Question

In Problems 15–24, x = 0is a regular singular point of the givendifferential equation. Show that the indicial roots of the singularity do not differ by an integer. Use the method of Frobenius to obtain two linearly independent series solutions about x= 0. Form the general solution on $\left(0,\infty \right)$.

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

Short answer

Therefore, the solution is:

$y\left(x\right)=C_1{x}^{\frac{1}{3}}\left(1+\frac{3}{2}x+\frac{3^2}{5\times 2^2}{x}^2+\frac{3^3}{5\times 3\times 2^5}{x}^3+L\right)+C_2{x}^{\frac{2}{3}}\left(1+\frac{3}{2^2}x+\frac{3^2}{7\times 2^3}{x}^2+\frac{3^3}{7\times 5\times 3\times 2^4}{x}^3+L\right)$

Step-by-step solution

Given Information.

The given differential equation is

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

Use Differentiation

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

Differentiate (*) we get

$y\text{'}=\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r-1} \dots \left(1\right)$

$y\text{'}\text{'}=\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-2} \dots \left(2\right)$

Substitute the equation.

Replacing the first and second equation in the initial differential equation, we get;

$$\begin{gathered} x^2\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-2}-\left(x-\frac{2}{9}\right)\sum _{n=0}^{\infty }{c}_n{x}^{n+r}=0 \\ \sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r}-\sum _{n=0}^{\infty }{c}_n{x}^{n+r+1}+\frac{2}{9}\sum _{n=0}^{\infty }{c}_n{x}^{n+r}=0 \\ \sum _{n=0}^{\infty }\left((n+r)(n+r-1)+\frac{2}{9}\right)c_n{x}^{n+r}-\sum _{n=0}^{\infty }{c}_n{x}^{n+r+1}=0 \\ {x}^r\left|r(r-1)c_0+\frac{2}{9}{c}_0+\sum _{n=0}^{\infty }\left((n+r)(n+r-1)+\frac{2}{9}\right)c_n{x}^n-\sum _{n=0}^{\infty }{c}_n{x}^{n+1}\right|=0 \end{gathered}$$

For the first summation k = n -1. For the second summation, we have k = n. Hence,

$$\begin{gathered} x^r\left|r(r-1)c_0+\frac{2}{9}{c}_0+\sum _{k=0}^{\infty }\left((k+1+r)(k+r)+\frac{2}{9}\right)c_{k+1}{x}^{k+1}-\sum _{k=0}^{\infty }{c}_k{x}^{k+1}\right|=0 \\ {x}^r\left|r(r-1)c_0+\frac{2}{9}{c}_0+\sum _{k=0}^{\infty }\left((k+1+r)(k+r)+\frac{2}{9}\right)c_{k+1}-c_k\right|=0 \end{gathered}$$

Identity Property

$$\begin{gathered} r(r-1)c_0+\frac{2}{9}{c}_0=0 \\ r(r-1)+\frac{2}{9}=0 \\ {r}^2-r+\frac{2}{9}=0\Rightarrow r_1=\frac{1}{3} \& r_2=\frac{2}{3} \end{gathered}$$

Now,

$$\begin{gathered} \left((k+1+r)(k+r)+\frac{2}{9}\right)c_{k+1}-c_k=0 \\ {c}_{k+1}=\frac{c_k}{(k+1+r)(k+r)+\frac{2}{9}} \end{gathered}$$

For r₁=0 ,we have:

$c_{k+1}=\frac{3c_k}{(3k+2)(k+1)},k=0,1,2\dots$

If, k = 0 then:

$\Rightarrow c_1=\frac{3c_0}{2}$

If, k = 1 then:

$$\begin{gathered} c_2=\frac{3c_1}{10} \\ =\frac{3^2{c}_0}{5·2^2} \end{gathered}$$

If, k = 2 then:

$$\begin{gathered} c_3=\frac{3c_2}{24} \\ =\frac{3^3{c}_0}{5·3·2^5} \end{gathered}$$

For r₂=2/3 we have;

$c_{k+1}=-\frac{3c_k}{(3k+4)(k+1)},k=0,1,2\dots$

If, k = 0 then:

$c_1=\frac{3c_0}{2^2}$

If, k = 1 then:

$$\begin{gathered} c_2=\frac{3c_1}{14} \\ =\frac{3^2{c}_0}{7·2^3} \end{gathered}$$

If, k = 2 then:

$$\begin{gathered} c_3=\frac{3c_2}{30} \\ =\frac{3^3{c}_0}{7·5·3·2^4} \end{gathered}$$

For the indicial root as r₁ = 0 we get the solution as:

$$\begin{gathered} y_1\left(x\right)=x^{\frac{1}{3}}\left(c_0+\frac{3}{2}{c}_{0}x+\frac{3^2}{5·2^2}{c}_0{x}^2+\frac{3^3}{5·3·2^5}{c}_0{x}^3+\cdots \right) \\ =C_1{x}^{\frac{1}{3}}\left(1+\frac{3}{2}x+\frac{3^2}{5·2^2}{x}^2+\frac{3^3}{5·3·2^5}{x}^3+\cdots \right) \end{gathered}$$

For the indicial root as r₂= 3/2 we get the solution as;

$$\begin{gathered} y_2\left(x\right)=x^{\frac{2}{3}}\left(c_0+\frac{3}{2^2}{c}_{0}x+\frac{3^2}{7·2^3}{c}_0{x}^2+\frac{3^3}{7·5·3·2^4}{c}_0{x}^3+\cdots \right) \\ =C_2{x}^{\frac{2}{3}}\left(1+\frac{3}{2^2}x+\frac{3^2}{7·2^3}{x}^2+\frac{3^3}{7·5·3·2^4}{x}^3+\cdots \right) \end{gathered}$$

Thus, the solution is:

$$\begin{gathered} y\left(x\right)=y_1\left(x\right)+y_2\left(x\right) \\ =C_1{x}^{\frac{1}{3}}\left(1+\frac{3}{2}x+\frac{3^2}{5·2^2}{x}^2+\frac{3^3}{5·3·2^5}{x}^3+\cdots \right)+C_2{x}^{\frac{2}{3}}\left(1+\frac{3}{2^2}x+\frac{3^2}{7·2^3}{x}^2+\frac{3^3}{7·5·3·2^4}{x}^3+\cdots \right) \end{gathered}$$