TITLE: Solving $f''(t) = g(f(t))$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: $$f''(t)=g(f(t))\implies 2f'(t)f''(t)=2g(f(t))f'(t),$$
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)