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$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$.

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

Short answer

The solution is:

$y=C_{1}x\left[1-\frac{1}{10}{x}^2+\frac{1}{180}{x}^4+\dots \right]+C_2{x}^{1/2}\left[1-\frac{1}{6}{x}^2+\frac{1}{168}{x}^4+\dots \right]$

Step-by-step solution

Given Information.

The given problem is

$2x^{2}y\text{'}\text{'}-xy\text{'}+\left(x^2+1\right)y=0$

Use Differentiation.

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

Differentiate the above equation, we get:

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

$y\text{'}\text{'}=\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2} \dots \left(3\right)$

Substitute the equation.

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=2x^2\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2}-x\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r-1}+\left(x^2+1\right)\sum _{n=0}^{\infty }{c}_n{x}^{n+r} \\ =\underset{a_0\to x^r}{\underbrace{\sum _{n=0}^{\infty }2c_n(n+r)(n+r-1)x^{n+r}}}-\underset{a_0\to x^r}{\underbrace{\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r}}}+\underset{a_0\to x^{r+2}}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+r+2}}}+\underset{a_0\to x^r}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+r}}} \end{gathered}$$

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left[2c_{0}r\right(r-1)x^r+2c_1(r+1\left)\right(r)x^{r+1}+\underset{a_2\to x^{r+2}}{\underbrace{\left.\sum _{n=2}^{\infty }2c_n(n+r)(n+r-1)x^{n+r}\right]}}-[c_{0}r{x}^r+c_1(r+1)x^{r+1} \\ +\underset{a_2\to x^{r+2}}{\underbrace{\sum _{n=2}^{\infty }{c}_n(n+r)x^{n+r}}}]+\underset{a_0\to x^{r+2}}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+r+2}}}+[c_0{x}^r+c_1{x}^{r+1}+\underset{a_2\to x^{r+2}}{\underbrace{\left.\sum _{n=2}^{\infty }{c}_n{x}^{n+r}\right]}} \end{gathered}$$

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left[2c_{0}r(r-1)x^r+2c_1(r+1)\left(r\right)x^{r+1}+\sum _{n=2}^{\infty }2c_n(n+r)(n+r-1)x^{n+r}\right] \\ -\left[c_{0}r{x}^r+c_1(r+1)x^{r+1}+\sum _{n=2}^{\infty }{c}_n(n+r)x^{n+r}\right]+\sum _{n-2=0}^{\infty }{c}_{n-2}{x}^{(n-2)+r+2} \\ +\left[c_0{x}^r+c_1{x}^{r+1}+\sum _{n=2}^{\infty }{c}_n{x}^{n+r}\right] \\ =\left[2c_{0}r(r-1)x^r+2c_1(r+1)\left(r\right)x^{r+1}+\sum _{n=2}^{\infty }2c_n(n+r)(n+r-1)x^{n+r}\right] \\ -\left[c_{0}r{x}^r+c_1(r+1)x^{r+1}+\sum _{n=2}^{\infty }{c}_n(n+r)x^{n+r}\right] \\ +\sum _{n=2}^{\infty }{c}_{n-2}{x}^{n+r}+\left[c_0{x}^r+c_1{x}^{r+1}+\sum _{n=2}^{\infty }{c}_n{x}^{n+r}\right] \end{gathered}$$

Solving the above result further, we get:

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left[2r\right(r-1)-r+1]c_0{x}^r+\left[2\right(r+1\left)\right(r)-(r+1)+1]c_1{x}^{r+1} \\ +\sum _{n=2}^{\infty }\left[2c_n(n+r)(n+r-1)-c_n(n+r)+c_{n-2}+c_n\right]x^{n+r} \\ =\left[2r^2-2r-r+1\right]c_0{x}^r+\left[2r^2+2r-r-1+1\right]c_1{x}^{r+1} \\ +\sum _{n=2}^{\infty }\left[c_n\left[2\right(n+r\left)\right(n+r-1)-(n+r)+1]+c_{n-2}\right]x^{n+r} \\ =0 \end{gathered}$$

Identity Property

$$\begin{gathered} 2r^2-2r-r+1=0 \\ 2r^2-3r+1=0 \\ (2r-1)(r-1)=0 \\ {r}_1=1,r_2=\frac{1}{2} \end{gathered}$$

Now,

$c_1=0 \dots \left(4\right)$

$$\begin{gathered} c_n\left[2\right(n+r\left)\right(n+r-1)-(n+r)+1]+c_{n-2}=0 \\ {c}_n=\frac{-c_{n-2}}{\left[2\right(n+r\left)\right(n+r-1)-(n+r)+1]} \end{gathered}$$

For r₁ = 1

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^{n+1} \\ =x\sum _{n=0}^{\infty }{c}_n{x}^n \end{gathered}$$

$c_n=\frac{-c_{n-2}}{2(n+1)n-(n+1)+1} \cdots \left(5\right)$

For n = 2

$$\begin{gathered} c_2=\frac{-c_0}{2·3·2-(2+1)+1} \\ =\frac{-c_0}{10} \end{gathered}$$

For, n = 3

$$\begin{gathered} \backslash c_3=\frac{-c_1}{2\left(4\right)·3-(3+1)+1} \\ =0 \end{gathered}$$

For, n = 4

$$\begin{gathered} c_4=\frac{-c_2}{2·5·4-5+1} \\ =\frac{-\left(-c_0/10\right)}{36} \\ =\frac{c_0}{360} \end{gathered}$$

$$\begin{gathered} y_1=x\left[c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \right] \\ =x\left[c_0+0·x-\frac{1}{10}{c}_0{x}^2+0·x^3+\frac{1}{180}{c}_0{x}^4+\dots \right] \\ =c_{0}x\left[1-\frac{1}{10}{x}^2+\frac{1}{180}{x}^4+\dots \right] \dots \left(6\right) \end{gathered}$$

Now, we have:

For r₂= 1/2

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^{n+1/2} \\ =x^{1/2}\sum _{n=0}^{\infty }{c}_n{x}^n \end{gathered}$$

$c_n=\frac{-c_{n-2}}{2(n+1/2)(n-1/2)-(n+1/2)+1}, n=2,3,4\dots \dots \left(7\right)$

For, n = 2

$$\begin{gathered} c_2=\frac{-c_0}{2·5/2·3/2-5/2+1} \\ =\frac{-c_0}{6} \end{gathered}$$

For, n = 3

C₃= 0

For n = 4

$$\begin{gathered} c_4=\frac{-c_2}{2·9/2·7/2-9/2+1} \\ =\frac{-\left(-c_0/6\right)}{28} \\ =\frac{c_0}{168} \end{gathered}$$

Therefore,

$$\begin{gathered} y_2=x^{1/2}\left[c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \right] \\ =x^{1/2}\left[c_0+0·x-\frac{1}{6}{c}_0{x}^2+0·x^3+\frac{1}{168}{c}_0{x}^4+\dots \right] \\ =c_0{x}^{1/2}\left[1-\frac{1}{6}{x}^2+\frac{1}{168}{x}^4+\dots \right] \dots \left(8\right) \end{gathered}$$

Hence, the answer is:

$y=C_{1}x\left[1-\frac{1}{10}{x}^2+\frac{1}{180}{x}^4+\dots \right]+C_2{x}^{1/2}\left[1-\frac{1}{6}{x}^2+\frac{1}{168}{x}^4+\dots \right]$