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$(0,\infty )$.

$2xy\text{'}\text{'}+5y\text{'}+xy=0$

Short answer

The solution is:

$y=C_1\left[1-\frac{1}{14}{x}^2+\frac{1}{616}{x}^4+\dots \right]+C_2{x}^{-3/2}\left[1-\frac{1}{2}{x}^2+\frac{1}{40}{x}^4+\dots \right]$

Step-by-step solution

Given Information.

The given problem is:

$2xy\text{'}\text{'}+5y\text{'}+xy=0$

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.

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=2x\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2}+5\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r-1}+x\sum _{n=0}^{\infty }{c}_n{x}^{n+r} \\ =\underset{a_0\to x^{r-1}}{\underbrace{\sum _{n=0}2c_n(n+r)(n+r-1)x^{n+r-1}}}+\underset{a_0\to x^{r-1}}{\underbrace{\sum _{n=0}5c_n(n+r)x^{n+r-1}}}+\underset{a_0\to x^{r+1}}{\underbrace{\sum _{n=0}{c}_n{x}^{n+r+1}}} \end{gathered}$$

$$\begin{gathered} \left(y\text{'}\text{'},y\text{'},y\right)=\left[2c_{0}r\right(r-1)x^{r-1}+2c_1(r+1\left)\right(r)x^r+\underset{a_2\to x^{r+1}}{\underbrace{\left.\sum _{n=2}^{\infty }2c_n(n+r)(n+r-1)x^{n+r-1}\right]}} \\ +[5c_{0}r{x}^{r-1}+5c_1(r+1)x^r+\underset{a_2\to x^{r+1}}{\underbrace{\left.\sum _{n=2}^{\infty }5c_n(n+r)x^{n+r-1}\right]}}+\underset{a_0\to x^{r+1}}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+r+1}}} \end{gathered}$$

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=2c_{0}r(r-1)x^{r-1}+2c_1(r+1)\left(r\right)x^r+\sum _{n=2}^{\infty }2c_n(n+r)(n+r-1)x^{n+r-1} \\ +5c_{0}r{x}^{r-1}+5c_1(r+1)x^r+\sum _{n=2}^{\infty }5c_n(n+r)x^{n+r-1}+\sum _{n-2=0}^{\infty }{c}_{n-2}{x}^{(n-2)+r+1} \end{gathered}$$

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=2c_{0}r(r-1)x^{r-1}+2c_1(r+1)\left(r\right)x^r+\sum _{n=2}^{\infty }2c_n(n+r)(n+r-1)x^{n+r-1}+5c_{0}r{x}^{r-1}+5c_1(r+1)x^r \\ +\sum _{n=2}^{\infty }5c_n(n+r)x^{n+r-1}+\sum _{n=2}^{\infty }{c}_{n-2}{x}^{n+r-1} \end{gathered}$$

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left[2r\right(r-1)+5r]c_0{x}^{r-1}+\left[2\right(r+1\left)\right(r)+5(r+1\left)\right]c_1{x}^r \\ +\sum _{n=2}^{\infty }\left[2c_n(n+r)(n+r-1)+5c_n(n+r)+c_{n-2}\right]x^{n+r} \end{gathered}$$

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left.\left[2r^2-2r+5r\right]c_0{x}^{r-1}+\left[2r^2+2r+5r+5\right)\right]c_1{x}^r \\ +\sum _{n=2}^{\infty }\left[c_n(n+r)[2n+2r-2+5]+c_{n-2}\right]x^{n+r} \end{gathered}$$

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left[r\right(2r+3\left)\right]c_0{x}^{r-1}+\left[\right(2r+5\left)\right(r+1\left)\right]c_1{x}^r \\ +\sum _{n=2}^{\infty }\left[c_n(n+r)[2n+2r+3]+c_{n-2}\right]x^{n+r} \end{gathered}$$

Identity Property

$f\left(y\text{'}\text{'},y\text{'},y\right)=0$

Therefore, we have;

$r(2r+3)=0\Rightarrow r_1=0,r_2=-\frac{3}{2}$

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

$$\begin{gathered} c_n(n+r)[2n+2r+3]+c_{n-2}=0 \\ {c}_n=\frac{-c_{n-2}}{(n+r)[2n+2r+3]} \end{gathered}$$

Now,

If, r₁=0 then:

$y=\sum _{n=0}^{\infty }{c}_n{x}^n$

$c_n=\frac{-c_{n-2}}{n[2n+3]} \dots \left(5\right)$

For n = 2,

$$\begin{gathered} c_2=\frac{-c_0}{2·7} \\ =\frac{-c_0}{14} \end{gathered}$$

For, n = 3

C₃= 0

For, n = 4

$$\begin{gathered} c_4=\frac{-c_2}{4·11} \\ =\frac{-\left(-c_0/14\right)}{44} \\ =\frac{c_0}{616} \end{gathered}$$

Therefore,

$$\begin{gathered} y_1=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \\ =c_0+0·x-\frac{1}{14}{c}_0{x}^2+0·x^3+\frac{1}{616}{c}_0{x}^4+\dots \\ =c_0\left[1-\frac{1}{14}{x}^2+\frac{1}{616}{x}^4+\dots \right] \dots \left(6\right) \end{gathered}$$

For, r₁=-3/2

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^{n-3/2} \\ =x^{-3/2}\sum _{n=0}^{\infty }{c}_n{x}^n \end{gathered}$$

$$\begin{gathered} c_n=\frac{-c_{n-2}}{(n-3/2)[2n+2(-3/2)+3]} \\ =\frac{-c_{n-2}}{n(2n-3)}\cdots \left(7\right) \end{gathered}$$

For, n = 2

$$\begin{gathered} c_2=\frac{-C_0}{2·1} \\ =\frac{-c_0}{2} \end{gathered}$$

For n = 3, C₃=0

For n = 4

$$\begin{gathered} c_4=\frac{-c_2}{4·5} \\ =\frac{-\left(-c_0/2\right)}{20} \\ =\frac{c_0}{40} \end{gathered}$$

$$\begin{gathered} y_2=x^{-3/2}\left[c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \right] \\ =x^{-3/2}\left[c_0+0·x-\frac{1}{2}{c}_0{x}^2+0·x^3+\frac{1}{40}{c}_0{x}^4+\dots \right] \\ =c_0{x}^{-3/2}\left[1-\frac{1}{2}{x}^2+\frac{1}{40}{x}^4+\dots \right] \dots \left(8\right) \end{gathered}$$

For, r₂=-3/2

$y=C_1\left[1-\frac{1}{14}{x}^2+\frac{1}{616}{x}^4+\dots \right]+C_2{x}^{-3/2}\left[1-\frac{1}{2}{x}^2+\frac{1}{40}{x}^4+\dots \right]$