Question

In problem 9 and 10 use (18) to find the general solution of the given differential equation on

$\left(0,\infty \right)$

Question 10:

${\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{\text{'}}\mathbf{-}\mathbf{(}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{64}\mathbf{)}\mathbf{=}\mathbf{0}$

Short answer

The differential equation of the given equation is

$y=C_1{I}_8\left(\sqrt{2}x\right)+c_2{k}_8\left(\sqrt{2}x\right)$

Step-by-step solution

Step 1: Definition of frobenius method

To simply the given equation, we have to use the Forbenius method.

The Frobenius technique may be used to find a power series solution to a differential equation if p(z) and q(z) are analytic at 0 or if both of their limits at 0 are analytic elsewhere (and are finite).

We have

$x^{2}y\text{'}\text{'}+xy\text{'}-(2x^2+64)=0$

Our aim is to find the differential equation using the “Forbenius method”

t = iax

Modified based function of order V

Here, we will use the Forbenius form and solve the equation.

$t^2\frac{d^{2}y}{d{t}^2}+t\frac{dy}{dt}+(t^2-v^2)y=0$

Forbinous form

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

Now we will consider the two roots $$\begin{gathered} r_1=v \\ {r}_2=-v \end{gathered}$$ in the recurrence relation.

Now here, we will used Modified based function of order V

$$\begin{gathered} I_v\left(ax\right)=i^{-v}{J}_v\left(iax\right) \\ {I}_{-v}\left(ax\right)=i^v{J}_{-v}\left(iax\right) \end{gathered}$$

Where,

$$\begin{gathered} J_v\left(x\right)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n!\Gamma (1+v+n)}{\left(\frac{x}{2}\right)}^{2n-v} \\ {J}_v\left(x\right)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n!\Gamma (1-v+n)}{\left(\frac{x}{2}\right)}^{2n-v} \end{gathered}$$

Step 3: Use Superposition principle for finding equation

Now, we will find the linear independent for $v\ne$interger

$$\begin{gathered} {\alpha }^2=2 \\ \alpha =\sqrt{2} \end{gathered}$$

$$\begin{gathered} r^2=64 \\ {r}_1=8 \\ {r}_2=-8 \end{gathered}$$

The two solution are

$$\begin{gathered} I_8\left(\sqrt{2}x\right)=i^{-8}{J}_8\left(\sqrt{2}i8\right) \\ {J}_{-8}\left(4x\right)=i^8{J}_{-8}\left(\sqrt{2}i8\right) \end{gathered}$$

By using Superposition principle, the general solution is

$y=C_1{I}_8\left(\sqrt{2}x\right)+c_2{k}_8\left(\sqrt{2}x\right)$

Therefore, the differential equation is

$y=C_1{I}_8\left(\sqrt{2}x\right)+c_2{k}_8\left(\sqrt{2}x\right)$