Question

Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.)

Answer 1

$y=\displaystyle\sum_{n\ge0}a_nx^n=a_0+\sum_{n\ge1}a_nx^n$

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n=a_1+\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n$

Notice that $y(0)=-3=a_0$, and $y'(0)=2=a_1$.

Substitute these series into the ODE:

$(x-1)y''-xy'+y=0$

$\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^{n+1}-\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n-\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}+\sum_{n\ge0}a_nx^n=0$

Shift the indices to get each series to include a $x^n$ term.

$\displaystyle\sum_{n\ge1}n(n+1)a_{n+1}x^n-\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n-\sum_{n\ge1}na_nx^n+\sum_{n\ge0}a_nx^n=0$

Remove the first term from both series starting at $n=0$ to get all the series starting on the same index $n=1$:

$\displaystyle-2a_2+a_0+\sum_{n\ge1}n(n+1)a_{n+1}x^n-\sum_{n\ge1}(n+1)(n+2)a_{n+2}x^n-\sum_{n\ge1}na_nx^n+\sum_{n\ge1}a_nx^n=0$

$\displaystyle-2a_2+a_0+\sum_{n\ge1}\bigg[n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-(n-1)a_n\bigg]x^n=0$

The coefficients are given recursively by

$\begin{cases}a_0=-3\\a_1=2\\\\a_n=\dfrac{(n-2)(n-1)a_{n-1}-(n-3)a_{n-2}}{n(n-1)}&\text{for }n>1\end{cases}$

Let's see if we can find a pattern to these coefficients.

$a_2=\dfrac{a_0}2=-\dfrac32=-\dfrac3{2!}$

$a_3=\dfrac{2a_2}{3\cdot2}=-\dfrac12=-\dfrac3{3!}$

$a_4=\dfrac{2\cdot3a_3-a_2}{4\cdot3}=-\dfrac18=-\dfrac3{4!}$

$a_5=\dfrac{3\cdot4a_4-2a_3}{5\cdot4}=-\dfrac1{40}=-\dfrac3{5!}$

$a_6=\dfrac{4\cdot5a_5-3a_4}{6\cdot5}=-\dfrac1{240}=-\dfrac3{6!}$

and so on, suggesting that

$a_n=-\dfrac3{n!}$

which is also consistent with $a_0=3$. However,

$a_1=2\neq-\dfrac3{1!}=-3$

but we can adjust for this easily:

$y(x)=-3+2x-\dfrac3{2!}x^2-\dfrac3{3!}x^3-\dfrac3{4!}x^4+\cdots$

$y(x)=5x-3-3x-\dfrac3{2!}x^2-\dfrac3{3!}x^3-\dfrac3{4!}x^4+\cdots$

Now all the terms following $5x$ resemble an exponential series:

$y(x)=5x-3\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

$\implies\boxed{y(x)=5x-3e^x}$