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.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

The general solution of the given differential equation is .$\mathrm{y}=\mathrm{Asinx}+\mathrm{Bcosx}$

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$n$.A Bessel equation of the first kind,indicated by$J_n\left(x\right)$, 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$\mathit{x}\mathbf{=}\mathbf{0}$, this solution is regular. The second solution, which is singular at$\mathit{x}\mathbf{=}\mathbf{0}$, 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$\mathrm{y}\text{'}\text{'}+\mathrm{y}=0$.

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,

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

Find the value of constants.

Compare the equations (1) and (2).

Solve for$\mathrm{a}$ :

Solve for $\mathrm{c}$:

$\begin{array}{c}2\mathrm{c}-2=0\\ \mathrm{c}=1\end{array}$

Solve for $\mathrm{b}$:

$\begin{array}{c}{\mathrm{b}}^2{\mathrm{c}}^2=1\\ \mathrm{b}=1\end{array}$

Solve for $\mathrm{p}$:

Obtain the general solution

There are two series which are linearly independent.

$\begin{array}{c}{\mathrm{y}}_1={\mathrm{x}}^{1/2}{\mathrm{J}}_{1/2}\left(\mathrm{x}\right)\\ {\mathrm{y}}_2={\mathrm{x}}^{1/2}{\mathrm{J}}_{-1/2}\left(\mathrm{x}\right)\end{array}$

The general solution by using superposition principle is,

$\begin{array}{c}\mathrm{y}={\mathrm{C}}_1{\mathrm{x}}^{1/2}{\mathrm{J}}_{1/2}\left(\mathrm{x}\right)+{\mathrm{C}}_2{\mathrm{x}}^{1/2}{\mathrm{J}}_{-1/2}\left(\mathrm{x}\right)\\ ={\mathrm{x}}^{1/2}\left[{\mathrm{C}}_1{\mathrm{J}}_{1/2}\left(\mathrm{x}\right)+{\mathrm{C}}_2{\mathrm{J}}_{-1/2}\left(\mathrm{x}\right)\right]\\ ={\mathrm{x}}^{1/2}\left[{\mathrm{C}}_1\sqrt{\frac{2}{\mathrm{\pi x}}}\mathrm{sinx}+{\mathrm{C}}_2\sqrt{\frac{2}{\mathrm{\pi x}}}\mathrm{cosx}\right]\\ ={\mathrm{x}}^{1/2}\left[{\mathrm{C}}_1{\mathrm{x}}^{-1/2}\sqrt{\frac{2}{\mathrm{\pi }}}\mathrm{sinx}+{\mathrm{C}}_2{\mathrm{x}}^{-1/2}\sqrt{\frac{2}{\mathrm{\pi }}}\mathrm{cosx}\right]\\ ={\mathrm{C}}_1\sqrt{\frac{2}{\mathrm{\pi }}}\mathrm{sinx}+{\mathrm{C}}_2\sqrt{\frac{2}{\mathrm{\pi }}}\mathrm{cosx}\\ \mathrm{y}=\mathrm{Asinx}+\mathrm{Bcosx}\end{array}$

Where $\mathrm{A}={\mathrm{C}}_1\sqrt{\frac{2}{\mathrm{\pi }}}$and $\mathrm{B}={\mathrm{C}}_2\sqrt{\frac{2}{\mathrm{\pi }}}$.