Question

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

$xy\text{'}\text{'}+3y\text{'}+x^{3}y=0$

Short answer

The general solutions of the given differential equation are

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

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 usingFrobinous 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 Y_n(x) 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$xy\text{'}\text{'}+3y\text{'}+x^{3}y=0$, 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,

$$\begin{gathered} \frac{x}{x}y\text{'}\text{'}+\frac{3x}{x}y\text{'}+\frac{x^3}{x}y=0 \\ y\text{'}\text{'}+\frac{3}{x}y\text{'}+\left(x^2+\frac{0}{x^2}\right)y=0........2 \end{gathered}$$

Find the value of constants.

Compare the equations (1) and (2).

Solve for a:

$$\begin{gathered} 1-2a=3 \\ a=-1 \end{gathered}$$

Solve for:

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

Solve for:

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

Solve for:

$$\begin{gathered} a^2-p^2{c}^2=0 \\ {p}_1=\frac{1}{2},p_2=\frac{-1}{2} \end{gathered}$$

Obtain the general solution.

There are two series which are linearly independent.

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

The general solution by using superposition principle is,

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

There is another general solution obtained from the Bessel’s equation of second order. (i.e.) $y=x^{-1}\left(C_1{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{Y}_{1/2}\left(\frac{1}{2}{x}^2\right)\right)$.

Hence, the general solutions are $y=x^{-1}\left(C_1{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{J}_{-1/2}\left(\frac{1}{2}{x}^2\right)\right)andy=x^{-1}\left(C_1{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{Y}_{1/2}\left(\frac{1}{2}{x}^2\right)\right)$and .