I am looking at a SDOF system where the spring stiffness, $K$, is a function of the displacement, $x$, such that $K=x^2$. Using the dynamic equilibrium, the problem can be written as:
$$Mx' ' + Kx = 0 \mid K=x^2$$
$$Mx' ' + x^3 = 0 $$
This is differentiated with respect to time.
I am looking for a valid solution to this problem, specifically, the natural frequency.
You have $$mx''=-x^{3}$$ Multiply by $x'$ $$mx'x''=-x'x^{3}$$ Integrate $$(x')^{2}+\frac{x^{4}}{2m}=c$$ Than $$x'=\pm\sqrt{c-\frac{x^{4}}{2m}}$$ This gives $$t-t_{0}=\pm\int_{x_{0}}^{x}\frac{dy}{\sqrt{c-\frac{y^{4}}{2m}}}$$ We now have to evaluate $$I=\int_{x_{0}}^{x}\frac{dy}{\sqrt{c-\frac{y^{4}}{2m}}}$$ We make the substitution $y=\big(\frac{2m}{c}\big)^{1/4}z$ $$I=\frac{(2m)^{1/4}}{c^{3/4}}\int_{\big(\frac{c}{2m}\big)^{1/4}x_{0}}^{\big(\frac{c}{2m}\big)^{1/4}x}\frac{dz}{\sqrt{1-z^{4}}}=\frac{(2m)^{1/4}}{c^{3/4}}\Big[\mathfrak{F}\Big{(}\arcsin\Big{\{}\big(\frac{c}{2m}\big)^{1/4}x\Big{\}}|-1\Big{)}-\mathfrak{F}\Big{(}\arcsin\Big{\{}\big(\frac{c}{2m}\big)^{1/4}x_{0}\Big{\}}|-1\Big{)}\Big]$$ Where $\mathfrak{F}(a|b)$ is the elliptic integral of the first kind http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html