Question

Bessel’s Equation

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

13.$\mathbf{xy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

The general solution of the given differential equation is

$\mathrm{y}={\mathrm{x}}^{-1/2}\left[{\mathrm{C}}_1{\mathrm{J}}_1\left(4{\mathrm{x}}^{1/2}\right)+{\mathrm{C}}_2{\mathrm{Y}}_1\left(4{\mathrm{x}}^{1/2}\right)\right]$

Step-by-step solution

Bessel’s equation of order v

The special form of equation (20) encountered by Bessel eventually carried out a systematic study of the properties of the solutions because many differential equations fit into its form by appropriate choices of the parameters in the general equation.

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

We have the form in

$xy\text{'}\text{'}+2y\text{'}+4y=0$

Rewriting as equation (1),

$\Rightarrow \frac{x}{x}y\text{'}\text{'}+\frac{2}{x}y\text{'}+\frac{4}{x}y=0$

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

Use Forbinous methodto find the general solution of (1)

$y=\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

Step 2: Find the first and second derivative

$y=x^{-1/2}\left[C_1{J}_1\left(4x^{1/2}\right)+C_2{Y}_1\left(4x^{1/2}\right)\right]$Substituting the given differential equation ends up with two roots,

r₁ = p and r₂ = - p

Identify the solution to have the form of Bessel function of First kind of order p.

The two series solutions are:

$$\begin{gathered} y_1=x^a{J}_p\left(b{x}^c\right) \\ {y}_2=x^a{J}_{-p}\left(b{x}^c\right) \end{gathered}$$

which are linearly independent for p is not equal to integer

The two solutions are not linearly independent for p = integer.

Using Bessel function of Second kind,obtain another solution linearly independent with y₁,

$$\begin{gathered} y_3=x^a{Y}_b\left(b{x}^c\right) \\ =x^a\left[\frac{\cos p\pi J_p\left(b{x}^c\right)-J_{-p}\left(b{x}^c\right)}{\sin p\pi }\right]...........\left(3\right) \end{gathered}$$

Now, we find the values of constants a, b, c and p of the given differential equation. By

comparing (1) and (2), we get

$$\begin{gathered} 1-2a=2\Rightarrow a=\frac{-1}{2} \\ 2c-2=-1\Rightarrow c=\frac{1}{2} \\ {b}^2{c}^2=4\Rightarrow b=4 \end{gathered}$$

$a^2-p^2{c}^2=0,p=1andp=-2$

Therefore, we have two series solutions which are not linearly independent.

$$\begin{gathered} y_1=x^{-1/2}{J}_1\left(4x^{1/2}\right) \\ {y}_2=x^{-1/2}{J}_{-1}\left(4x^{1/2}\right) \end{gathered}$$

We can obtain another solution which is linearly independent with , using Bessel function of Second kind,which is

$y_3\left(x\right)=x^{-1/2}{Y}_1\left(4x^{1/2}\right)$

By using the Superposition Principle,the general solution is

$y=x^{-1/2}\left[C_1{J}_1\left(4x^{1/2}\right)+C_2{Y}_1\left(4x^{1/2}\right)\right]$

Where and are arbitrary constants

Therefore, the general solution of the given differential equation$xy\text{'}\text{'}+2y\text{'}+4y=0$ on is