Question

In Problems, 7-18 find two power series solutions of the given differential equation about the ordinary point $\left(x^2+2\right){\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

Therefore, the solution is:

$$\begin{gathered} y_1\left(x\right)=c_0\left[1+\frac{1}{4}{x}^2-\frac{7}{4·4!}{x}^4+\frac{23·7}{8·6!}{x}^6-\cdots \right] \\ {y}_2\left(x\right)=c_1\left[x-\frac{1}{6}{x}^3+\frac{14}{2·5!}{x}^5-\frac{34·14}{4·7!}{x}^7-\cdots \right] \end{gathered}$$

Step-by-step solution

Given Information

The given differential equation is;

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

Plug into the differential equation

$y\left(x\right)=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\cdots =\sum _{n=0}^{\infty }{c}_n{x}^n\text{(*)}$

Differentiating and translating the index by one and two respectively, we obtain

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

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

Replacing the(1) and equation (2) in the given differential equation, we obtain

$$\begin{gathered} \left(x^2+2\right)\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+3x\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}-\sum _{n=0}^{\infty }{c}_n{x}^n=0 \\ {x}^2\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+2\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+3x\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}-\sum _{n=0}^{\infty }{c}_n{x}^n=0 \\ \sum _{n=2}^{\infty }n(n-1)c_n{x}^n+2\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+3\sum _{n=1}^{\infty }n{c}_n{x}^n-\sum _{n=0}^{\infty }{c}_n{x}^n=0 \\ \sum _{n=2}^{\infty }n(n-1)c_n{x}^n+2·2·1c_2{x}^0+2·3·2c_{3}x+2\sum _{n=4}^{\infty }n(n-1)c_n{x}^{n-2}+3·1c_{1}x+3\sum _{n=2}^{\infty }n{c}_n{x}^n-c_0-c_{1}x-\sum _{n=2}^{\infty }{c}_n{x}^n=0 \\ \sum _{n=2}^{\infty }n(n-1)c_n{x}^n+4c_2+12c_{3}x+2\sum _{n=4}^{n}n(n-1)c_n{x}^{n-2}+3c_{1}x+3\sum _{n=2}^{\infty }n{c}_n{x}^n-c_0-c_{1}x-\sum _{n=2}^{n=2}{c}_n{x}^n=0 \\ 4c_2-c_0+\left(12c_3+3c_1-c_1\right)x+\sum _{n=2}^{\infty }n(n-1)c_n{x}^n+2\sum _{n=4}^{\infty }n(n-1)c_n{x}^{n-2}+3\sum _{n=2}^{\infty }n{c}_n{x}^{n-1}-\sum _{n=2}^{\infty }{c}_n{x}^n=0 \end{gathered}$$

Identity Property

$$\begin{gathered} 4c_2-c_0=0 \\ {c}_2=\frac{c_0}{4} \\ 12c_3+2c_1=0 \\ {c}_3=-\frac{c_1}{6} \\ k(k-1)c_k+2(k+2)(k+1)c_{k+2}+3k{c}_k-c_k=0 \\ {c}_{k+2}=\frac{c_k\left(-k^2-2k+1\right)}{2(k+2)(k+1)} \\ k=2\Rightarrow c_4=\frac{c_2(-7)}{4·3·2}=\frac{-7c_0}{4·4!}\backslash \mathrm{hfill} \\ k=3\Rightarrow c_5=-\frac{14c_3}{2·5·4}=\frac{14c_1}{2·5!}\backslash \mathrm{hfill} \\ k=5\Rightarrow c_6=-\frac{23c_4}{2·6·5}=\frac{23·7c_1}{8·6!}\backslash \mathrm{hfill} \\ k=6\Rightarrow c_7=-\frac{34c_5}{2·7·6}=-\frac{34·14c_1}{4·7!}\backslash \mathrm{hfill} \end{gathered}$$

Continuing the same procedure and replacing in $\left(*\right),$we obtain

$$\begin{gathered} y\left(x\right)=c_0+c_{1}x+\frac{1}{4}{c}_0{x}^2-\frac{1}{6}{c}_1{x}^3-\frac{7}{4·4!}{c}_0 \\ =c_0\left[1+\frac{1}{4}{x}^2-\frac{7}{4·4!}{x}^4+\frac{23·7}{8·6!}{x}^6-\cdots \right] \\ +c_1\left[x-\frac{1}{6}{x}^3+\frac{14}{2·5!}{x}^5-\frac{34·14}{4·7!}{x}^7-\cdots \right]x^4 \\ +\frac{14}{2·5!}{c}_1{x}^5+\frac{23·7}{8·6!}{c}_1{x}^6-\frac{34·14}{4·7!}{c}_1{x}^7+\cdots \end{gathered}$$

Therefore, the solution is:

$$\begin{gathered} y_1\left(x\right)=c_0\left[1+\frac{1}{4}{x}^2-\frac{7}{4·4!}{x}^4+\frac{23·7}{8·6!}{x}^6-\cdots \right]\backslash \mathrm{hfill} \\ {y}_2\left(x\right)=c_1\left[x-\frac{1}{6}{x}^3+\frac{14}{2·5!}{x}^5-\frac{34·14}{4·7!}{x}^7-\cdots \right]\backslash \mathrm{hfill} \end{gathered}$$