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 $a_n$:

$a_n=\dfrac{a_{n-3}}n\text{ for }n\ge3$

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

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

$k=1\implies n=4\implies a_4=\dfrac{a_1}4=0$

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

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

$k=1\implies n=3\implies a_3=\dfrac{a_0}3$

$k=2\implies n=6\implies a_6=\dfrac{a_3}6=\dfrac{a_0}{3\cdot6}=\dfrac{a_0}{3^2(2\cdot1)}$

$k=3\implies n=9\implies a_9=\dfrac{a_6}9=\dfrac{a_0}{3^3(3\cdot2\cdot1)}a_0$

and so on, with the general pattern

$a_{3k}=\dfrac{a_0}{3^kk!}$

Then the series solution is

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

$y=\displaystyle a_0\sum_{k\ge0}\frac{x^{3k}}{3^kk!}$

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

whose first four terms are

$\boxed{a_0\left(1+\dfrac{x^3}3+\dfrac{x^6}{18}+\dfrac{x^9}{162}\right)}$