Question

Bessel’s Equation

In Problems 1-6 use (1) to find the general solution of the given differential equation on $\left(0,\infty \right)$.

6. $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left[xy\text{'}\right]\mathbf{+}\left(x-\frac{4}{x}y\right)\mathbf{=}\mathbf{0}$

Short answer

The general solution of the given differential equation is

$\mathrm{y}={\mathrm{C}}_1{\mathrm{J}}_2\left(\mathrm{X}\right)+{\mathrm{C}}_2{\mathrm{Y}}_2\left(\mathrm{X}\right)$

Step-by-step solution

Bessel’s equation of order v

The special form of equation (1) encountered by Bessel eventually carried out a systematic study of the properties of the solutions of the general equation.

$x^{2}y\text{'}\text{'}+x{y}^{\text{'}}+\left(x^2-v^2\right)y=0$

We have the form in

$$\begin{gathered} \frac{d}{dx}\left[xy\text{'}\right]+\left(x-\frac{4}{x}y\right)=0 \\ \frac{d}{dx}\left[x\right]y\text{'}+x\frac{d}{dx}\left[y\text{'}\right]+\left(x-\frac{4}{x}y\right)=0 \\ y\text{'}+xy\text{'}\text{'}+\left(x-\frac{4}{x}y\right)=0 \end{gathered}$$

multiplying by on both sides,

$$\begin{gathered} \left[y\text{'}+xy\text{'}\text{'}+\left(x-\frac{4}{x}y\right)\right]\left(x\right)=0\left(x\right) \\ \Rightarrow x{y}^{\text{'}}+x^{2}y+\left(x^2-4\right)y=0 \end{gathered}$$

UseForbinous methodto solve this differential equation

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

Step 2: Find the first and second derivative

Substituting the given differential equation ends up with two roots,

r₁ = v and r₂ = -v

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

The two solutions are:

$$\begin{gathered} y_1\left(x\right)=J_v\left(x\right) \\ {J}_V\left(X\right)=\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(1+v+n\right)}{\left(\frac{x}{2}\right)}^{2n+v} \\ {y}_2\left(x\right)=J_{-v}\left(x\right) \\ {J}_{-V}\left(X\right)=\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(1-v+n\right)}{\left(\frac{x}{2}\right)}^{2n-v} \end{gathered}$$

which are linearly independent for v = non-integer

The two solutions are not linearly independent and constant multiple of each other, for v = integer.

Using Bessel function of Second kind,obtain another solution linearly independent with J_v(X) ,

$$\begin{gathered} y_3\left(x\right)=Y_v\left(x\right) \\ {Y}_v\left(x\right)=\frac{\cos v\pi J_v\left(x\right)-J_{-v}\left(x\right)}{\sin v\pi } \end{gathered}$$

For the given differential equation, the two roots are:

v²= 4

v = 2 and v= -2

The two solutions are:

$$\begin{gathered} y_1\left(x\right)=J_2\left(x\right) \\ {y}_2\left(x\right)=J_{-2}\left(x\right) \end{gathered}$$

which are not linearly independent.

We can obtain another solution, which is linearly independent with y₁, usingBessel function of Second kind.

$y_3\left(x\right)=Y_2\left(x\right)$

By using theSuperposition Principle, the general solution is

$y=C_1{J}_2\left(X\right)+C_2{Y}_2\left(X\right)$

Where C₁ and C₂ are arbitrary constants

Therefore, the general solution of the given differential equation $\frac{d}{dx}\left[xy\text{'}\right]+\left(x-\frac{4}{x}y\right)=0$ on $\left(0,\infty \right)$is $y=C_1{J}_2\left(X\right)+C_2{Y}_2\left(X\right)$