Question

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

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

Short answer

Thus, the eigenfunctions set that solves the eigenvalue problem is:

$y_n\left(x\right)=\sin \left(\frac{(2n-1)\pi }{2}\ln x\right)$

Step-by-step solution

Given Information

The given value is

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

The following is how to convert the equation into a BVP eigenvalue problem

We begin by putting the equation into a form of BVP eigenvalue problem as follows:

$\left(x^2{D}^2+xD\right)y=-\lambda y$

where D is defined as the differential operator as follows:

$D=\frac{d}{dx}$

Here, the eigenfunction is y while the eigen value of the problem is$-\lambda .$

This is a Cauchy-Euler differential equation which could be solved by obtaining an auxiliary equation as follows:

Let$y=x^m$whereis the auxiliary parameter, then

Then,

$m(m-1)+m+\lambda =0=m^2+\lambda$

Since $x^m$can't be equal to $0$.The above equation is called the auxiliary equation. Then,$m=\pm i\sqrt{\lambda }$

Where

$y=c_1{x}^{m_1}+c_2{x}^{m_2}$

Apply the boundary condition

$y=c_1{x}^{i\sqrt{\lambda }}+c_2{x}^{-i\sqrt{\lambda }}$

$=c_1{e}^{i\sqrt{\lambda }\ln x}+c_2{e}^{-i\sqrt{\lambda }\ln x}$

localid="1664126746709" $=c_1\left(\cos \right(\sqrt{\lambda }\ln x)+i\sin (\sqrt{\lambda }\ln x\left)\right)+c_2\left(\cos \right(\sqrt{\lambda }\ln x)-i\sin (\sqrt{\lambda }\ln x\left)\right)$

$=\cos \left(\sqrt{\lambda }\ln x\right)\left(c_1+c_2\right)+\sin \left(\sqrt{\lambda }\ln x\right)\left(i{c}_1-i{c}_2\right)$

$=A\cos \left(\sqrt{\lambda }\ln x\right)+B\sin \left(\sqrt{\lambda }\ln x\right)$

Where$A=c_1+c_2$and$A=c_1+c_2$
Now we apply the boundary conditions:

$y\left(1\right)=0$and$y^{\text{'}}\left(e\right)=0$

Hence,

$y^{\text{'}}\left(x\right)=\frac{1}{x}(-A\sqrt{\lambda }\sin (\sqrt{\lambda }\ln x)+B\sqrt{\lambda }\cos (\sqrt{\lambda }\ln x\left)\right)$

Then,

$y^{\text{'}}\left(e\right)=\frac{1}{e}(-A\sqrt{\lambda }\sin (\sqrt{\lambda }\ln e)+B\sqrt{\lambda }\cos (\sqrt{\lambda }\ln e\left)\right)$

$$\begin{gathered} =\frac{1}{e}(-A\sqrt{\lambda }\sin (\sqrt{\lambda })+B\sqrt{\lambda }\cos (\sqrt{\lambda }\left)\right)\backslash \mathrm{hfill} \\ =0 \end{gathered}$$
Then,

$-A\sin \left(\sqrt{\lambda }\right)+B\cos \left(\sqrt{\lambda }\right)=0$

Also,

localid="1668509477642" $$\begin{gathered} y\left(1\right)=A\cos \left(\sqrt{\lambda }\ln 1\right)+B\sin \left(\sqrt{\lambda }\ln 1\right) \\ =A \\ =0 \end{gathered}$$

Hence,

$0=\cos \left(\sqrt{\lambda }\right)$

Then,

$\sqrt{\lambda }=\frac{(2n-1)\pi }{2},\forall n=1,3\dots$

Thus, the eigenfunctions set that solves the eigenvalue problem is:

$y_n\left(x\right)=\sin \left(\frac{(2n-1)\pi }{2}\ln x\right)$