Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathit{\lambda }\mathbf{+}\mathbf{1}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution is${\lambda }_n=n^2{\pi }^2/25,y_n=e^{-x}sin\left(n\pi x/5\right),n=1,2,...$

Step-by-step solution

Given information

The given value is:

$y\text{'}\text{'}+2y\text{'}+(\lambda +1)y=0,y\left(0\right)=0,y\left(5\right)=0$

Three cases will be considered

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

The auxiliary equation is$m^2+2m+(\lambda +1)=0$

Discriminant is$-4\lambda$

Case 1: for$\lambda =0,$

$\lambda =0$is the solution of $y\text{'}\text{'}+2y\text{'}+y=0$is $y=(c_{1}x+c_2)e^{-x}$. The condition $y\left(0\right)=0$makes $c_2=0$and $y\left(5\right)=0$makes $y\left(5\right)=0$and which $y\left(x\right)=0$is trivial.

Case 2: for$\lambda <0$

$\lambda =-{\alpha }^2$, the positive number . The auxiliary equation is$m^2+2m+\left(1-{\alpha }^2\right)=0$.

The roots are$-1\pm \alpha$

The general solution is$y=c{e}^{\left(-1+\alpha \right)t}+c_2{e}^{\left(-1-\alpha \right)t}$

The two conditions$y\left(0\right)=0$and$y\left(5\right)=0$.claim$c_1=c_2=0$

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

Case 3: for$\lambda >0$

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

The auxiliary equation is$m^2+2m+\left({\alpha }^2+1\right)=0$

The complex roots$-1\pm \alpha i$

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

$y\left(0\right)=0$yields$c_1=0$

$y\left(x\right)=c_2{e}^{-x}sin\alpha x$

The last condition $y\left(5\right)=0$and $c_2{e}^{-5}sin5\alpha =0$

$c_2\ne 0$, we get $sin5\alpha =0$and hence $5\alpha =n\pi$every integer n. So that $\alpha =n\pi /5$and ${\lambda }_n=n^2{\pi }^2/25$. The real nonzero $c_2$, the solution of problem is .

$y_n\left(x\right)=c_2{e}^{-x}sin\left(n\pi x/5\right)$.$c_2=1$

The nontrivial eigenvalues are

${\lambda }_n=n^2{\pi }^2/25$

Corresponding eigenfunctions

$y_n=e^{-x}sin\left(n\pi x/5\right)$

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

The final solution is ${\lambda }_n=n^2{\pi }^2/25,y_n=e^{-x}sin\left(n\pi x/5\right),n=1,2,...$