Question

Bessel’s Equation

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

1.

$x^{2}y\text{'}\text{'}+x{y}^{\text{'}}+\left(x^2-\frac{1}{9}\right)y=0$

Short answer

The general solution of the given differential equation is

$y=c_1\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(4/3+n\right)}{\left(\frac{x}{2}\right)}^{2n+1/3}+c_2\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(2/3+n\right)}{\left(\frac{x}{2}\right)}^{2n-1/3}$

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

$x^{2}y\text{'}\text{'}+x{y}^{\text{'}}+\left(x^2-\frac{1}{9}\right)y=0$

Use Forbinous methodto solve this differential equation

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

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:

$J_V\left(X\right)=\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(1+v+n\right)}{\left(\frac{x}{2}\right)}^{2n+v}$

$J_{-V}\left(X\right)=\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(1-v+n\right)}{\left(\frac{x}{2}\right)}^{2n-v}$

The two roots are:

$v^2=\frac{1}{9}$

$\Rightarrow v=\frac{1}{3}$and $v=-\frac{1}{3}$

Substituting the values of v:

$$\begin{gathered} J_{1/3}\left(X\right)=\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(1+1/3+n\right)}{\left(\frac{x}{2}\right)}^{2n+1/3} \\ =\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(4/3+n\right)}{\left(\frac{x}{2}\right)}^{2n+1/3} \\ {J}_{-1/3}\left(X\right)=\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(1-1/3+n\right)}{\left(\frac{x}{2}\right)}^{2n-1/3} \\ =\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(2/3+n\right)}{\left(\frac{x}{2}\right)}^{2n-1/3} \end{gathered}$$

By using the Superposition Principle, $y=C_1{J}_{1/3}\left(X\right)+C_2{J}_{-1/3}\left(X\right)$

$y=c_1\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(4/3+n\right)}{\left(\frac{x}{2}\right)}^{2n+1/3}+c_2\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(2/3+n\right)}{\left(\frac{x}{2}\right)}^{2n-1/3}$

Where C₁ and are C₂ arbitrary constants

Therefore, the general solution of the given differential equation

$x^{2}y\text{'}\text{'}+x{y}^{\text{'}}+\left(x^2-\frac{1}{9}\right)y=0$on $\left(0,\infty \right)$is

$y=c_1\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(4/3+n\right)}{\left(\frac{x}{2}\right)}^{2n+1/3}+c_2\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\Gamma \left(2/3+n\right)}{\left(\frac{x}{2}\right)}^{2n-1/3}$