Asymptotic behavior of solution to nonlinear ODE I am trying to determine the asymptotic behavior as $t \to \infty$ of the solution to the IVP:
$$y''(t) - y(t) + \frac{1}{[y(t)]^3} = 0\\y(0) = 1;\, y'(0) = 1$$
Without the nonlinearity, we have solutions of the form $e^{\pm t}$.  I think the nonlinear solution has similar behavior, but I'm not sure how to show this or determine the precise asymptotic form.
 A: Let's see if we can solve this exactly.
Inverting the function gives us:
$$-\frac{t''}{t'^3} - y + \frac{1}{y^3} = 0 \\ t''+\left(y-\frac{1}{y^3} \right) t'^3=0$$
$$t'(y)=f(y)$$
$$f'=\left(\frac{1}{y^3}-y \right) f^3$$
$$-\frac{1}{2 f^2}=-\frac{1}{2 y^2}-\frac{y^2}{2}+C$$
$$\frac{1}{f^2}=\frac{1}{y^2}+y^2+C$$
$$f^2=\frac{y^2}{1+Cy^2+y^4}$$
$$f=\frac{y}{\sqrt{1+2C_1y^2+y^4}}$$
$$t(y)= \int \frac{y ~dy}{\sqrt{1+2C_1y^2+y^4}}=\frac{1}{2} \int \frac{du}{\sqrt{1+2C_1 u+u^2}}=\frac{1}{2} \log \left(C_1+u+\sqrt{1+2C_1 u+u^2} \right)+C_2$$
So we have:
$$t(y)=\frac{1}{2} \log \left(C_1+y^2+\sqrt{1+2C_1 y^2+y^4} \right)+C_2$$
Substituting the first condition gives us:
$$C_2=-\frac{1}{2} \log \left(C_1+1+\sqrt{2(C_1+1)) } \right)$$
For the second condition we find:
$$t'(1)=\frac{1}{\sqrt{2(C_1+1)}}=1$$
$$C_1=-\frac{1}{2}$$
$$C_2=-\frac{1}{2} \log \frac{3}{2}$$
So we get finally the exact implicit solution:
$$t(y)=\frac{1}{2} \log \left(y^2-\frac12+\sqrt{y^4-y^2+1} \right)-\frac{1}{2} \log \frac{3}{2}$$

We can find $y(t)$ by solving the above equation, which would reduce to a quadratic one:
$$y^2-\frac12+\sqrt{y^4-y^2+1}=\frac{3}{2} e^{2t}$$
$$\sqrt{y^4-y^2+1}=\frac{3}{2} e^{2t}+\frac12 -y^2$$
$$y^4-y^2+1=y^4-\left(1+3 e^{2t} \right)y^2+\left(\frac{1}{2}+\frac{3}{2} e^{2t} \right)^2 $$
$$3 e^{2t} y^2=\left(\frac{1}{2}+\frac{3}{2} e^{2t} \right)^2 -1$$
$$y(t)=  \frac{e^{-t}}{2\sqrt{3}} \sqrt{\left(1+3 e^{2t} \right)^2 -4}$$

$$y(t)=  \frac{e^{-t}}{2} \sqrt{3 e^{4t}+2e^{2t}  -1}$$

This is the exact solution, which can be checked by direct substitution. From this we can find the asymptotic.
For $t \to +\infty$ we have:
$$y(t) \asymp \frac{\sqrt{3} }{2}e^{t}$$
The asymptotic gives a very good approximation for $t>2$, see the plot (blue is the exact solution, orange the asymptotic):

