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$\left(0,\infty \right)$.

$\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

Therefore, the solution is:

$y=C_1{x}^{1/2}\left[1-\frac{2}{5}x+\frac{2}{35}{x}^2-\frac{4}{945}{x}^3+\dots \right]+C_2{x}^{-1}\left[1+2x-2x^2+\frac{4}{9}{x}^3+\dots \right]$

Step-by-step solution

Given Information.

The given problem is:

$2x^{2}y\text{'}\text{'}+3xy\text{'}+(2x-1)y=0$

Use Differentiation.

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

Differentiate (*) we get;

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

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

Substitute the equation.

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

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left[2c_{0}r\right(r-1)x^r+\sum _{n=1}^{\infty }\underset{a_1\to x^{r+1}}{\underbrace{2c_n(n+r)(n+r-1)x^{n+r}}}++[3c_{0}r{x}^r+\sum _{n=1}^{\infty }\underset{a_1\to x^{r+1}}{\underbrace{3c_n(n+r)x^{n+r}}}+\sum _{n=1}^{\infty }2c_n{x}^{n+r+1} \\ -[c_0{x}^r+\sum _{n=1}^{\infty }\underset{a_1\to x^{r+1}}{\underbrace{c_n{x}^{n+r}}} \end{gathered}$$

localid="1664863478850" $$\begin{gathered} \mathrm{f}\left(\mathrm{y}\text{'}\text{'},\mathrm{y}\text{'},\mathrm{y}\right)=2{\mathrm{c}}_0\mathrm{r}(\mathrm{r}-1){\mathrm{x}}^{\mathrm{r}}+\sum _{}^{\infty }2{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1){\mathrm{x}}^{\mathrm{n}+\mathrm{r}}+3{\mathrm{c}}_0{\mathrm{rx}}^{\mathrm{r}}+\sum _{\mathrm{n}=1}^{\infty }3{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r}){\mathrm{x}}^{\mathrm{n}+\mathrm{r}} \\ +\sum _{\mathrm{n}-1=0}^{\infty }2{\mathrm{c}}_{\mathrm{n}-1}{\mathrm{x}}^{(\mathrm{n}-1)+\mathrm{r}+1}-{\mathrm{c}}_0{\mathrm{x}}^{\mathrm{T}}-\sum _{\mathrm{n}=1}^{\infty }{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+\mathrm{r}} \\ =2{\mathrm{c}}_0\mathrm{r}(\mathrm{r}-1){\mathrm{x}}^{\mathrm{r}}+\sum _{}^{\infty }2{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1){\mathrm{x}}^{\mathrm{n}+\mathrm{r}}+3{\mathrm{c}}_0{\mathrm{rx}}^{\mathrm{r}}+\sum _{\mathrm{n}=1}^{\infty }3{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r}){\mathrm{x}}^{\mathrm{n}+\mathrm{r}}\underset{\mathrm{n}=1}{\overset{\infty }{+\sum }}2{\mathrm{c}}_{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}+\mathrm{r}} \\ -{\mathrm{c}}_0{\mathrm{x}}^{\mathrm{r}}-\sum _{\mathrm{n}=1}^{\infty }{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+\mathrm{r}} \end{gathered}$$

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

Identity Property.

$2r^2+r-1=0\Rightarrow (2r-1)(r+1)=0$

Therefore,

$r_1=\frac{1}{2} \& r_2=-1$

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

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

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

For, n = 1 therefore

$c_1=\frac{-2c_0}{6-1}$

For n = 2

$$\begin{gathered} c_2=\frac{-2c_1}{15-1} \\ =\frac{-2\left(-2c_0/5\right)}{14} \\ =\frac{2c_0}{35} \end{gathered}$$

For n = 3

$$\begin{gathered} c_3=\frac{-2c_2}{28-1} \\ =\frac{-2\left(2c_0/35\right)}{27} \\ =\frac{-4c_0}{945} \end{gathered}$$

Thus, the solution is:

$$\begin{gathered} y_1=x^{1/2}\left[c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \right] \\ =x^{1/2}\left[c_0-\frac{2}{5}{c}_{0}x+\frac{2}{35}{c}_0{x}^2-\frac{4}{945}{c}_0{x}^3+\dots \right] \\ =c_0{x}^{1/2}\left[1-\frac{2}{5}x+\frac{2}{35}{x}^2-\frac{4}{945}{x}^3+\dots \right] \end{gathered}$$

For r2 = -1

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

Therefore,

localid="1664863493997" $\begin{array}{rcl}{\mathrm{c}}_{\mathrm{n}}& =& \frac{-2{\mathrm{c}}_{\mathrm{n}-1}}{(\mathrm{n}-1)(2\mathrm{n}-2+1)-1}\\ & =& \frac{-2{\mathrm{c}}_{\mathrm{n}-2}}{(\mathrm{n}-1)(2\mathrm{n}-1)-1}\end{array}$

For n = 1

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

For n = 2

role="math" $\begin{array}{rcl}{\mathrm{c}}_2& =& \frac{-2{\mathrm{c}}_1}{3-1}\\ & =& \frac{-2\left(2{\mathrm{c}}_0\right)}{2}\\ & & -2{\mathrm{c}}_0\end{array}$

For n = 3

$$\begin{gathered} c_3=\frac{-2c_2}{10-1} \\ =\frac{-2\left(-2c_0\right)}{9} \\ =\frac{4c_0}{9} \end{gathered}$$

Thus, the solution is:

$$\begin{gathered} y_2=x^{-1}\left[c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \right] \\ =x^{-1}\left[c_0+2c_{0}x-2c_0{x}^2+\frac{4}{0}{c}_0{x}^3+\dots \right] \\ =c_0{x}^{-1}\left[1+2x-2x^2+\frac{4}{9}{x}^3+\dots \right] \end{gathered}$$

$y=C_1{x}^{1/2}\left[1-\frac{2}{5}x+\frac{2}{35}{x}^2-\frac{4}{945}{x}^3+\dots \right]+C_2{x}^{-1}\left[1+2x-2x^2+\frac{4}{9}{x}^3+\dots \right]$

Thus, the solution is;

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