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 $\mathbf{0}\mathbf{,}\mathbf{\infty }$.$\mathbf{2}\mathbf{xy}\mathbf{"}\mathbf{-}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Short answer

Therefore, the solution is:

$y\left(x\right)=C_1\left[1+2x-2x^2+\frac{2^2}{3\times 3!}{x}^3-....\right]+C_2{x}^{\frac{3}{2}}\left[1-\frac{2}{5}x+\frac{2^2}{7\times 5\times 2}{x}^2-\frac{2^3}{9\times 7\times 5\times 3!}{x}^3+.....\right]$

Step-by-step solution

Step 1:Given Information.

The given problem is:

$2xy\text{'}\text{'}-y\text{'}+2y=0$

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;

$$\begin{gathered} 2x\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-2}-\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r-1}+2\sum _{n=0}^{\infty }{c}_n{x}^{n+r}=0 \\ 2\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-1}-\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r-1}+2\sum _{n=0}^{\infty }{c}_n{x}^{n+r}=0 \\ \sum _{n=0}^{\infty }\left[2(n+r)(n+r-1)c_n-c_n(n+r)\right]x^{n+r-1}+2\sum _{n=0}^{\infty }{c}_n{x}^{n+r}=0 \\ \left(2r^2-3r\right)c_0{x}^{r-1}+\sum _{n=1}^{\infty }\left[2(n+r)(n+r-1)c_n-c_n(n+r)\right]x^{n+r-1}+2\sum _{n=0}^{\infty }{c}_n{x}^{n+r}=0 \\ {x}^r\left[\left(2r^2-3r\right)c_0{x}^{-1}+\sum _{n=1}^{\infty }\left[2(n+r)(n+r-1)c_n-c_n(n+r)\right]x^{n-1}+2\sum _{n=0}^{\infty }{c}_n{x}^n\right]=0 \end{gathered}$$

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

$$\begin{gathered} x^r\left[\left(2r^2-3r\right)c_0{x}^{-1}+\sum _{k=0}^{\infty }\left[2(k+1+r)(k+r)c_{k+1}-c_{k+1}(k+1+r)\right]x^k+2\sum _{k=0}^{\infty }{c}_k{x}^k\right]=0 \\ {x}^r\left[\left(2r^2-3r\right)c_0{x}^{-1}+\sum _{k=0}^{\infty }\left[2(k+1+r)(k+r)c_{k+1}-c_{k+1}(k+1+r)+2c_k\right]x^k\right]=0 \end{gathered}$$

Identity Property.

$$\begin{gathered} \left(2r^2-3r\right)c_0=0 \\ r(2r-3)=0 \end{gathered}$$

Therefore,

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

For we have;

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

For k = 0,

$$\begin{gathered} c_1=-\frac{2c_0}{-1} \\ =2c_0 \end{gathered}$$

For, k = 1

$$\begin{gathered} c_2=-\frac{2c_1}{2} \\ =-2c_0 \end{gathered}$$

For, k = 2

$$\begin{gathered} c_3=-\frac{2c_2}{9} \\ =\frac{2^3{c}_0}{3\times 3!} \end{gathered}$$

For r₂ = 3/2 we have;

$c_{k+1}=-\frac{2c_k}{(2k+5)(k+1)},k=0,1,2\dots$

For, k = 0

$c_1=-\frac{2c_0}{5}$

For, k = 1

$$\begin{gathered} c_2=-\frac{2c_1}{14} \\ =\frac{2^2{c}_0}{7\times 5\times 2} \end{gathered}$$

For, k = 2

$$\begin{gathered} c_3=-\frac{2c_2}{27} \\ =-\frac{2^3{c}_0}{9\times 7\times 5\times 3!} \end{gathered}$$

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

$$\begin{gathered} y_1\left(x\right)=x^0\left(c_0+2c_{0}x-2c_0{x}^2+\frac{2^2}{3\times 3!}{c}_0{x}^3-\dots \right) \\ =C_1\left(1+2x-2x^2+\frac{2^2}{3\times 3!}{x}^3-\dots \right) \end{gathered}$$

For the indicial root as we get the solution as

$$\begin{gathered} y_2\left(x\right)=x^{\frac{3}{2}}\left(c_0-\frac{2}{5}{c}_{0}x+\frac{2^2}{7\times 5\times 2}{c}_0{x}^2+\frac{2^3}{9\times 7\times 5\times 3!}{c}_0{x}^3+....\right) \\ =C_2{x}^{\frac{3}{2}}\left(1-\frac{2}{5}x+\frac{2^2}{7\times 5\times 2}{x}^2-\frac{2^3}{9\times 7\times 5\times 3!}{x}^3+....\right) \end{gathered}$$

Thus,

$$\begin{gathered} y\left(x\right)=y_1\left(x\right)+y_2\left(x\right) \\ y\left(x\right)=C_1\left[1+2x-2x^2+\frac{2^2}{3\times 3!}{x}^3-....\right]+C_2{x}^{\frac{3}{2}}\left[1-\frac{2}{5}x+\frac{2^2}{7\times 5\times 2}{x}^2-\frac{2^3}{9\times 7\times 5\times 3!}{x}^3+.....\right] \end{gathered}$$