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$\left|0,\infty \right|$.

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

Short answer

Therefore, the solution is:

$y=C_1{x}^{5/2}\left[1+\frac{4}{7}x+\frac{32}{693}{x}^2+\dots \right]+C_2\left[1+\frac{1}{3}x-\frac{1}{6}{x}^2-\frac{1}{6}{x}^3+\dots \right]$

Step-by-step solution

Given Information.

The given problem is:

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

Use Differentiation.

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

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.

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=2x\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2}-(3+2x)\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r-1}+\sum _{n=0}^{\infty }{c}_n{x}^{n+r} \\ =\underset{a_0\to x^{r-1}}{\underbrace{\sum _{n=0}^{\infty }2c_n(n+r)(n+r-1)x^{n+r-1}}}-\underset{a_0\to x^{r-1}}{\underbrace{\sum _{n=0}^{\infty }3c_n(n+r)x^{n+r-1}}}-\underset{a_0\to x^r}{\underbrace{\sum _{n=0}^{\infty }2c_n(n+r)x^{n+r}}}+\underset{a_0\to x^r}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+r}}} \end{gathered}$$

Now, we do the shift for the third and fourth series from n to n-1 yields;

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left[2c_{0}r(r-1)x^{r-1}+\sum _{n=1}^{\infty }2c_n(n+r)(n+r-1)x^{n+r-1}\right] \\ -\left[3c_{0}r{x}^{r-1}+\sum _{n=1}^{\infty }3c_n(n+r)x^{n+r-1}\right]-\sum _{n-1=0}^{\infty }2c_{n-1}\left(\right(n-1)+r)x^{(n-1)+r}+\sum _{n-1=0}^{\infty }{c}_{n-1}{x}^{(n-1)+r} \\ =2c_{0}r(r-1)x^{r-1}+\sum _{n=1}^{\infty }2c_n(n+r)(n+r-1)x^{n+r-1} \\ -3c_{0}r{x}^{r-1}-\sum _{n=1}^{\infty }3c_n(n+r)x^{n+r-1}-\sum _{n=1}^{\infty }2c_{n-1}(n+r-1)x^{n+r-1}+\sum _{n=1}^{\infty }{c}_{n-1}{x}^{n+r-1} \end{gathered}$$

The coefficient of terms

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

Identity Property

$r(2r-5)=0\Rightarrow r_1=\frac{5}{2} \& r_2=0$

For , r₁ = 5/2 we have:

$$\begin{gathered} c_n(n+r)(2n+2r-5)-c_{n-1}(2n+2r-3)=0 \\ {c}_n=\frac{c_{n-1}(2n+2r-3)}{(n+r)(2n+2r-5)} \end{gathered}$$

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^{n+5/2} \\ =x^{5/2}\sum _{n=0}^{\infty }{c}_n{x}^n \end{gathered}$$

Now,

$$\begin{gathered} c_n=\frac{c_{n-1}(2n+2r-3)}{(n+r)(2n+2r-5)} \\ =\frac{c_{n-1}(2n+2(5/2)-3)}{(n+5/2)(2n+2(5/2)-5)} \\ =\frac{c_{n-1}2(n+1)}{(n+5/2)\left(2n\right)} \\ =\frac{2c_{n-1}(n+1)}{(2n+5)n} \end{gathered}$$

Now,

For $n=1,c_1=\frac{4c_0}{7}$

For n = 2

$$\begin{gathered} c_2=\frac{2c_1·3}{9·2} \\ =\frac{\left(4c_0/7\right)}{3} \\ =\frac{4c_0}{21} \end{gathered}$$

For, n = 3

$$\begin{gathered} c_3=\frac{2c_2·4}{11·3} \\ =\frac{8\left(4c_0/21\right)}{33} \\ =\frac{32c_0}{44} \\ =\frac{32c_0}{693} \end{gathered}$$

For r = 0 we have:

$c_n=\frac{c_{n-1}(2n-3)}{n(2n-5)}$

For, n = 1

$$\begin{gathered} c_1=\frac{c_0·-1}{-3} \\ =\frac{c_0}{3} \end{gathered}$$

For n = 2

$$\begin{gathered} c_2=\frac{c_1·1}{2·-1} \\ =\frac{\left(c_0/3\right)}{-2} \\ =\frac{-c_0}{6} \end{gathered}$$

For n = 3

$$\begin{gathered} c_3=\frac{c_2·3}{3·1} \\ =\frac{-c_0}{6} \end{gathered}$$

For the indicial root as r₁= 5/2 ,we get the solution as:

$$\begin{gathered} y_1=x^{5/2}\left[c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \right] \\ =x^{5/2}\left[c_0+\frac{4}{7}{c}_{0}x+\frac{32}{693}{c}_0{x}^2+\dots \right] \\ =c_0{x}^{5/2}\left[1+\frac{4}{7}x+\frac{32}{693}{x}^2+\dots \right] \end{gathered}$$

For the indicial root as r₂ = 0 we get the solution as

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

Thus, the solution is:

$$\begin{gathered} y\left(x\right)=y_1\left(x\right)+y_2\left(x\right) \\ =C_1{x}^{5/2}\left[1+\frac{4}{7}x+\frac{32}{693}{x}^2+\dots \right]+C_2\left[1+\frac{1}{3}x-\frac{1}{6}{x}^2-\frac{1}{6}{x}^3+\dots \right] \end{gathered}$$