Question

Solve this differential equation using power series and indicial roots about (0,0):

Answer 1

Let $y=\displaystyle\sum_{k\ge0}a_kx^k$, so that

$y'=\displaystyle\sum_{k\ge1}ka_kx^{k-1}$
$y''=\displaystyle\sum_{k\ge2}k(k-1)a_kx^{k-2}$

Substituting into the ODE gives

$\displaystyle3x\sum_{k\ge2}k(k-1)a_kx^{k-2}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0$
$\displaystyle3\sum_{k\ge2}k(k-1)a_kx^{k-1}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0$

The first series starts with a linear term, while the other two start with a constant. Extract the first term from each of the latter two series:

$\displaystyle6\sum_{k\ge1}ka_kx^{k-1}=6\sum_{k\ge2}ka_kx^{k-1}+6a_1$
$\displaystyle\sum_{k\ge0}a_kx^k=\sum_{k\ge1}a_kx^k+a_0$

Finally, to get the series to start at the same index, shift the index of the first two series by replacing $k$ with $k+1$. Then the ODE becomes

$\displaystyle3\sum_{k\ge1}k(k+1)a_{k+1}x^k+6\sum_{k\ge1}(k+1)a_{k+1}x^k+\sum_{k\ge1}a_kx^k+6a_1+a_0=0$

which can be consolidated to get

$\displaystyle\sum_{k\ge1}\bigg[(3k(k+1)+6(k+1))a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0$
$\displaystyle\sum_{k\ge1}\bigg[3(k+1)(k+2)a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0$

You're fixing the solution so that it contains the origin, which means

$y(0)=\displaystyle\sum_{k\ge0}a_kx^k=a_0=0$

which in turn means $a_1=0$. With the given recurrence, it follows that $a_k=0$ for all $k\ge2$, so the solution would be $y=0$. This is to be expected, since $x=0$ is clearly a singular point for the ODE.