Question

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

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

Short answer

Therefore, the numbers are${\lambda }_n=n^2,y_n=\sin \left(n\log x\right),n=1,2,\cdots$

Step-by-step solution

Given Information

The given value is

$x^{2}y\text{'}\text{'}+xy\text{'}+\lambda y=0,y\left(1\right)=0,y\left(e^{\pi }\right)=0$

Using the transformation method

We get $x=e^t,$ by applying the transformation.

$xy\text{'}=\frac{dy}{dt}$and

$x^2{y}^{\text{'}\text{'}}=\frac{d^{2}y}{1+2}-\frac{dy}{x}$

We get the following differential equation

$\frac{d^{2}y}{d{t}^2}+\lambda y=0,y\left(0\right)=0,y\left(\pi \right)=0$

We shall consider three cases:

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

Solve the boundary value problem

Case I: For $\lambda =0$ the solution of$y\text{'}\text{'}=0$ is ${\text{y =c}}_{1}t+c_2.$The conditions$y\left(0\right)=0$ and $y\left(\pi \right)=0$makes$c_1=c_2=0$ and so$y\left(t\right)=0$ which is trivial.

Case II: For$\lambda <0,$ say$\lambda =-{\alpha }^2,$ for some positive number $\alpha$So that the characteristic equation is$m^2-{\alpha }^2=0$ whose roots are$\pm \alpha .$ Thus the general solution is

$y=c_1{e}^{\alpha t}+c_2{e}^{-\alpha t}$

Since $y\left(0\right)=y\left(\pi \right)=0.$

Hence,$c_1=c_2=0$ to get the trivial solution$y\left(t\right)=0.$

Case III: For $\lambda >0$we write $\lambda ={\alpha }^2,$where is a positive number. The characteristic equation is $m^2+{\alpha }^2=0$which has complex roots $\pm \alpha i.$The general solution is

$y\left(x\right)=c_1\cos \alpha t+c_2\sin \alpha t$

As before, $y\left(0\right)=0$yields$c_1=0$ and so

$y\left(x\right)=c_2\sin \alpha t$

Now the last condition$y\left(\pi \right)=0$ and so

$c_2\sin \alpha \pi =0$

Choosing $c_2\ne 0,$we get $\sin \alpha \pi =0$and hence $\alpha \pi =n\pi$for every integer n. So that $\alpha =n$and ${\lambda }_n=n^2.$Therefore for any real nonzero$c_2,y_n\left(t\right)=c_2\sin \left(nx\right)$ is a solution of the problem for each positive integer Because the differential equation is homogeneous, any constant multiple of a solution is also a solution, so we may, if desired, simply take$c_2=1.$

Thus the nontrivial eigenvalues are

${\lambda }_n=n^2$

With corresponding eigenfunctions

$y_n\left(x\right)=\sin \left(n\log x\right)$

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

Therefore, the numbers are${\lambda }_n=n^2,y_n=\sin \left(n\log x\right),n=1,2,\cdots$