Question

In Problems 15–24 x= 0,is 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$\left(0,\infty \right)$.

$\mathbf{9}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

Therefore, the solution is:

$y=x^{1/3}{C}_1\left(1-\frac{1}{2}x+\frac{1}{5}{x}^2-\frac{7}{120}{x}^3+\dots \right)+x^{2/3}{C}_2\left(1-\frac{1}{2}x+\frac{5}{28}{x}^2-\frac{1}{21}{x}^3+\dots \right)$

Step-by-step solution

Given Information.

The given problem is;

$0=9x^{2}y\text{'}\text{'}+9x^{2}y\text{'}+2y$

Use Differentiate

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

Differentiate (*) we get;

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

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

Substitute the equation.

$0=9x^2\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2}+9x^2\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r-1}+2{}^2\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

$0=9x^2\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r}+9x^2\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r+1}+2{}^2\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

$0=9x^2\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2}+2c_n{x}^{n+r}\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r-1}+2{}^2\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

$0=\left|9c_{0}r(r-1)+2c_0\right|x^r+\sum _{n=1}^{\infty }\left[9c_n(n+r)(n+r-1)+2c_n\right]x^{n+r}+9\sum _{n=1}^{\infty }{c}_n(n+r)x^{n+r+1}$

$0=\left(9r^2-9r+2\right)c_0{x}^r+\sum _{n=1}^{\infty }\left[9c_n(n+r)(n+r-1)+2c_n\right]x^{n+r}+9\sum _{n=1}^{\infty }{c}_n(n+r)x^{n+r+1}$

k = n

k = n + 1

k = n - 1

$$\begin{gathered} =\left(9r^2-9r+2\right)c_0{x}^r+\sum _{k=1}^{\infty }\left[9c_k(k+r)(k+r-1)+2c_k\right]x^{k+r}+9\sum _{k=1}^{\infty }{c}_{k-1}(k+r-1)x^{k+\tau } \\ =\left(9r^2-9r+2\right)c_0{x}^r+\sum _{k=1}^{\infty }\left[9c_k(k+r)(k+r-1)+2c_k+9c_{k-1}(k+r-1)\right]x^{k+r} \end{gathered}$$

Identity Property.

$$\begin{gathered} 0=\left(9r^2-9r+2\right)c_0 \\ 0=9r^2-9r+2 \\ 0=9r^2-6r-3r+2 \\ 0=3r(3r-2)-1(3r-2) \end{gathered}$$

$(3r-1)(3r-2)=0\Rightarrow r=\frac{1}{3},r=\frac{2}{3}$

$$\begin{gathered} 0=9c_k(k+r)(k+r-1)+2c_k+9c_{k-1}(k+r-1) \\ 0=\left[9\right(k+r\left)\right(k+r-1)+2]c_k+9c_{k-1}(k+r-1) \\ {c}_k\left[9\right(k+r\left)\right(k+r-1)+2]=-9c_{k-1}(k+r-1) \\ {c}_k=\frac{-9c_{k-1}(k+r-1)}{9(k+r)(k+r-1)+2}fork⩾1 \end{gathered}$$

For; r = 2/3

$$\begin{gathered} c_k=\frac{-9c_{k-1}\left(k-\frac{2}{3}\right)}{9\left(k+\frac{1}{3}\right)\left(k-\frac{2}{3}\right)+2} \\ {c}_k=\frac{-9c_{k-1}\left(k-\frac{2}{3}\right)}{9\left(k^2-\frac{1}{3}k-\frac{2}{9}\right)+2} \\ {c}_k=\frac{-9c_{k-1}\left(k-\frac{2}{3}\right)}{3k(3k-1)} \\ {c}_k=\frac{-3c_{k-1}\left(k-\frac{2}{3}\right)}{k(3k-1)} \\ {c}_k=\frac{-c_{k-1}(3k-2)}{k(3k-1)} \end{gathered}$$

k = 2

$c_2=-\frac{4}{10}$

$c_1=-\frac{2}{5}{c}_1$

$=-\frac{2}{5}\left(-\frac{c_0}{2}\right)$

=c₀ / 5

r = 2/3

$c_k=\frac{-9c_{k-1}\left(k-\frac{1}{3}\right)}{9\left(k+\frac{2}{3}\right)\left(k-\frac{1}{3}\right)+2}$

$$\begin{gathered} c_k=\frac{-9c_{k-1}\left(k-\frac{1}{3}\right)}{9\left(k^2+\frac{1}{3}k-\frac{2}{9}\right)+2} \\ {c}_k=\frac{-9c_{k-1}\left(k-\frac{1}{3}\right)}{3k(3k+1)} \end{gathered}$$

$c_k=\frac{-c_{k-1}(3k-1)}{k(3k+1)}$

k = 1

C₁= -1/2C₀

k = 2

C₂= -5/14

$$\begin{gathered} c_1=-\frac{5}{14}\left(-\frac{1}{2}{c}_0\right) \\ =\frac{5}{28}{c}_0 \end{gathered}$$

$$\begin{gathered} c_3=-\frac{8}{30}{c}_2 \\ =-\frac{4}{15}\left(\frac{5}{28}{c}_0\right) \\ =-\frac{1}{21}{c}_0 \end{gathered}$$