Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\left(\lambda +1\right)\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution is${\lambda }_n=n^2{\pi }^2-1,y_n\left(x\right)=cos\left(n\pi x\right),n=0,1,2....$

Step-by-step solution

Given information

The given value is:

$y\text{'}\text{'}+\left(\lambda +1\right)y=0,y\text{'}\left(0\right)=0,y\text{'}\left(1\right)=0$

consider three cases

$\lambda =-1,\lambda <-1,\lambda >-1$

Case 1: for$\lambda =-1,$the solution of $y\text{'}\text{'}=0$is$y=c_{1}x+c_{2}x$. The condition$y\text{'}\left(0\right)=0$makes$c_1=0$and$y\left(1\right)=0$makes again$c_1=0$.$y\left(x\right)=1$for the eigen values -1.

Case 2: for$\lambda <-1$say$\lambda =-1-{\alpha }^2$the positive number$\alpha$.

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

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

The condition$y\text{'}\left(0\right)$gives$c_1-c_2=0$and$y\text{'}\left(1\right)=0$gives$c_1{e}^{\alpha }-c_2{e}^{-\alpha }=0$

Previous equations imply$c_1=c_2=0$. The trivial solution is$y\left(x\right)=0$.

Case 3: for$\lambda >-1$

$\lambda =-1+{\alpha }^2$, the positive number$\alpha .$

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

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 $c_2=0$,$y\left(x\right)=c_{1}cos\alpha x$

The last condition$y\text{'}\left(1\right)=0$,$-c_1\alpha sin\alpha =0$

$c_2\ne 0$, we get $sin\alpha =0$.hence$\alpha =n\pi$every integer n . ${\lambda }_n=n^2{\pi }^2-1$So for every$n=1,2,3...$the any real nonzero $c_2$, the solution of the problem $y_n\left(x\right)=c_2\cos \left(n\pi x\right)$

$c_2=1$.

The nontrivial eigenvalues are

${\lambda }_n=n^2{\pi }^2-1$

Corresponding eigenfunctions

$y_n\left(x\right)=\cos \left(n\pi x\right)$

For every $n=1,2,3...$

The final solution is$\mathrm{\backslash [}{\lambda }_n=n^2{\pi }^2-1,y_n\left(x\right)=cos\left(n\pi x\right),n=0,1,$