General solution of a differential equation $x''+{a^2}x+b^2x^2=0$ How do you derive the general solution of this equation:$$x''+{a^2}x+b^2x^2=0$$where a and b are constants.
Please help me to derive solution thanks a lot.

First make the substitution:
$x=-\,{\frac {6y}{{b}^{2}}}-\,{\frac {{a}^{2}}{{2b}^{2}}}$ 
This will give you the differential equation:
$y^{''} =6y^{2}-\frac{a^{4}}{24}$ which is to be compared with the second order differential equation for the Weierstrass elliptic function ${\wp}(t-\tau_{0},g_2,g_3)$: 
${\wp}^{''} =6{\wp}^{2}-\frac{g_{2}}{2}$ 
Where $g_{2}$, $g_3$ are known as elliptic invariants. It then follows that the solution is given by: 
$y={\wp}(t-\tau_{0},\frac{a^{4}}{12},g_3)$ 
$x=\,-{\frac {6}{{b}^{2}}}{\wp}(t-\tau_{0},\frac{a^{4}}{12},g_3)-\,{\frac {{a}^{2}}{{2b}^{2}}}$ 
Where $\tau_0$ and $g_3$ are constants determined by the initial conditions and $t$ is the function variable.
