# Find the Eigenvalues and Eigenfunctions for the Boundary problem

I recently found this answer to a similar problem I'm currently working on. The problem is the following...

Find the eigenvalues and eigenfunctions for

$$y^{\prime \prime}+\lambda y=0$$

with the boundary conditions

$$y^{\prime} (0)=0$$ , $$y^{\prime} (1)=0$$

In the answer linked above, the Ansatz for

• $$\lambda <0$$ is $$y(x)=C_1e^{\sqrt{\lambda} x}+C_2e^{-\sqrt{\lambda} x}$$
• $$\lambda >0$$ is $$y(x)=C_1\cos(\sqrt{\lambda}x)+C_2\sin(\sqrt{\lambda}x)$$

But I don't really get why. Has someone got some background information/explanation why $$y(x)$$ must be of this form?

The case for $$\lambda =0$$ is clear to me.

Thanks in advance.

• Try to solve the characteristic polynomial $r^2+\lambda =0$ For $\lambda > 0$ and $\lambda <0$. Jan 24 at 0:21
• Sorry for the basic question but why exactly $r^2+\lambda$? It makes sense because I get the Ansatz for $y(x)$ above but is this always the characteristic polynomial I've to look for? Jan 24 at 0:24
• When you solve a differential equation with constant coefficients yes you need the polynomial characteristiic. Maybe you should read the wiki page on differential linear equations. Jan 24 at 0:32
• $y''+by'+cy=0$ has the characteristic polynomial as $r^2+br+c=0$ You find $r$ then the solution is $y=ke^{rx}$ Maybe you should read this en.wikipedia.org/wiki/Linear_differential_equation Jan 24 at 0:33

## 1 Answer

$$y^{\prime \prime}+\lambda y=0$$ We suppose that the solution is on the form $$y=e^{rx}$$ then you get: $$r^2e^{rx}++\lambda e^{rx}=0$$ $$e^{rx}(r^2+\lambda)=0$$ $$\implies r^2+ \lambda=0$$ Solve the characteristic polynomial. You have three cases: $$\lambda >0, \lambda =0, \lambda<0$$. Then the solution is: $$\lambda <0 \implies y=c_1e^{r_1x}+c_2e^{r_2x}$$ $$\lambda >0 \implies y=c_1 \cos ({r_1x)}+c_2\sin (r_2x)$$ Where $$r_1,r_2$$ are solution of the polynomial characteristic. Apply the initial conditions you'll find the value of the constants $$c_1,c_2$$.

• Makes sense now, thanks a lot! Jan 24 at 0:41
• You are welcome. The wiki page is also interesting. Jan 24 at 0:42