Question

Use the indicated change of variables to find the general solution of the given differential equations on$\left(0,\infty \right)$

${\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\left({\alpha }^2{x}^2-{\nu }^2+\frac{1}{4}\right)\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{=}\sqrt{\mathbf{x}}\mathbf{\nu }\left(x\right)$

Short answer

The general solution of the given differential equation using the Forbinous Method is

$y\left(x\right)=x^{\frac{1}{2}}{C}_1{J}_{\nu }\left(\alpha x\right)+C_2{Y}_{\nu }\left(\alpha x\right)$

Step-by-step solution

Definition of Forbinous method for differential equations.

The Forbinous method is a method for finding an infinite series solution to a second-order ordinary differential equation with and near the regular singular point.

We have the differential equation as,

$x^2\frac{d^{2}y}{d{x}^2}+\left({\alpha }^2{x}^2-{\nu }^2+\frac{1}{4}\right)y=0andy=x^{\frac{1}{2}}\nu \left(x\right)\to \left(1\right)$

The goal is to represent the given differential equation in terms of v rather than y. And then solve for v. We apply the indicated variable change as follows:

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}{x}^{1/2}\nu \left(x\right) \\ \frac{dy}{dx}=\frac{d}{dx}{x}^{1/2}\nu \left(x\right)+x^{1/2}\frac{d}{dx}\nu \left(x\right) \\ \frac{dy}{dx}=\frac{1}{2}{x}^{-1/2}\nu \left(x\right)+x^{1/2}{\nu }^{\text{'}}\left(x\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\frac{1}{2}{x}^{-1/2}\nu \left(x\right)+x^{1/2}{\nu }^{\text{'}}\left(x\right) \end{gathered}$$

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\frac{1}{2}{x}^{-1/2}\nu \left(x\right)+\frac{d}{dx}{x}^{1/2}{\nu }^{\text{'}}\left(x\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\frac{1}{2}{x}^{-1/2}\nu \left(x\right)+\frac{1}{2}{x}^{-1/2}\frac{d}{dx}\nu \left(x\right)+\frac{d}{dx}{x}^{1/2}{\nu }^{\text{'}}\left(x\right)+x^{1/2}\frac{d}{dx}{\nu }^{\text{'}}\left(x\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{-1}{4}{x}^{-3/2}\nu \left(x\right)+\frac{1}{2}{x}^{-1/2}{\nu }^{\text{'}}\left(x\right)+\frac{1}{2}{x}^{-1/2}{\nu }^{\text{'}}\left(x\right)+x^{1/2}{\nu }^{\text{'}\text{'}}\left(x\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{-1}{4}{x}^{-3/2}\nu \left(x\right)+x^{-1/2}{\nu }^{\text{'}}\left(x\right)+x^{1/2}{\nu }^{\text{'}\text{'}}\left(x\right)\to \left(2\right) \end{gathered}$$

Substituting the equation in differential equation

We substitute the and into the given differential equation, yields,

$$\begin{gathered} x^2\frac{-1}{4}{x}^{-3/2}\nu \left(x\right)+x^{-1/2}{\nu }^{\text{'}}\left(x\right)+x^{1/2}{\nu }^{\text{'}\text{'}}\left(x\right)+\left({\alpha }^2{x}^2-{\nu }^2+\frac{1}{4}\right)x^{1/2}\nu \left(x\right)=0 \\ \frac{-1}{4}{x}^{1/2}\nu \left(x\right)+x^{3/2}{\nu }^{\text{'}}\left(x\right)+x^{5/2}{\nu }^{\text{'}\text{'}}\left(x\right)+\left({\alpha }^2{x}^{5/2}-x^{1/2}{\nu }^2+\frac{1}{4}{x}^{1/2}\right)\nu \left(x\right)=0 \\ \frac{-1}{4}{x}^{1/2}\nu \left(x\right)+x^{3/2}{\nu }^{\text{'}}\left(x\right)+x^{5/2}{\nu }^{\text{'}\text{'}}\left(x\right)+\left({\alpha }^2{x}^{5/2}-x^{1/2}{\nu }^2\right)\nu \left(x\right)+\frac{1}{4}{x}^{1/2}\nu \left(x\right)=0 \\ {x}^{3/2}{\nu }^{\text{'}}\left(x\right)+x^{5/2}{\nu }^{\text{'}\text{'}}\left(x\right)+\left({\alpha }^2{x}^{5/2}-x^{1/2}{\nu }^2\right)\nu \left(x\right)=0 \end{gathered}$$

Assuming the form of the general solution

We multiply by x^-1/2, yields,

$$\begin{gathered} x^{3/2-1/2}{\nu }^{\text{'}}\left(x\right)+x^{5/2-1/2}{{\nu }^{\text{'}}}^{\text{'}}\left(x\right)+\left({\alpha }^2{x}^{5/2-1/2}-x^{1/2-1/2}{\nu }^2\right)\nu \left(x\right)=0 \\ x{\nu }^{\text{'}}\left(x\right)+x^2{{\nu }^{\text{'}}}^{\text{'}}\left(x\right)+\left({\alpha }^2{x}^2-{\nu }^2\right)\nu \left(x\right)=0 \\ {x}^2{\nu }^{\text{'}\text{'}}\left(x\right)+x{\nu }^{\text{'}}\left(x\right)+\left({\alpha }^2{x}^2-{\nu }^2\right)\nu \left(x\right)=0 \end{gathered}$$

Which has a regular singular point at x = 0 and has the form

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

Our aim is to find the general solution for the given differential equation . We can use Forbinous Method to solve this differential equation.

First we need change the variable of differential equation to t instead of x,

t = ax

And you will end up with the

$t^2\frac{d^2\nu }{d{t}^2}+t\frac{d\nu }{dt}+\left(t^2-p^2\right)\nu =0$

Assume the solution, according to Forbinous, to be in the form,

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

Using Bessel Function of First Kind and Second Kind

Find the first and second derivative, substitute it into the given differential equation and you would end up with two roots, which is r₁ = 1 and r₂ =2 . Then you can find the two series solution for each case, based on the recurrence relation. Then you will identify the solution to have the form of Bessel Function of First Kind with a little modification and the series solutions are,

$$\begin{gathered} {\nu }_1\left(x\right)=J_{\nu }\left(\alpha x\right) \\ {\nu }_2\left(x\right)=J_{-\nu }\left(\alpha x\right) \end{gathered}$$

And you will find that the two solutions are linearly independent for integer. For , we can obtain another solution, which is linearly independent with , using Bessel Function of Second Kind, which is,

$$\begin{gathered} {\nu }_3\left(x\right)=Y_{\nu }\left(\alpha x\right) \\ =\frac{\cos \nu \pi J_{\nu }\left(\alpha x\right)-J_{-\nu }\left(\alpha x\right)}{\sin \nu \pi }\to \left(3\right) \end{gathered}$$

Generalized equation of the differential equation

By using the Superposition Principle, the general solution is

$$\begin{gathered} \nu \left(x\right)=C_1{J}_{\nu }\left(\alpha x\right)+C_2{Y}_{\nu }\left(\alpha x\right) \\ {x}^{-1/2}y\left(x\right)=C_1{J}_{\nu }\left(\alpha x\right)+C_2{Y}_{\nu }\left(\alpha x\right) \\ y\left(x\right)=x^{1/2}{C}_1{J}_{\nu }\left(\alpha x\right)+C_2{Y}_{\nu }\left(\alpha x\right) \end{gathered}$$

Where C1 and C2 are arbitrary constants.

Therefore the general solution of the given differential equation is

$y\left(x\right)=x^{1/2}{C}_1{J}_{\nu }\left(\alpha x\right)+C_2{Y}_{\nu }\left(\alpha x\right)$