Question

In Problems 13 and 14, x= 0 is a regular singular point of the given differential equation. Use the general form of the indicial equation in (14) to ­nd the indicial roots of the singularity. Without solving, discuss the number of series solutions you would expect to ­nd using the method of Frobenius.

${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\left(\frac{5}{3}x+x^2\right)\mathit{y}\mathbf{\text{'}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

r₁=1/3 and r₂=-1. We have two solutions for the two values.

Step-by-step solution

To Find differential equation

We state, without a proof, that if we have a differential equation, with the following the standard form;

$y^{\text{'}\text{'}}+P\left(x\right)y^{\text{'}}+Q\left(x\right)=0$

which has a regular singular point at x=x₀, then we can use Forbenious Method to solve this differential equation. In order for the differential equation to have a regular singular point at x=x₀, two conditions must be satisfied;

$p\left(x\right)=(x-x_0)p\left(x\right)$must have a around that point.

$q\left(x\right)=(x-x_0)q\left(x\right)$must have a around that point.

If the two conditions are satisfied, then we can use the following form to solve the differential equation.

$y=\sum _{n=0}^{\infty }{c}_n{\left(x-x_0\right)}^{n+r}\dots \left(1\right)$

Now, we have

$x^{2}y\text{'}\text{'}+\left(\frac{5}{3}x+x^2\right)y^{\text{'}}-\frac{1}{3}y=0\dots \left(2\right)$

which has a regular singular point at $x=0$. We put (2) in the standard form.

$$\begin{gathered} y^{\text{'}\text{'}}+\frac{x\left(\frac{5}{3}+x\right)}{x^2}{y}^{\text{'}}-\frac{1}{3x^2}y=0 \\ {y}^{\text{'}\text{'}}+\frac{\backslash \mathrm{not}x\left(\frac{5}{3}+x\right)}{x^2}{y}^{\text{'}}-\frac{1}{3x^2}y=0 \\ {y}^{\text{'}\text{'}}+\underset{P\left(x\right)}{\underbrace{\frac{5/3+x}{x}}}{y}^{\text{'}}-\underset{Q\left(x\right)}{\underbrace{\frac{1}{3x^2}}}y=0 \end{gathered}$$

By using (l) and the Taylor series expansion for p(x) and q(x), we can find the indicial quadratic equation of r, upon which we will know, what is the final solution will look like. We will not go through proofs, regarding this situation. The quadratic equation, you would get, is

$r(r-1)+a_{0}r+b_0=0$

where a₀ and b₀ are the first term of Taylor series of q(x) and p(x), respectively. We write down the Taylor series expansion of the two functions as follows:

$$\begin{gathered} p\left(x\right)=xP\left(x\right) \\ =x\left(\frac{5/3+x}{x}\right) \\ =5/3+x \\ q\left(x\right)=x^{2}Q\left(x\right) \\ =x^2\left(\frac{-1}{3x^2}\right) \\ =\frac{-1}{3} \end{gathered}$$

The quadratic equation

Therefore, a₀=5/3 and b₀₌ -1/3. And the quadratic equation of is reduced to the following form.

$$\begin{gathered} r^2-r+\frac{5}{3}r-\frac{1}{3}=0 \\ {r}^2+\frac{2}{3}r-\frac{1}{3}=0 \\ 3r^2+2r-1=0 \\ \left(3r-1\right)\left(r+1\right)=0 \end{gathered}$$

Thus, we have two distinct roots.

$r_1=\frac{1}{3}and{r}_2=-1$

Since they don't differ by an integer. Therefore, we have two linearly independent solutions, resulting from using of Forbenius Method; one for r₁=1/3 and the other for r₂= -1

Final Answer

Since they don't differ by an integer. Therefore, we have two linearly independent solutions, resulting from using of Forbenius Method; one for r₁=1/3 and the other for r₂=-1.

r₁=1/3 and r₂=-1 . We have two solutions for the two values.