Question

In Problems 15–24, x = 0is a regular singular point of the given dif-

differential 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$(0,\infty )$.

$3xy\text{'}\text{'}+\left(2-x\right)y\text{'}-y=0$

Short answer

Therefore, the solution is:

$y\left(x\right)=C_1\left(1+\frac{1}{2}x+\frac{1}{5·2}{x}^2+\frac{1}{8·5·2}{x}^3+\cdots \right)+C_2{x}^{\frac{1}{3}}\left(1+\frac{1}{3}x+\frac{1}{3^2·2}{x}^2+\frac{1}{3^3·3!}{x}^3+\cdots \right)$

Step-by-step solution

Given Information.

The given problem is:

$3xy\text{'}\text{'}+(2-x)y\text{'}-y=0$

Use Differentiation.

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

Differentiate the above equation, 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 equations in the initial differential equation, we get:

$$\begin{gathered} 3x\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-2}+(2-x)\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r-1}-\sum _{n=0}^{\infty }{c}_n{x}^{n+r}=0 \\ 3\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-1}+2\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r-1}-\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r}-\sum _{n=0}^{\infty }{c}_n{x}^{n+r}=0 \\ 3\sum _{n=0}^{\infty }(n+r)(n+r-1)+2(n+r)]c_n{x}^{n+r-1}-\sum _{n=0}^{\infty }(n+r+1)c_n{x}^{n+r}=0 \end{gathered}$$

$x^r\left(r\right(3r-1)c_0{x}^{-1}+\sum _{n=0}^{\infty }(n+r)(3n+3r-1)]c_n{x}^{n-1}-\sum _{n=0}^{\infty }(n+r+1)c_n{x}^n)=0$

For the first summation . For the second summation, we . Hence,

$$\begin{gathered} x^r\left(r\right(3r-1)c_0{x}^{-1}+\sum _{n=0}^{\infty }(n+r)(3n+3r-1)]c_n{x}^{n-1}-\sum _{n=0}^{\infty }(n+r+1)c_n{x}^n)=0 \\ {x}^r(r(3r-1)c_0{x}^{-1}+\sum _{n=0}^{\infty }(n+r)(3n+3r-1\left)\right]c_n{x}^{n-1}-(n+r+1)c_n{x}^n)=0 \end{gathered}$$

 Step 4: Identity Property

$r(3r-1)c_0=0\Rightarrow r(3r-1)=0\Rightarrow r_1=0 \& r_2=\frac{1}{3}$

$$\begin{gathered} (k+1+r)(3k+3r+2)c_{k+1}-(k+r+1)c_k=0 \\ {c}_{k+1}=\frac{(k+r+1)c_k}{(k+1+r)(3k+3r+2)} \\ {c}_{k+1}=\frac{c_k}{3k+3r+2} \end{gathered}$$

For r₁= 0 we have

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

For , k = 0

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

k = 1

$$\begin{gathered} \Rightarrow c_2=\frac{c_1}{5} \\ =\frac{c_0}{5·2} \end{gathered}$$

For k = 2

$$\begin{gathered} c_3=\frac{c_2}{8} \\ =\frac{c_0}{8·5·2} \end{gathered}$$

For r₂ = 1/3 we have

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

For, k = 0

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

For ; k = 1

$$\begin{gathered} c_2=\frac{c_1}{6} \\ =\frac{c_0}{3^2·2} \end{gathered}$$

For , k = 2

$$\begin{gathered} c_3=\frac{c_2}{9} \\ =\frac{c_0}{3^3·3!} \end{gathered}$$