Question

In Problems, 19-22 use the power series method to solve the given initial-value problem.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

Therefore, the general solution is reduced to the following form;

$y=2-x+\left[-1-\frac{2}{2}\right]x^2+\left[\frac{-2}{6}\right]x^3+\left[\frac{-(-1)}{4}+\frac{2}{8}\right]x^4+\dots$

Step-by-step solution

Given Information

The given differential equation is;

$y\text{'}\text{'}-2xy\text{'}+8y=0,y\left(0\right)=3,y\text{'}\left(0\right)=0$

Plug into the differential equation

We have

$f\left(y\text{'}\text{'},y\text{'},y\right)=y\text{'}\text{'}-2xy\text{'}+8y=0$

And $y\left(0\right)=3$and $y\text{'}\left(0\right)=0$

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;

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

Thus, the given differential equation has a solution of the following form

$y=\sum _{n=0}^{\infty }{c}_n{x}^n \dots \left(1\right)$, where$x_0=0$

We need to find the two linearly independent series solutions for the given differential equation and then solve the initial value problem. We find the first and second derivatives of (1) as follows;

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

We shift the summation index to n = 1 for y' like the first term of the power series equals to zero, and n = 2 for y":

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

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

Substituting the given differential equation

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

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-2x\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}+8\sum _{n=0}^{\infty }{c}_n{x}^n \\ =\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-\sum _{n=1}^{\infty }2x{c}_{n}n{x}^{n-1}+\sum _{n=0}^{\infty }8c_n{x}^n \\ =\underset{a_2\to x^0}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}}}-\underset{a_1\to x^1}{\underbrace{\sum _{n=1}^{\infty }2c_{n}n{x}^n}}+\underset{a_0\to x^0}{\underbrace{\sum _{n=0}^{\infty }8c_n{x}^n}} \end{gathered}$$

The make the summation index and the power of x

Now, we shift the summation index for the first three power series. Note that, this wouldn't affect the terms of the series, the shift yields the same terms.

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\sum _{n+2=3}^{\infty }{c}_{n+2}(n+2)\left(\right(n+2)-1)x^{(n+2)-2}-\sum _{n=1}^{\infty }2c_{n}n{x}^n+\sum _{n=1}^{\infty }8c_n{x}^n+\left[2c_2+8c_0\right] \\ =\sum _{n=1}^{\infty }{c}_{n+2}(n+2)(n+1)x^n-\sum _{n=1}^{\infty }2c_{n}n{x}^n+\sum _{n=1}^{\infty }8c_n{x}^n+\left[2c_2+8c_0\right] \\ =\sum _{n=1}^{\infty }\left[c_{n+2}(n+1)(n+2)-2c_{n}n+8c_n\right]x^n+\left[2c_2+8c_0\right] \\ =\sum _{n=1}^{\infty }\left[c_{n+2}(n+1)(n+2)-2c_n(n-4)\right]x^n+\underset{\text{coefficients of}{x}^0}{\underbrace{\left[2c_2+8c_0\right]}} \\ =0 \end{gathered}$$

Therefore, we have;

$c_{n+2}(n+1)(n+2)-2c_n(n-4)=0$

And

$$\begin{gathered} 2c_2+8c_0=0 \\ {c}_2=-4c_0 \dots \left(4\right) \end{gathered}$$

Find the coefficients of the power series in terms

Now, we have the recurrence relation.

$$\begin{gathered} c_{n+2}(n+1)(n+2)=2c_n(n-4) \\ {c}_{n+2}=\frac{2c_n(n-4)}{(n+2)(n+1)} \dots \left(5\right) \end{gathered}$$

By using (4) and (5), we can find the coefficients of the power series in terms of $c_0$and$c_1$

At n = 1,$$\begin{gathered} c_3=\frac{2c_1(-3)}{3.2} \\ =-c_1 \end{gathered}$$

If n = 2 ,then:

$$\begin{gathered} n=2,c_4=\frac{2c_2(-2)}{4·3} \\ =\frac{-\left(-4c_0\right)}{3} \\ =\frac{\left(4c_0\right)}{3} \end{gathered}$$

Now we expand the power series in (1),

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^n \\ =c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \dots \left(8\right) \end{gathered}$$

Now, using the initial values for y and y' we can determine$c_0$ and$c_1$ from (8), we get $y\text{'}\left(x\right)=0+c_1+2x+\dots$,therefore,

$$\begin{gathered} y\left(0\right)=c_0 \\ =3 \\ y\text{'}\left(0\right)=c_1 \\ =0 \end{gathered}$$

Therefore, the general solution is reduced to the following form

$$\begin{gathered} y=c_0+c_{1}x+\left[-4c_0\right]x^2+\left[-c_1\right]x^3+\left[\frac{4}{3}{c}_0\right]x^4+\dots \\ =3+0·x+[-4·3]x^2+\left[0\right]x^3+\left[\frac{4}{3}·3\right]x^4+\dots \\ =3-12x^2+4x^4+\dots \end{gathered}$$

The general solution has the form:

$y=3-12x^2+4x^4+\dots$