Question

In Problems 13-20use (20)to find the general solution of the given differential equation on$\left(0,\infty \right)$.

$9x^{2}y\text{'}\text{'}+9xy\text{'}+\left(x^6-36\right)y=0$

Short answer

The general solutions of the given differential equation are

$y=C_1{J}_{2/3}\left(\frac{1}{9}{x}^3\right)+C_2{J}_{-2/3}\left(\frac{1}{9}{x}^3\right)andy=C_1{J}_{2/3}\left(\frac{1}{9}{x}^3\right)+C_2{Y}_{2/3}\left(\frac{1}{9}{x}^3\right)$

Step-by-step solution

Define Bessel’s equation.

Let the Bessel equation be ${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\left(x^2-n^2\right)\mathit{y}\mathbf{=}\mathbf{0}$. This equation has two linearly independent solutions for a fixed value of n. A Bessel equation of the first kind, indicated by J_n(x), is one of these solutions that may be derived using Frobinous approach.

$$\begin{gathered} {\mathit{y}}_{\mathbf{1}}\mathbf{=}{\mathit{x}}^{\mathbf{a}}{\mathit{J}}_{\mathbf{p}}\left(b{x}^c\right) \\ {\mathit{y}}_{\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{a}}{\mathit{J}}_{\mathbf{-}\mathbf{p}}\left(b{x}^c\right) \end{gathered}$$

At x = 0 , this solution is regular. The second solution, which is singular at x = 0, is represented by and is called a Bessel function of the second kind.

$y_3=x^a\left[\frac{cosp\pi J_p\left(b{x}^c\right)-J_{-p}\left(b{x}^c\right)}{sinp\pi }\right]$

Determine the general form of the Bessel’s equation.

Let the given differential equation be $9x^{2}y\text{'}\text{'}+9xy\text{'}+\left(x^6-36\right)y=0$, that has a singular point at x = 0.

The equation becomes in the following form:

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

That yields,

$$\begin{gathered} \frac{9x^2}{9x^2}y\text{'}\text{'}+\frac{9x}{9x^2}y\text{'}+\frac{x^6-36}{9x^2}y=0 \\ y\text{'}\text{'}+\frac{1}{x}y\text{'}+\left(\frac{1}{9}{x}^4-\frac{4}{x^2}\right)y=0 \end{gathered}$$

Find the value of constants.

Compare the equations (1) and (2).

Solve for a:

$$\begin{gathered} 1-2a=1 \\ a=0 \end{gathered}$$

Solve for c:

$$\begin{gathered} 2c-2=4 \\ c=3 \end{gathered}$$

Solve for b:

$$\begin{gathered} b^2{c}^2=\frac{1}{9} \\ b=\frac{1}{9} \end{gathered}$$

Solve for p:

$$\begin{gathered} a^2-p^2{c}^2=-4 \\ {p}_1=\frac{2}{3},p_2=\frac{-2}{3} \end{gathered}$$

Obtain the general solution.

There are two series which are linearly independent.

$$\begin{gathered} y_1=x^0{J}_{2/3}\left(\frac{1}{9}{x}^3\right) \\ =J_{2/3}\left(\frac{1}{9}{x}^3\right) \\ {y}_2=J_{-2/3}\left(\frac{1}{9}{x}^3\right) \end{gathered}$$

The general solution by using superposition principle is,

$y=C_1{J}_{2/3}\left(\frac{1}{9}{x}^3\right)+C_2{J}_{-2/3}\left(\frac{1}{9}{x}^3\right)$

There is another general solution obtained from the Bessel’s equation of second order. (i.e.)

$y=C_1{J}_{2/3}\left(\frac{1}{9}{x}^3\right)+C_2{Y}_{2/3}\left(\frac{1}{9}{x}^3\right)$