Solving $f''(t) = g(f(t))$ A recent question asked for a solution to $f'(t)= k f^2(t)$.  I could do this.  In fact, it seems that a large class of similar differential equations $f'(t) = g(f(t))$ are not hard to solve.  
This led me to consider $f''(t) = g(f(t))$.  
I started with a specific example: $f''(t)= k f^2(t)$.  I could solve this but I did so by just guessing that there would be a solution of the form $a x^b$.  I found $f(x) = \frac{6}{k x^2}$ as a specific solution. Obviously, for this case and others of this class, translations of a solution as also solutions. 
However, as $g$ becomes more complex, guesswork will probably fail me.  Is there any good way to approach equations of this form?
 A: Here is an attempt.  Unfortunately, without knowing $g$ exactly, it is difficult to write a solution $f(t)$ to $$f''(t)=g\big(f(t)\big)$$ as a function of $t$ explicitly.
Write $x(t):=f(t)$ and $v(t):=f'(t)$.  Then, you have 
$$v\,\frac{\text{d}v}{\text{d}x}=\frac{\text{d}^2x}{\text{d}t^2}=g(x)\,.$$
This implies
$$\frac{1}{2}v^2=G(x)+c\,,$$
where $G$ is an antiderivative of $g$ and $c$ is some constant. That is,
$$\frac{\text{d}x}{\text{d}t}=v=\pm\sqrt{2\,G(x)+2\,c}\,,$$
or
$$\pm \sqrt{2}\,(t-\tau)=H_{c}(x)\,,$$
where $H_{c}$ is an antiderivative of $\dfrac{1}{\sqrt{G+c}}$ and $\tau$ is a constant.  That is, by continuity,
$$x=H_{c}^{-1}\big(+\sqrt{2}\,(t-\tau)\big)\text{ for }+\sqrt{2}\,(t-\tau)\in\text{Range}(H_c)\tag{1}$$
or
$$x=H_{c}^{-1}\big(-\sqrt{2}\,(t-\tau)\big)\text{ for }-\sqrt{2}\,(t-\tau)\in\text{Range}(H_c)\,,\tag{2}$$
given that $H_c$ is injective.

An example with $g(u)=k\,u^2$ for some constant $k\neq 0$.  Then, we can take $G(u)=\dfrac{k}{3}\,u^3$.  That is, we can also take
$$H_c(u)=-\int_{u}^\infty\,\frac{1}{\sqrt{\frac{k}{3}\,s^3+c}}\,\text{d}s\,.$$
I do not know the explicit form of $H_c$ for an arbitrary $c$, but
$$H_0(u)=-\int_u^\infty\,\frac{1}{\sqrt{\frac{k}{3}\,s^3}}\,\text{d}s=-\frac{2\sqrt{3}}{\sqrt{k\,u}}\,.$$
Thus, the range of $H_0$ is $(-\infty,0)$.  On $(-\infty,0)$,
$$H_0^{-1}(z)=\frac{12}{k\,z^2}\text{ for }z<0\,.$$
Combining (1) and (2), a solution is given by
$$x(t)=H_0^{-1}\big(-\sqrt{2}\,|t-\tau|\big)=\frac{6}{k\,(t-\tau)^2}$$
for all $t\in\mathbb{R}\setminus\{\tau\}$.

On the other hand, it is easier to solve for $f$ from $$f'(t)=g\big(f(t)\big)\,,$$
but you will still end up with some implicit relation.  With the same notation as before, we have $$\frac{\text{d}x}{\text{d}t}=g(x)\,,$$
so
$$\gamma(x)=t-\tau$$
for some constant $\tau$, and $\gamma$ is an antiderivative of $\dfrac{1}{g}$.  Therefore, 
$$x=\gamma^{-1}(t-\tau)\text{ for }t-\tau\in\text{Range}(\gamma)\,,$$
provided that $\gamma$ is injective.
For $g(u)=k\,u^2$ with $k\neq 0$, we can take 
$$\gamma(u):=-\int_u^\infty\,\frac{1}{k\,s^2}\,\text{d}s=-\frac{1}{k\,u}\,.$$
That is, $$\gamma^{-1}(z)=-\frac{1}{k\,z}\text{ for }z\neq 0\,,$$
and we get a solution
$$x(t)=-\frac{1}{k\,(t-\tau)}\text{ for all }t\in\mathbb{R}\setminus\{\tau\}\,.$$
A: $$f''(t)=g(f(t))\implies 2f'(t)f''(t)=2g(f(t))f'(t),$$
then
$$f'^2(t)=2\int g(f(t))f'(t)dt+C.$$
Next,
$$\frac{f'(t)}{\sqrt{2\displaystyle\int g(f(t))f'(t)dt+C}}=\pm1$$
and
$$\int\frac{f'(t)}{\sqrt{2\displaystyle\int g(f(t))f'(t)dt+C}}dt=\pm t+C''.$$

With $g(t)=kt^2$,
$$\int\frac{f'(t)}{\sqrt{2\displaystyle\int kf^2(t))f'(t)dt+C}}dt=\int\frac{f'(t)}{\sqrt{\dfrac{2k}3f^3(t)+C}}dt=\pm t+C''.$$
For $C\ne0$, there is no elementary solution.
https://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt(2k%2F3+x%5E3%2BC)
