Question

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

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

Short answer

the solution is

${\lambda }_n=n^2,y_n=sinnx,n=1,2...$and${\lambda }_n={\left(n+\frac{1}{2}\right)}^2,y_n=cos\left(n+\frac{1}{2}\right)x$

Step-by-step solution

given information

The given value is:

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

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 $y=c_{1}x+c_2$.

$y(-\pi )=0$and $y\left(\pi \right)=0$makes $-\pi c_1+c_2=0$and $\pi c_1+c_2=0$.$c_1=c_2=0$

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

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 two conditions $y(-\pi )=0$and $y\left(\pi \right)=0$and $c_{1}e-\alpha \pi +c_2{e}^{\alpha \pi }=0$ and

$c_{1}e\alpha \pi +c_2{e}^{-\alpha \pi }=0$

$c_1=c_2=0$

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

Case 3: for$\lambda >0$

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

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

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

$y(-\pi )=0$yields$c_{1}cos\alpha \pi -c_{2}sin\alpha \pi =0$and $y\left(\pi \right)=0$gives$c_{1}cos\alpha \pi +c_{2}sin\alpha \pi =0$

$c_1\ne 0$and $c_2=0$, we get $cos\alpha \pi =0$and so $\alpha =n+\frac{1}{2}$ implies ${\lambda }_n={\left(n+\frac{1}{2}\right)}^2$corresponding eigenfunctions $y_n\left(x\right)=cos\left(n+\frac{1}{2}\right)x$

$c_2\ne 0$and $c_1=0$we get $sin\alpha \pi =0$and$\alpha =n$implies${\lambda }_n=n^2$with corresponding eigenfunctions $y_n\left(x\right)=sinnx$

If $c_1,c_2\ne 0,$we get$sin\alpha \pi =cos\alpha \pi =0$. It is impossible.

The final solution is

${\lambda }_n=n^2,y_n=sinnx,n=1,2...$and${\lambda }_n={\left(n+\frac{1}{2}\right)}^2,y_n=cos\left(n+\frac{1}{2}\right)x$