Question

Use a CAS to approximate the eigenvalues ${\lambda }_1,$${\lambda }_2,{\lambda }_3,$ and ${\lambda }_4$ of the boundary-value problem:

$y^{\text{'}\text{'}}+\lambda y=0,y\left(0\right)=0,y\left(1\right)-\frac{1}{2}{y}^{\text{'}}\left(1\right)=0.$Give the corresponding approximate eigenfunctions $y_1\left(x\right),y_2\left(x\right)$, $y_3\left(x\right)$ , and $y_4\left(x\right).$

Short answer

$\begin{array}{rr}{\text{λ}}_{\text{1}}{\text{=-3.667,y}}_{\text{1}}\text{(x)=}& \text{sinh(1.915x)}\\ {\text{λ}}_{\text{2}}{\text{=18.2738,y}}_{\text{2}}\text{(x)=}& \text{sin(4.27487x)}\\ {\text{λ}}_{\text{3}}{\text{=57.7075,y}}_{\text{3}}\text{(x)=}& \text{sin(7.59655xx)}\\ {\text{λ}}_{\text{4}}{\text{=116.9139,y}}_{\text{4}}\text{(x)=}& \text{sin(10.8127x)}\end{array}$

Step-by-step solution

To Find the solution of the differential equation

In the case when $\lambda =-{\alpha }^2<0$, the solution of the differential equation is $y=c_{1}cosh\alpha x+c_{2}sinh\alpha x$. The condition $y\left(0\right)=0$ gives $c_1=0.$ The condition $y\left(1\right)-\frac{1}{2}{y}^{\text{'}}\left(1\right)=0$applied to $y=c_{2}sinh\alpha x$ gives $c_2\left(sinh\alpha -\frac{1}{2}\alpha cosh\alpha \right)=0$ or $tanh\alpha =\frac{1}{2}\alpha .$ As can be seen from the figure, the graphs of $y=\tan hx$ and $y=\frac{1}{2}x$intersect at a single point with approximate $x-$ coordinate ${\alpha }_1=1.915$. Thus, there is a single negative eigenvalue ${\lambda }_1=-{\alpha }_1^2\approx -3.667$and the corresponding eigenfunction is $y_1\left(x\right)=sinh1.915x$. For $\lambda =0$ the only solution of the boundary-value problem is $y=0.$

For $\lambda ={\alpha }^2>0$ the solution of the differential equation is $y=c_{1}cos\alpha x+c_{2}sin\alpha x$. The condition $y\left(0\right)=0$ gives $c_1=0$, so $y=c_{2}sin\alpha x.$The condition $y\left(1\right)-\frac{1}{2}{y}^{\text{'}}\left(1\right)=0$ gives $c_2\left(sin\alpha -\frac{1}{2}cos\alpha \right)=0$ , so the eigenvalues are ${\lambda }_n={\alpha }_n^2$ when $a_n,n=2,3,4,....$ are the positive roots of $\tan \alpha =\frac{1}{2}\alpha$.

Using a CAS we find that the first three values of $\alpha$ are ${\alpha }_2=4.27487,{\alpha }_3=7.59655$ and ${\alpha }_4=10.8127$ The first three eigenvalues are then ${\lambda }_2={{\alpha }^2}_2=18.2738,{\lambda }_3={{\alpha }^2}_3=57.7075$ and ${\lambda }_4={{\alpha }^2}_4=116.9139$with corresponding eigenfunctions ,$y_2=\sin 4.27487x,y_3=\sin 7.59655x,$ and $y_4=\sin 10.8127x$.

Final proof

$$\begin{gathered} {\lambda }_1=-3.667,y_1\left(x\right)=\sin h\left(1.915x\right) \\ {\lambda }_2=18.2738,y_2\left(x\right)=\sin \left(4.27487x\right) \\ {\lambda }_3=57.7075,y_3\left(x\right)=\sin \left(7.59655xx\right) \\ {\lambda }_4=116.9139,y_4\left(x\right)=\sin \left(10.8127x\right) \end{gathered}$$