General solution of a differential equation $x''+{a^2}x+b^2x^2=0$

Question

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.

Answer 1

Note that the derivative of $(x')^2$ (respectively, $x^2$, $x^3$) is $2x'x''$ (respectively, $2xx'$, $3x^2x'$). Multiplying by $6x'$, we get $$(3)(2x'x'')+(3a^2)(2xx')+(2b^2)(3x^2x')=0$$ and integrating $$3(x')^2+3a^2x^2+2b^2x^3=C$$ Then $x'=\sqrt{C-3a^2x^2-2b^2x^3}/3$, which is variables-separable. But the integral it leads to is not elementary; as Graham notes, it leads to elliptic functions.

Answer 2

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.