If $\int {yy''dx} = 3xy$ is it possible to find y'?

Disclaimer: I managed to get a recursive definition of the coefficients of the Maclaurin series, but not a closed form.

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From the ODE $$yy'' - (3xy)' = 0$$

This expands to $$yy'' - 3xy' -3y = 0$$

Let $y = \sum_{n=0}^{\infty} a_nx^n$. Then the relation is $$\left(\sum_{n=0}^{\infty} a_nx^n\right)\left(\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}\right) + \sum_{n=0}^{\infty} (-3na_n-3a_n)x^{n} = 0$$

Expanding this, it becomes

$$\sum_{n=0}^{\infty} x^n \left(-3na_n - 3a_n + \sum_{m=0}^n (m+2)(m+1)a_{m+2}a_{n-m}\right) = 0$$

For it to be $0$, each of the coefficients of $x^n$ must be $0$, so it is $$(-3n-3)a_n + \sum_{m=0}^n (m+2)(m+1)a_{m+2}a_{n-m} = 0$$

for all nonnegative integer $n$. Assume that $y(0) = a_0 \not = 0$ and $y'(0) = a_1$ are fixed at some values. Then solving for $a_{n+2}$, it is $$a_{n+2} = \frac{(3n+3)a_n - \sum_{m=0}^{n-1} (m+2)(m+1)a_{m+2}a_{n-m}}{(n+2)(n+1)a_0}$$

For the first few values, this comes out to $a_2 = \frac{3}{2}$, $a_3 = \frac{a_1(3-a_2)}{3a_0} = \frac{a_1}{2a_0}$, $a_4 = \frac{9a_{2}-2a_{2}a_{2}-6a_{3}a_{1}}{12a_{0}} = \frac{3a_{0}-a_{1}^{2}}{4a_{0}^{2}}$.