1 Answer

Tony Bogdanov · Answer 1 · 2018-02-23T10:10:51+0000

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

with

. 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