Question

Bessel’s Equation

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

$\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{xy}}^{\mathbf{\text{'}}}\mathbf{+}\left(4x^2-25\right)\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

The general solution of the given differential equation is

$\mathrm{y}={\mathrm{C}}_1{\mathrm{J}}_{5/2}\left(\mathrm{X}\right)+{\mathrm{C}}_2{\mathrm{J}}_{-5/2}\left(\mathrm{X}\right)$

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

$4x^{2}y\text{'}\text{'}+4x{y}^{\text{'}}+\left(4x^2-25\right)y=0$

$\Rightarrow \frac{4}{4}{x}^{2}y\text{'}\text{'}+\frac{4}{4}x{y}^{\text{'}}+\frac{\left(4x^2-25\right)}{4}y=0$

$\Rightarrow x^{2}y\text{'}\text{'}+x{y}^{\text{'}}+\left(x^2-\frac{25}{4}\right)y=0$

Use Forbinous method to 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 role="math" localid="1663959557915" $v=non-integer$

The two solutions are not linearly independent and one of the solution is constant multiple of the other, for $v=integer$.

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

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

$Y_v\left(x\right)=\frac{\cos v\pi J_v\left(x\right)-J_{-v}\left(x\right)}{\sin v\pi }$

The two roots for the given differential equation are:

$v^2=\frac{25}{4}$

v = 5/2 and v = -5/2

Substituting the values of v, the two linearly independent solutions are:

$$\begin{gathered} y_1\left(x\right)=J_{5/2}\left(x\right) \\ {y}_2\left(x\right)=J_{-5/2}\left(x\right) \end{gathered}$$

By using the Superposition Principle,

$y=C_1{J}_{5/2}\left(X\right)+C_2{J}_{-5/2}\left(X\right)$

Where C₁ and C₂ are arbitrary constants

Therefore, the general solution of the given differential equation

$4x^{2}y\text{'}\text{'}+4x{y}^{\text{'}}+\left(4x^2-25\right)y=0on\left(0,\infty \right)isy=C_1{J}_{5/2}\left(X\right)+C_2{J}_{-5/2}\left(X\right)$