Question

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

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

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;

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

Plug into the differential equation

We have,

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

Which has an ordinary point at$x=0$We state that, without proof, if the differential equation has an ordinary point$x=x_0,$atthen 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 $y\text{'}$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

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-1}-\left[c_2·2(2-1)x^0+\sum _{n=3}^{\infty }{c}_{n}n(n-1)x^{n-2}\right]-\sum _{n=1}^{\infty }{c}_{n}n{x}^n+\left[c_0{x}^0+\sum _{n=1}^{\infty }{c}_n{x}^n\right] \\ =\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-1}-2c_2-\sum _{n=3}^{\infty }{c}_{n}n(n-1)x^{n-2}-\sum _{n=1}^{\infty }{c}_{n}n{x}^n+c_0+\sum _{n=1}^{\infty }{c}_n{x}^n \\ =\underset{a_2\to x^1}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-1}}}-\underset{a_3\to x^1}{\underbrace{\sum _{n=3}^{\infty }{c}_{n}n(n-1)x^{n-2}}}-\underset{a_1\to x^1}{\underbrace{\sum _{n=1}^{\infty }{c}_{n}n{x}^n}}+\underset{a_1\to x^1}{\underbrace{\sum _{n=1}^{\infty }{c}_n{x}^n}}+\left[-2c_2+c_0\right] \end{gathered}$$

The shift by substituting the summation index.

Now, we do the shift by replacing the summation index, for the first power series, from $n$to $n+1$, and for the second power series, from $n$to $n+2$

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

This implies that,

$-2c_2+c_0=0\Rightarrow c_2=1/2c_0 \dots \left(4\right)$

And

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

Find the coefficients of the power series in terms

Therefore,

$$\begin{gathered} c_{n+2}(n+2)(n+1)=c_{n+1}(n+1)n-c_n(n-1) \\ {c}_{n+2}=\frac{c_{n+1}(n+1)n-c_n(n-1)}{(n+2)(n+1)}, where n>0 \end{gathered}$$

By using the recurrence relation and (4), we can find the coefficients of the power series in terms.

At$n=1$

$$\begin{gathered} c_3=\frac{c_2(1+1)-c_1·0}{(1+2)(1+1)} \\ =\frac{2c_2}{6} \\ =\frac{2·1/2c_0}{6} \\ =\frac{c_0}{6} \\ =\frac{c_0}{3\times 2\times 1} \dots \left(5\right) \end{gathered}$$

For,$n=2$

$$\begin{gathered} c_4=\frac{c_3(2+1)·2-c_2(2-1)}{(2+2)(2+1)} \\ =\frac{6c_3-c_2}{12} \\ =\frac{6·c_0/6-1/2c_0}{12} \\ =\frac{c_0-1/2c_0}{12} \\ =\frac{c_0}{4\times 3\times 2\times 1} \dots \left(6\right) \end{gathered}$$

Now, using the initial values for $y$and $y\text{'}$we can determine ${\mathrm{C}}_0$and ${\mathrm{C}}_0$from (1), our solution can be expanded as follows:

$$\begin{gathered} y\left(x\right)=c_0+c_{1}x+c_2{x}^2+\dots \\ y\text{'}\left(x\right)=0+c_1+2x+\dots \end{gathered}$$

Therefore,

$$\begin{gathered} y\left(0\right)=c_0 \\ =-2 \\ y\text{'}\left(0\right)=c_1 \\ =6 \end{gathered}$$

And we can write the solution as follows

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^n \\ =c_0+c_{1}x+\sum _{n=2}^{\infty }{c}_n{x}^n \\ =c_0+c_{1}x+\sum _{n=2}^{\infty }\frac{c_0}{n!}{x}^n \\ =\underset{y_1}{\underbrace{c_{1}x}}+\underset{y_2}{\underbrace{c_0\left[1+\sum _{n=2}^{\infty }\frac{1}{n!}{x}^n\right]}} \\ =6x-2\left[1+\sum _{n=2}^{\infty }\frac{1}{n!}{x}^n\right] \end{gathered}$$