Question

In Problems, 7-18 find two power series solutions of the given differential equation about the ordinary point$\left(x^2+1\right)y^{\text{'}\text{'}}-6y=0$

Short answer

Therefore, the solution is:

$$\begin{gathered} y_1\left(x\right)=c_0\left[1+3x^2+x^4-\frac{1}{5}{x}^6+\frac{3}{35}{x}^8+\dots \right] \\ {y}_2\left(x\right)=c_1\left[x+x^3\right] \end{gathered}$$

Step-by-step solution

Given Information.

The given differential equation is:

$\left(x^2+1\right)y\text{'}\text{'}-6y=0$

Plug into the differential equation.

$\begin{array}{rcl}\left(x^2+1\right)y\text{'}\text{'}-6y& =& 0\\ \left(x^2+1\right)\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-6\sum _{n=0}^{\infty }{c}_n{x}^n& =& 0\\ \sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n+\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-6\sum _{n=0}^{\infty }{c}_n{x}^n& =& 0\\ \sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n+\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-6\sum _{n=0}^{\infty }{c}_n{x}^n& =& 0\\ & & \end{array}$

Extract thefirst and second terms from the two series on the right and increase their indices by $2.$

$\begin{array}{rcl}{c}_2·2\left(1\right)x^0+c_3·3\left(2\right)x^1-6c_0{x}^0-6c_1{x}^1+\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n+\sum _{n=4}^{\infty }{c}_{n}n(n-1)x^{n-2}-6\sum _{n=2}^{\infty }{c}_n{x}^n& =& 0\\ 2c_2+6c_{3}x-6c_0-6c_{1}x+\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n+\sum _{n=4}^{\infty }{c}_{n}n(n-1)x^{n-2}-6\sum _{n=2}^{\infty }{c}_n{x}^n& =& 0\\ 2c_2-6c_0+\left(c_3-c_1\right)6x+\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n+\sum _{n=4}^{\infty }{c}_{n}n(n-1)x^{n-2}-6\sum _{n=2}^{\infty }{c}_n{x}^n& =& 0\\ & & \end{array}$

For the series on the left and right, let, for the series in the middle, let $k=n-2,n=k+2$

$\begin{array}{rcl}2c_2-6c_0+\left(c_3-c_1\right)6x+\sum _{k=2}^{\infty }{c}_{k}k(k-1)x^k+\sum _{k+2=4}^{\infty }{c}_{k+2}(k+2)(k+2-1)x^k-6\sum _{k=2}^{\infty }{c}_k{x}^k& =& 0\\ 2c_2-6c_0+\left(c_3-c_1\right)6x+\sum _{k=2}^{\infty }{c}_{k}k(k-1)x^k+\sum _{k=2}^{\infty }{c}_{k+2}(k+2)(k+1)x^k-6\sum _{k=2}^{\infty }{c}_k{x}^k& =& 0\\ & & \end{array}$

Now combine all three series and factor out $x^k$.

$\begin{array}{rcl}2c_2-6c_0+\left(c_3-c_1\right)6x+\sum _{k=2}^{\infty }\left[c_{k}k(k-1)+c_{k+2}(k+2)(k+1)-6c_k\right]x^k& =& 0\\ 2c_2-6c_0+\left(c_3-c_1\right)6x+\sum _{k=2}^{\infty }\left[c_{k+2}(k+2)(k+1)+c_k\left(k^2-k\right)-6c_k\right]x^k& =& 0\\ 2c_2-6c_0+\left(c_3-c_1\right)6x+\sum _{k=2}^{\infty }\left[c_{k+2}(k+2)(k+1)+c_k\left(k^2-k-6\right)\right]x^k& =& 0\\ & & \end{array}$

Identity property

$\begin{array}{rcl}2c_2-6c_0& =& 0\\ c_2& =& 3c_0\\ & & \end{array}$

$\begin{array}{rcl}{c}_3-c_1& =& 0\\ c_3& =& c_1\\ & & \end{array}$

$\begin{array}{rcl}{c}_{k+2}(k+2)(k+1)+c_k\left(k^2-k-6\right)& =& 0\\ c_{k+2}(k+2)(k+1)+c_k(k-3)(k+2)& =& 0\\ c_{k+2}& =& -\frac{c_k(k-3)(k+2)}{(k+2)(k+1)}\\ c_{k+2}& =& -\frac{c_k(k-3)}{k+1} for k⩾2\\ & & \end{array}$

$\begin{array}{rcl}k& =& 2:c_4=-\frac{c_2(-1)}{3}=c_0\\ k& =& 3:c_5=-\frac{c_3\left(0\right)}{3}=0\\ k& =& 4:c_6=-\frac{c_4\left(1\right)}{5}=-\frac{1}{5}{c}_0\\ k& =& 5:c_7=-\frac{c_5\left(2\right)}{6}=0\\ k& =& 6:c_8=-\frac{c_6\left(3\right)}{7}=\frac{3}{35}{c}_0\\ k& =& 7:c_9=-\frac{c_7\left(4\right)}{8}=0\\ & & \end{array}$

plugging back into the original power series for we get

$\begin{array}{rcl}y& =& \sum _{n=0}^{\infty }{c}_n{x}^n\\ y& =& c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+c_5{x}^5+c_6{x}^6+c_7{x}^7+c_8{x}^8+c_9{x}^9+\dots \\ y& =& c_0+c_{1}x+3c_0{x}^2+c_1{x}^3+c_0{x}^4+\left(0\right)x^5-\frac{1}{5}{c}_0{x}^6+\left(0\right)x^7+\frac{3}{35}{c}_0{x}^8+\left(0\right)x^9+\dots \\ y& =& \left(c_0+3c_0{x}^2+c_0{x}^4-\frac{1}{5}{c}_0{x}^6+\frac{3}{35}{c}_0{x}^8+\dots \right)+c_{1}x+c_1{x}^3\\ y& =& c_0\left(1+3x^2+x^4-\frac{1}{5}{x}^6+\frac{3}{35}{x}^8+\dots \right)+c_1\left(x+x^3\right)\\ & & \end{array}$

Therefore, the solution is:

$\begin{array}{rcl}{y}_1\left(x\right)& =& c_0\left[1+3x^2+x^4-\frac{1}{5}{x}^6+\frac{3}{35}{x}^8+\dots \right]\\ y_2\left(x\right)& =& c_1\left[x+x^3\right]\\ & & \end{array}$