Question

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

Short answer

Therefore, the two power series solution of the given differential equation is:

$$\begin{gathered} y_1=c_0\left[1-\frac{1}{2}{x}^2-\frac{1}{8}{x}^4+\dots \right] \\ {y}_2=c_{3}x \end{gathered}$$

Step-by-step solution

Given Information

The given differential equation is:

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

Plug into the differential equation

We have,

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=\left(x^2-1\right)y^{\text{'}\text{'}}+x{y}^{\text{'}}-y \\ =0 \end{gathered}$$

Which has an ordinary point at x = 0 We state that, without proof, if the differential equation has an ordinary point at $x=x_0,$then it has a power series solution, with two linearly independent solutions, which has the following form,

$y=\sum _{n=0}^{\infty }{c}_n{\left(x-x_0\right)}^n$

$y\text{'}\text{'}=\sum _{n=0}^{\infty }{c}_{n}n(n-1)x^{n-2}$

We shift the summation index to n = 1, for$y^{\text{'}},$ like the first term of the power series equals to zero, and n = 2, for$y\text{'}\text{'}$

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

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

Substituting the given differential equation.

By substituting (1), (2), and (3) into the given differential equation, yields

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n-\left[c_2·2(2-1)x^0+c_3·3(3-1)x+\sum _{n=4}^{\infty }{c}_{n}n(n-1)x^{n-2}\right]+ \\ \left[c_{1}x+\sum _{n=2}^{\infty }{c}_{n}n{x}^n\right]-\left[c_0{x}^0+c_{1}x+\sum _{n=2}^{\infty }{c}_n{x}^n\right] \\ =\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n-\left[2c_2+6c_{3}x+\sum _{n=4}^{\infty }{c}_{n}n(n-1)x^{n-2}\right]+\left[c_{1}x+\sum _{n=2}^{\infty }{c}_{n}n{x}^n\right]-\left[c_0+c_{1}x+\sum _{n=2}^{\infty }{c}_n{x}^n\right] \\ =\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n-2c_2-6c_{3}x-\sum _{n=4}^{\infty }{c}_{n}n(n-1)x^{n-2}+c_{1}x+\sum _{n=2}^{\infty }{c}_{n}n{x}^n-c_0-c_{1}x-\sum _{n=2}^{\infty }{c}_n{x}^n \\ =\underset{a_2\to x^2}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n}}-\underset{a_4\to x^2}{\underbrace{\sum _{n=4}^{\infty }{c}_{n}n(n-1)x^{n-2}}}+\underset{a_2\to x^2}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n{x}^n}}-\underset{a_2\to x^2}{\underbrace{\sum _{n=2}^{\infty }{c}_n{x}^n}}+\left[-2c_2-c_0\right]-6c_{3}x \end{gathered}$$

For the second power series, the shift by substituting the summation index.

Now, we do the shift by replacing the summation index, for the second power series, from n to n + 2 as indicated below. Note that, this wouldn't affect the terms of the series, the shift yields the same terms.

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n-\sum _{n+2=4}^{\infty }{c}_{n+2}(n+2)\left(\right(n+2)-1)x^{(n+2)-2}+\sum _{n=2}^{\infty }{c}_{n}n{x}^n-\sum _{n=2}^{\infty }{c}_n{x}^n+\left[-2c_2-c_0\right]-6c_{3}x \\ =\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n-\sum _{n=2}^{\infty }{c}_{n+2}(n+2)(n+1)x^n+\sum _{n=2}^{\infty }{c}_{n}n{x}^n-\sum _{n=2}^{\infty }{c}_n{x}^n+\left[-2c_2-c_0\right]-6c_{3}x \\ =\sum _{n=2}^{\infty }\left[c_{n}n(n-1)-c_{n+2}(n+2)(n+1)+c_{n}n-c_n\right]x^n+\left[-2c_2-c_0\right]-6c_{3}x \\ =\sum _{n=2}^{\infty }\left[c_n(n-1)(n+1)-c_{n+2}(n+2)(n+1)\right]x^n+\left[-2c_2-c_0\right]-6c_{3}x \\ =\sum _{n=2}^{\infty }\left[c_n(n-1)-c_{n+2}(n+2)\right](n+1)x^n+\underset{\text{coefficients of}{x}^0}{\underbrace{\left[-2c_2-c_0\right]}}-6c_{3}x \\ =0 \end{gathered}$$

This implies that;

$c_n(n-a)-c_{n+2}(n+2)=0\Rightarrow c_{n+2}=\frac{c_n(n-1)}{(n+2)}----\left(4\right)$Where $n⩾2$

$$\begin{gathered} -2c_2-c_0=0 \\ {c}_2=\frac{-1}{2}{c}_0 \dots \left(5\right) \end{gathered}$$

$$\begin{gathered} -6c_3=0 \\ {c}_3=0 \dots \left(6\right) \end{gathered}$$

Find the coefficient of the power series in terms

By using (4), (5), and (6)

At n = 2,

$$\begin{gathered} c_4=\frac{c_2(2-1)}{2+2} \\ =\frac{c_2}{4} \\ =\frac{-c_0/2}{4} \\ =\frac{-c_0}{8} \end{gathered}$$

At n = 3,

$$\begin{gathered} c_5=\frac{c_3}{5} \\ =0 \end{gathered}$$

Expand the power series

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^n \\ =c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \\ =c_0+c_{1}x-\frac{c_0}{2}{x}^2+0·x^3-\frac{c_0}{8}{x}^4+0·x^5+\dots \\ =\underset{y_1}{\underbrace{c_0\left[1-\frac{1}{2}{x}^2-\frac{1}{8}{x}^4+\dots \right]}}+\underset{y_2}{\underbrace{c_{1}x}} \end{gathered}$$

Hence, $y_1$and $y_2$are the two series solution of the given partial differential equation.

Note that the coefficient of the power series $y_2$is all zero except $c_3$. Therefore, the two power series solution of the given differential equation is:

role="math" localid="1663954248797" $$\begin{gathered} y_1=c_0\left[1-\frac{1}{2}{x}^2-\frac{1}{8}{x}^4+\dots \right] \\ {y}_2=c_{3}x \end{gathered}$$