# How to solve this Sturm-Liouville problem $y''+2y'+\lambda y = 0$?

I have the following Sturm-Liouville problem:

\left\{ \begin{aligned} & y''+2y'+\lambda y = 0, 0

I have tried to reduce it to Sturm-Liouville form, got this: \left\{ \begin{aligned} & (e^{2x}y')'+e^{2x}\lambda y = 0, 0

Then, I checked whether there exist negative lambdas via: $$p(x)y(x)y'(x)\Big|_0^a$$ where $$p(x)=e^{2x}$$

So it evaluated 0, so we know that for $$\lambda < 0$$ there is no non-trivial solutions.

But reducing didn't help much, since I anyway had to find the general solution of the equation.

Solving, I got $$\alpha_{1,2}=-1\pm\sqrt{1-\lambda}\\ y(x)=C_1e^{\alpha_1}+C_2e^{\alpha_2}$$

So I then checked for $$\lambda=0$$ and got: $$y(x) = C_1 + C_2e^{-x}\\ y'(x) = -C_2e^{-x}$$ And substituting the initial conditions: \left\{ \begin{aligned} & C_1 + C_2 = 0\\ & -C_2e^{-a} = 0 \end{aligned}\right. \iff\left\{ \begin{aligned} & C_1 = 0\\ & C_2 = 0 \end{aligned} \right. So also no nontrivial solutions for this case.

Then I checked $$\lambda=1$$: $$y(x) = C_1e^{-x} + C_2xe^{-x}\\ y'(x) = -C_1e^{-x} + C_2e^{-x}-C_2xe^{-x}$$ And with initial conditions this evaluates to: \left\{ \begin{aligned} & C_1=0\\ & C_2-C_2a=0 \end{aligned} \right. So if $$a=1$$ we have non-trivial solutions, where $$C_1=0$$ and $$C_2$$ is any non-zero number. So we can pick 1.

So if a = 1: $$y(x)=xe^{-x}$$

Now all that's left is to check $$\lambda>1$$: $$\alpha_{1,2}=-1\pm i \sqrt{\lambda-1}\\ y(x)=e^{-x}C_1\cos(\sqrt{\lambda-1}x)+e^{-x}C_2\sin(\sqrt{\lambda-1}x)\\ y'(x)=-C_1\sqrt{\lambda-1}e^{-x}\sin(\sqrt{\lambda-1}x)-C_2e^{-x}\sin(\sqrt{\lambda-1}x) +C_2\sqrt{\lambda-1}e^{-x}\cos(\sqrt{\lambda-1}x)-C_1e^{-x}\sin(\sqrt{\lambda-1}x)$$ With initial conditions we got: \left\{ \begin{aligned} & C_1=0\\ & C_2e^{-a}(\sqrt{\lambda-1}\cos(\sqrt{\lambda-1}a)-\sin(\sqrt{\lambda-1}a))=0 \end{aligned} \right. And here I stuck, because i don't know how to solve: $$\sqrt{\lambda-1}\cos(\sqrt{\lambda-1}a)-\sin(\sqrt{\lambda-1}a) = 0$$

Where did I go wrong or what could I do to solve this?

• Your eigenvalue equation is right but it cannot be solve by hand. It can be solved graphically/numerically only. Commented Oct 21, 2021 at 17:38

The equation $$y''+2y'+\lambda y = 0$$ can be transformed by multiplying by $$e^x$$: $$e^x y''+2e^x y'+\lambda e^x y =0 \\ (e^x y)''+(\lambda-1) e^x y = 0$$ The solutions are $$e^x y = A\cos(\sqrt{\lambda-1}x)+B\frac{\sin(\sqrt{\lambda-1}x)}{\sqrt{\lambda-1}} \\ y = Ae^{-x}\cos(\sqrt{\lambda-1}x)+Be^{-x}\frac{\sin(\sqrt{\lambda-1}x)}{\sqrt{\lambda-1}}.$$ The limiting form as $$\lambda\rightarrow 1$$ is $$y=Ae^{-x}+Bxe^{-x}.$$ For all $$\lambda$$, these solutions satisfy $$y(0)=A,\;\; y'(0)=(B-A)$$ You want the solutions to satisfy $$y(0)=0$$, which gives $$y=Be^{-x}\frac{\sin(\sqrt{\lambda-1}x)}{\sqrt{\lambda-1}}.$$ In order for $$y'(a)=0$$ to hold, it is necessary and sufficient that $$\lambda$$ satisfy the eigenvalue equation: $$0=y'(a)=-Be^{-a}\frac{\sin(\sqrt{\lambda-1}a)}{\sqrt{\lambda-1}}+Be^{-a}\cos(\sqrt{\lambda-1}a) \\ \frac{\sin(\sqrt{\lambda-1}a)}{\sqrt{\lambda-1}}=\cos(\sqrt{\lambda-1}a)$$ $$\lambda=1$$ is a solution iff $$a=1$$. Otherwise, the eigenvalue equation is transcendental: $$\tan(\sqrt{\lambda-1}a)=\sqrt{\lambda-1}.$$