Power Series solution of $y''+x^3y'+3x^2y=0$.

Question

Power Series solution of the following differential equation: $$y''+x^3y'+3x^2y=0$$ I started with $y = \sum_{n=0}^{\infty}a_nx^n$, then $y' = \sum_{n=1}^{\infty}na_nx^{n-1}$ and $y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$, then, $$y''+x^3y'+3x^2y= \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}+\sum_{n=1}^{\infty}na_nx^{n+2}+3\sum_{n=0}^{\infty}a_nx^{n+2}$$ but unable to proceed further held needed.

I have also observed that $y''+x^3y'+3x^2y= y''+(x^3y)'$.

Answer 1

Collect like terms \begin{align} y''+x^3y'+3x^2y&= \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}+\sum_{n=1}^{\infty}na_nx^{n+2}+3\sum_{n=0}^{\infty}a_nx^{n+2} \\&= \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n} +\sum_{n=3}^{\infty}(n-2)a_{n-2}x^{n} +3\sum_{n=2}^{\infty}a_{n-2}x^{n} \\ &= \big[2a_2\big] + \big[6a_3\big]x + \big[12a_4+3a_0\big]x^2 +\sum_{n=3}^\infty \big[(n+2)(n+1)a_{n+2} +(n+1)a_{n-2}\big]x^n \end{align}

So to get $0$ as the result, all coefficients must be zero: \begin{align} 2a_2 &= 0 \\ 6a_3 &= 0 \\ 12a_4+3a_0 &= 0 \\ (n+2)(n+1)a_{n+2} +(n+1)a_{n-2} &= 0\quad\text{for }n\ge 3 \end{align} So $a_0, a_1$ are arbitrary, $a_2 = a_3 = 0$, $a_4=-a_0/4$, $$ a_k = -\frac{1}{k}a_{k-4}\quad\text{for }k \ge 5 $$

Answer 2

Another way to do it.

When I was a student, we were taught to never change the index. So, the way we did it :

$$y = \sum_{n=0}^{\infty}a_nx^n\qquad y' = \sum_{n=0}^{\infty}na_nx^{n-1}\qquad y'' = \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}$$ $$y''+x^3y'+3x^2y=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}+\sum_{n=0}^{\infty}na_nx^{n+2}+3\sum_{n=0}^{\infty}a_nx^{n+2}$$ that is to say $$y''+x^3y'+3x^2y=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}+\sum_{n=0}^{\infty}(n+3)a_nx^{n+2}$$

Now, to have the coefficient for $x^m$, the first summation requires $n-2=m$ that is to say $n=m+2$ and the second summation requires $n+2=m$ that is to say $n=m-2$. So, $$(m+2)(m+1)a_{m+2}+(m+1) a_{m-2}=0\implies a_{m+2}=-\frac {a_{m-2}} {m+2}$$

Now, make $m+2=k$ that is to say $m=k-2$ to get $$a_k=-\frac {a_{k-4}} k$$ For sure, you need to identify the very first coefficients $a_0,a_1,a_2,a_3,a_4$ but this is simple as already shown in GEdgar's answer.