Question

Find the eigenvalues and eigenfunctions for the given boundary-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{\lambda }\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathit{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

the solution is${\lambda }_n=n^2,y_n=cosnx,n=0,1,2,....$

Step-by-step solution

Given information

The given value is:

$y\text{'}\text{'}+\lambda y=0,y\text{'}\left(0\right)=0,y\text{'}\left(\pi \right)=0$

Three cases will be considered

$\lambda =0,\lambda <0,$and$\lambda >0$

Case 1: for$\lambda >0$

$\lambda =0$is the solution of$y\text{'}\text{'}=0$ is $y=c_{1}x+c_2$.

The condition$y\text{'}\left(0\right)=0$ makes$c_1=0$ and$y=c_2$ and $y\text{'}\left(x\right)=0$.the$c_2$ is arbitrary. the eigenvalue$\lambda =0,$ with eigenfunction$y\left(x\right)=1$

Case 2: for$\lambda <0$

$\lambda =0$, the positive number$\alpha$

The auxiliary equation is $m^2-{\alpha }^2=0$.the roots are $\pm \alpha$.

The general solution is$y=c_1{e}^{\alpha x}+c_2{e}^{-\alpha x}$

$y\text{'}\left(0\right)=y\text{'}\left(\pi \right)=0$. Hence$c_1-c_2=0$ and $c_1{e}^{\pi \alpha }-c_2{e}^{-\pi \alpha }=0$.$c_1=c_2=0$

The trivial solution is$y\left(x\right)=0$

Case 3: for$\lambda >0$

$\lambda ={\alpha }^2$the positive number

The auxiliary equation is $m^2+{\alpha }^2=0$. The complex roots are$\pm \alpha i$

The general solution is$y\left(x\right)=c_{1}cos\alpha x+c_{2}sin\alpha x$

$y\text{'}\left(0\right)=0$yields,$c_2=0$,$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathit{x}$

The last condition $y\text{'}\left(\pi \right)=0$,is $-c_1\alpha sin\alpha \pi =0$.

$c_2\ne 0$,we get $sin\alpha \pi =0$.hence$\alpha \pi =n\pi$ every integer .

$\alpha =n$and ${\lambda }_n=n^2$. The real non zero $c_2$,

The solution of the problem $y_n\left(x\right)=c_{1}cos\left(nx\right)$.

$c_2=1$,

The nontrivial eigenvalues are

${\lambda }_n=n^2$

Corresponding eigenfunctions

$y_n=cosnx$

For every$n=1,2,....$

The final solution is${\lambda }_n=n^2,y_n=cosnx,n=0,1,2,....$