It has to be solved with Laplace transform and then converted to Bessel equation.
$L(xy'') = -\frac{dL(y'')}{ds}$
$L(4xy) = -\frac{4dL(y)}{ds}$
$L(y'') = s²L(y) - sy(0) - y'(0) = s²L(y) -3s$
$L(y') = sL(y) - sy(0) - y(0) = sL(y) - 3$
$-\frac{d(s²L(y)-3s)}{ds} + sL(y)-\frac{4dL(y)}{ds} =0$
$-\frac{d(s²L(y))}{ds} + 3 + sL(y) - 3 -\frac{4dL(y)}{ds} =0$
$-\frac{s²L(y)}{ds} + sL(y) -\frac{4dL(y)}{ds} =0$
$-\frac{L(y)(s²+4)}{ds} + sL(y) =0$ (1)
$\frac{dL(y)}{L(y)} = \frac{sds}{s²+4}$
Integrating both sides
$ln(L(y)) = \frac{ln(s²+4)}{2} + c$
$L(y) = c\sqrt{s²+4}$
which won't lead me to the right answer.
I realized that if at (1) I use $\frac{L(y)(s² + 4)}{ds} + sL(y) =0$ instead I'll get the right answer according to wolfram, but I can't see what I'm doing wrong to end up with that negative sign.
Managed to find my mistake, I solved as if s² wasn't part of the derivative at $-\frac{d(s²L(y))}{ds}$
$-\frac{d(s²L(y))}{ds} = -2sL(y) - \frac{s²d(L(y))}{ds}$ $-2sL(y) - \frac{s²d(L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0$ $ - \frac{s²d(L(y))}{ds} - sL(y) -\frac{4dL(y)}{ds} =0$ $ \frac{s²d(L(y))}{ds} + sL(y) +\frac{4dL(y)}{ds} =0$ $ \frac{d(L(y))(s²+4)}{ds} + sL(y) =0$