satisfying the boundary conditions $$y(0)=y(\pi)=0.$$

Using the integrating factor $$\mu(x)=e^{2x}$$, the Sturm-Liouville form of the ODE is $$(e^{2x}y')'+(e^{2x}+\lambda e^{2x})y=0.$$ For the solutions satisfying the boundary conditions, I considered three cases:

If $$\lambda=0$$, then $$y''+2y'+y=0$$. This has the general solution $$y(x)=Ae^{-x}+Bxe^{-x}$$ and upon applying the boundary conditions, leads to a trivial solution.

If $$\lambda=w^2<0$$, then $$y''+2y'+(-w^2+1)y=0$$. This has the general solution $$y(x)=Ce^{(-1+w)x}+De^{(-1-w)x}$$ and upon applying the boundary conditions, leads to a trivial solution.

If $$\lambda=w^2>0$$, then $$y''+2y'+(w^2+1)y=0$$ This has the general solution $$y(x)=e^{-x}\left(E\cos(wx)+F\sin(wx)\right)$$ and upon applying the boundary conditions, a non-trivial solution occurs when $$w=n, \ n\in\mathbb{Z^+}$$.

Does this mean that the eigenfunctions of this ODE are $$\phi_n(x)=e^{-x}\sin(nx)$$ with corresponding eigenvalues $$\lambda_n=n^2$$? I don't fully understand the terms eigenfunctions and eigenvalues.

Furthermore, I wish to show that the eigenfunctions corresponding to distinct eigenvalues are orthogonal: \begin{align} \int_0^\pi w(x)y_n(x)y_m(x) \ dx&=\int_0^\pi e^{2x}(e^{-2x}\sin(xn)\sin(xm)) \ dx \\ &=\frac{1}{2}\left(\frac{sin(\pi(n-m))}{n-m}-\frac{\sin(\pi(n+m))}{n+m}\right) \end{align}

• Yes, you are correct – Dylan Mar 12 at 8:37
• Thank you. But how come when I try to show by direct integration that eigenfunctions corresponding to different eigenvalues are orthogonal? $$\text{e.g.} \ \int_{0}^{\pi} \phi_1(x)\phi_2(x)=0.$$ Clearly this does not work. – Stuart-James Burney Mar 12 at 8:58
• They are orthogonal, but it's not straightforward. There's a weighting factor you'll need to find – Dylan Mar 12 at 8:59
• Ah, I forgot that. We can use the ODE in Sturm-Liouville form to see that the weighting function is $e^{2x}$. Thus the resulting integral reduces to zero. – Stuart-James Burney Mar 12 at 9:00

Eigenfunctions $$y_n=e^{-x}\sin(nx)$$ and eigenvalues $$\lambda_n=n^2$$ is solutions of eigenvalue problem $$Ly=\lambda y, \quad y(0)=y(\pi)=0,$$ here $$Ly=-(y''+2y'+y).$$