The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for \(y\)
\(\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0\)
which indeed gives the recurrence you found,
\(a_{n+3}=-\dfrac{n-3}{n+3}a_n\)
but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that \(a_2=0\), and substituting this into the recurrence, you find that \(a_2=a_5=a_8=\cdots=a_{3k-1}=0\) for all \(k\ge1\).
Next, the linear term tells you that \(6a_0+6a_3=0\), or \(a_3=a_0\).
Now, if \(a_0\) is the first term in the sequence, then by the recurrence you have
\(a_3=a_0\)
\(a_6=-\dfrac{3-3}{3+3}a_3=0\)
\(a_9=-\dfrac{6-3}{6+3}a_6=0\)
and so on, such that \(a_{3k}=0\) for all \(k\ge2\).
Finally, the quadratic term gives \(6a_1-12a_4=0\), or \(a_4=\dfrac12a_1\). Then by the recurrence,
\(a_4=\dfrac12a_1\)
\(a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1\)
\(a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1\)
\(a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1\)
and so on, such that
\(a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\)
for all \(k\ge2\).
Now, the solution was proposed to be
\(y=\displaystyle\sum_{n\ge0}a_nx^n\)
so the general solution would be
\(y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots\)
\(y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)\)
\(y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)\)
