Question

In Problems $\mathbf{23}\mathbf{-}\mathbf{26}$first use$\left(20\right)$to express the general solution of the given differential equation in terms of Bessel functions. Then use$\mathbf{\left(}\mathbf{26}\mathbf{\right)}$and$\mathbf{\left(}\mathbf{27}\mathbf{\right)}$to express the general solution in terms of elementary functions.

$\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{16}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

The general solution of the given differential equation is $\mathrm{y}={\mathrm{x}}^{1/2}[\mathrm{Asin}2\mathrm{x}+\mathrm{Bcos}2\mathrm{x}]$.

Step-by-step solution

Define Bessel’s equation.

Let the Bessel equation be${\mathrm{x}}^2\mathrm{y}\text{'}\text{'}+\mathrm{xy}\text{'}+({\mathrm{x}}^2-{\mathrm{n}}^2)\mathrm{y}=0$. This equation hastwo linearly independent solutionsfor a fixed value of$\mathrm{n}$.A Bessel equation of the first kind,indicated by${\mathrm{J}}_{\mathrm{n}}\left(\mathrm{x}\right)$, is one of these solutions that may be derived usingFrobinous approach.

$\begin{array}{c}{\mathrm{y}}_1={\mathrm{x}}^{\mathrm{a}}{\mathrm{J}}_{\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)\\ {\mathrm{y}}_2={\mathrm{x}}^{\mathrm{a}}{\mathrm{J}}_{-\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)\end{array}$

At role="math" localid="1667981314426" $\mathrm{x}=0$, this solution is regular. The second solution, which is singular at$\mathrm{x}=0$, is represented by${\mathrm{Y}}_{\mathrm{n}}\left(\mathrm{x}\right)$and is called a Bessel function of the second kind.

${\mathrm{y}}_3={\mathrm{x}}^{\mathrm{a}}\left[\frac{{\mathrm{cosp\pi J}}_{\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)-{\mathrm{J}}_{-\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)}{\mathrm{sinp\pi }}\right]$

The elementary functions are,

$\begin{array}{c}{\mathrm{J}}_{1/2}\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\mathrm{sinx}\\ {\mathrm{J}}_{-1/2}\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\mathrm{cosx}\end{array}$

Determine the general form of the Bessel’s equation.

Let the given differential equation be$4x^{2}y\text{'}\text{'}-4xy\text{'}+(16x^2+3)y=0$.

The equation becomes in the following form:

$\mathrm{y}\text{'}\text{'}+\frac{1-2\mathrm{a}}{\mathrm{x}}\mathrm{y}\text{'}+\left({\mathrm{b}}^2{\mathrm{c}}^2{\mathrm{x}}^{2\mathrm{c}-2}+\frac{{\mathrm{a}}^2-{\mathrm{p}}^2{\mathrm{c}}^2}{{\mathrm{x}}^2}\right)\mathrm{y}=0$… (1)

That yields,

$\frac{4{\mathrm{x}}^2}{4{\mathrm{x}}^2}\mathrm{y}\text{'}\text{'}-\frac{4\mathrm{x}}{4{\mathrm{x}}^2}\mathrm{y}\text{'}+\frac{(16{\mathrm{x}}^2+3)}{4{\mathrm{x}}^2}\mathrm{y}=0$

$\mathrm{y}\text{'}\text{'}-\frac{1}{\mathrm{x}}\mathrm{y}\text{'}+\left(4{\mathrm{x}}^0+\frac{3/4}{{\mathrm{x}}^2}\right)\mathrm{y}=0$ … (2)

Find the value of constants.

Compare the equations (1) and (2).

Solve for $a$:

Solve for $c$:

Solve for $b$:

Solve for$p$ :

Obtain the general solution

There are two series which are linearly independent.

The general solution by using superposition principle is,

$\begin{array}{c}\mathrm{y}={\mathrm{C}}_1{\mathrm{x}}^{1/2}\sqrt{\frac{1}{\mathrm{\pi }}}\sin 2\mathrm{x}+{\mathrm{C}}_2{\mathrm{x}}^{1/2}\sqrt{\frac{1}{\mathrm{\pi }}}\cos 2\mathrm{x}\\ \mathrm{y}={\mathrm{x}}^{1/2}[\mathrm{Asin}2\mathrm{x}+\mathrm{Bcos}2\mathrm{x}]\end{array}$

Where $\mathrm{A}={\mathrm{C}}_1\sqrt{\frac{1}{\mathrm{\pi }}}$and $\mathrm{B}={\mathrm{C}}_2\sqrt{\frac{1}{\mathrm{\pi }}}$.