Question

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

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{\lambda }\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{(}\mathit{\pi }\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{0}$

Short answer

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

Step-by-step solution

Given information

The given value is:

$y^{\text{'}\text{'}}+\lambda y=0,y\left(0\right)=0,y\text{'}(\pi /2)=0$

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$y\left(0\right)=0$ makes$c_2=0$ and$y\text{'}(\pi /2)=0$ makes $c_1=0$

$y\left(x\right)=0$ it is trivial.

Case 2: for$\lambda <0$

$\lambda =-{\alpha }^2$, 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 is$y\left(0\right)=0$ gives$c_1+c_2=0$ and $y\text{'}(\pi /2)=0$gives$c_1{e}^{\alpha \pi /2}-c_2{e}^{-\alpha \pi /2}=0$

The previous equation implies$c_1=c_2=0$

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

Case 3: for$\lambda >0$

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

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

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

$y\left(0\right)=0$yields$c_1=0$ ,$y\left(x\right)=c_{2}sin\alpha x$

The last condition $y\text{'}(\pi /2)=0$,is $c_2\alpha cos\alpha \frac{\pi }{2}=0$

$c_2\ne 0$, we get$cos\alpha \frac{\pi }{2}=0$ and hence$\alpha \frac{\pi }{2}=\left(n+\frac{1}{2}\right)\pi$ every integer n.

$\alpha =\left(2n+1\right)$and ${\lambda }_n={\left(2n+1\right)}^2$the real non zero $c_2$,

The solution of problem is $y_n\left(x\right)=c_{2}sin{\left(2n+1\right)}^{2}x$,the nonnegative integer n

$c_2=1$,

The nontrivial eigenvalues are${\lambda }_n={\left(2n+1\right)}^2$

Corresponding eigenfunctions$y_n=sin{\left(2n+1\right)}^{2}x$

For every $n=0,1,2,....$

The final solution is ${\lambda }_n={\left(2n+1\right)}^2,y_n=\left(sin\left(2n+1\right)x\right),n=0,1,2...$