Question

Solve y' -x^2y = 0 using power series and write the first four terms of the power series

Answer 1

We're looking for a solution of the form

`y=\displaystyle\sum_(n\ge0)a_nx^n`

with derivative

`y'=\displaystyle\sum_(n\ge0)(n+1)a_(n+1)x^n`

Note that
`x=0\implies y(0)=a_0`.

Substituting into the ODE gives

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

The first series starts with a constant term, while the second starts with a quadratic term, so we should pull out the first two terms of the first series and have it start at
`n=2`, then shift the index on the second series to achieve the same effect, which allows us to condense the left side as

`a_1+2a_2x+\displaystyle\sum_(n\ge2)\bigg((n+1)a_(n+1)-a_(n-2)\bigg)x^n=0`

so that the series solution's coefficients are given according to the recurrence

`\begin{cases}a_0=a_0\\a_1=a_2=0\\(n+1)a_(n+1)-a_(n-2)=0&\text{for }n\ge2\end{cases}`

We can simplify the latter equation somewhat to get it in terms of
`a_n`:

`a_n=\frac{a_(n-3)}n\text{ for }n\ge3`

This shows dependency between coefficients that are 3 indices apart, so we check 3 cases:

• If
  `n=3k+1`, where
  `k\ge0` is an integer, then

`k=0\implies n=1\implies a_1=0`

`k=1\implies n=4\implies a_4=\frac{a_1}4=0`

and so on for all such
`n`, giving

`a_(3k+1)=0`

• If
  `n=3k+2`, then

`k=0\implies n=2\implies a_2=0`

and we get the same conclusion as before,

`a_(3k+2)=0`

• If
  `n=3k`, then

`k=0\implies n=0\impiles a_0=a_0`

`k=1\implies n=3\implies a_3=\frac{a_0}3`

`k=2\implies n=6\implies a_6=\frac{a_3}6=(a_0)/(3\cdot6)=(a_0)/(3^2(2\cdot1))`

`k=3\implies n=9\implies a_9=\frac{a_6}9=(a_0)/(3^3(3\cdot2\cdot1))a_0`

and so on, with the general pattern

`a_(3k)=(a_0)/(3^kk!)`

Then the series solution is

`y=\displaystyle\sum_(k\ge0)\bigg(a_(3k)x^(3k)+a_(3k+1)x^(3k+1)+a_(3k+2)x^(3k+2)\bigg)`

`y=\displaystyle a_0\sum_(k\ge0)(x^(3k))/(3^kk!)`

`y=\displaystyle a_0\sum_(k\ge0)\frac{\left(\frac{x^3}3\right)^k}{k!}`

whose first four terms are

`\boxed{a_0\left(1+\frac{x^3}3+(x^6)/(18)+(x^9)/(162)\right)}`