We have a velocity-dependent potential system $$(1 + \lambda x^2)\ddot{x}+ \lambda x \dot{x}^2+ \omega_0^2x = 0$$ for some constant $\lambda$.
Given the initial condition, $x(0) = A$ and $\dot{x}(0)=0$, how do we find the solution to this?
If $\lambda = 0$, this reduces to harmonic oscillator solution and hence I feel this one will be a modified form of $A \cos(f(\omega))+B\sin(f(\omega))$.
Multiply with $2\dot x$ and integrate, which is successful and gives $$ (1+λx^2)(2\dot x\ddot x)+(2λx\dot x)\dot x^2+ω^2_0(2x\dot x)=0 \\~\\ (1+λx^2)\dot x^2+ω^2_0x^2=C=ω^2_0R^2 $$ The second term $λ(2x\dot x)\dot x^2$ in the first equation combines with the first $(1+λx^2)(2\dot x\ddot x)$ as the derivative of the product $(1+λx^2)\dot x^2$ in the integrated equation. The integration constant, as sum of squares and otherwise positive factors, is positive. One can chose to write it as $0<C=ω_0^2R^2$ for later convenience.
This is a circle equation where the points $(\sqrt{1+λx^2}\dot x, ω_0x)$ lie on a circle of radius $ω_0R$. Parametrize by the angle to get $$\sqrt{1+λx^2}\dot x=ω_0R\cos(\theta(t)),~~ x=R\sin(\theta(t)),$$ then the derivative of the last equation has to be compatible with the first, $$ \dot x=R\cos(θ(t))\dot θ(t) \\~\\\implies \dot θ(t)=\frac{ω_0}{\sqrt{1+λx^2}}=\frac{ω_0}{\sqrt{1+λR^2\sin^2(\theta(t))}} $$ This now is a separable ODE for $θ(t)$ and thus a "more simple" quadrature problem. Also, for $λ=0$ one recovers the harmonic oscillation with $θ(t)=θ_0+ω_0t$.