Question

Solve this differential Equation by using power series y''-x^2y=o

Answer 1

We're looking for a solution

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

which has second derivative

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

Substituting these into the ODE gives

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

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

$\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty a_{n-2}x^n=0$

$\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty\bigg((n+2)(n+1)a_{n+2}-a_{n-2}\bigg)x^n=0$

Right away we see $a_2=a_3=0$, and the coefficients are given according to the recurrence

$\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_2=0\\a_3=0\\n(n-1)a_n=a_{n-4}&\text{for }n\ge4\end{cases}$

There's a dependency between terms in the sequence that are 4 indices apart, so we consider 4 different cases.

• If $n=4k$, where $k\ge0$ is an integer, then

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

$k=1\implies n=4\implies a_4=\dfrac{a_0}{4\cdot3}=\dfrac2{4!}a_0$

$k=2\implies n=8\implies a_8=\dfrac{a_4}{8\cdot7}=\dfrac{6\cdot5\cdot2}{8!}a_0$

$k=3\implies n=12\implies a_{12}=\dfrac{a_8}{12\cdot11}=\dfrac{10\cdot9\cdot6\cdot5\cdot2}{12!}a_0$

and so on, with the general pattern

$a_{4k}=\dfrac{a_0}{(4k)!}\displaystyle\prod_{i=1}^k(4i-2)(4i-3)$

• If $n=4k+1$, then

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

$k=1\implies n=5\implies a_5=\dfrac{a_1}{5\cdot4}=\dfrac{3\cdot2}{5!}a_1$

$k=2\implies n=9\implies a_9=\dfrac{a_5}{9\cdot8}=\dfrac{7\cdot6\cdot3\cdot2}{9!}a_1$

$k=3\implies n=13\implies a_{13}=\dfrac{a_9}{13\cdot12}=\dfrac{11\cdot10\cdot7\cdot6\cdot3\cdot2}{13!}a_1$

and so on, with

$a_{4k+1}=\dfrac{a_1}{(4k+1)!}\displaystyle\prod_{i=1}^k(4i-1)(4i-2)$

• If $n=4k+2$ or $n=4k+3$, then

$a_2=0\implies a_6=a_{10}=\cdots=a_{4k+2}=0$

$a_3=0\implies a_7=a_{11}=\cdots=a_{4k+3}=0$

Then the solution to this ODE is

$\boxed{y(x)=\displaystyle\sum_{k=0}^\infty a_{4k}x^{4k}+\sum_{k=0}^\infty a_{4k+1}x^{4k+1}}$