Question

Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a

y + (x - 2)y+y=0

Answer 1

I'll assume the ODE is actually

`y''+(x-2)y'+y=0`

Look for a series solution centered at
`x=2`, with

`y=\displaystyle\sum_(n\ge0)c_n(x-2)^n`

`\implies y'=\displaystyle\sum_(n\ge0)(n+1)c_(n+1)(x-2)^n`

`\implies y''=\displaystyle\sum_(n\ge0)(n+2)(n+1)c_(n+2)(x-2)^n`

with
`c_0=y(2)=2` and
`c_1=y'(2)=0`.

Substituting the series into the ODE gives

`\displaystyle\sum_(n\ge0)(n+2)(n+1)c_(n+2)(x-2)^n+\sum_(n\ge0)(n+1)c_(n+1)(x-2)^(n+1)+\sum_(n\ge0)c_n(x-2)^n=0`

`\displaystyle\sum_(n\ge0)(n+2)(n+1)c_(n+2)(x-2)^n+\sum_(n\ge1)nc_n(x-2)^n+\sum_(n\ge0)c_n(x-2)^n=0`

`\displaystyle2c_2+c_0+\sum_(n\ge1)(n+2)(n+1)c_(n+2)(x-2)^n+\sum_(n\ge1)nc_n(x-2)^n+\sum_(n\ge1)c_n(x-2)^n=0`

`\displaystyle2c_2+c_0+\sum_(n\ge1)\bigg((n+2)(n+1)c_(n+2)+(n+1)c_n\bigg)(x-2)^n=0`

`\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_(n+2)+c_n=0&\text{for }n>0\end{cases}`

• If
  `n=2k` for integers
  `k\ge0`, then

`k=0\implies n=0\implies c_0=c_0`

`k=1\implies n=2\implies c_2=-\frac{c_0}2=(-1)^1(c_0)/(2^1(1))`

`k=2\implies n=4\implies c_4=-\frac{c_2}4=(-1)^2(c_0)/(2^2(2\cdot1))`

`k=3\implies n=6\implies c_6=-\frac{c_4}6=(-1)^3(c_0)/(2^3(3\cdot2\cdot1))`

and so on, with

`c_(2k)=(-1)^k(c_0)/(2^kk!)`

• If
  `n=2k+1`, we have
  `c_(2k+1)=0` for all
  `k\ge0` because
  `c_1=0` causes every odd-indexed coefficient to vanish.

So we have

`y(x)=\displaystyle\sum_(k\ge0)c_(2k)(x-2)^(2k)=\sum_(k\ge0)(-1)^k((x-2)^(2k))/(2^(k-1)k!)`

Recall that

`e^x=\displaystyle\sum_(n\ge0)(x^k)/(k!)`

The solution we found can then be written as

`y(x)=\displaystyle2\sum_(k\ge0)\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k`

`\implies\boxed{y(x)=2e^(-(x-2)^2/2)}`