Differential Equation $f''(x)+\frac{(n-1)(f'(x))^2}{\sinh(x)}=0 $ How do I solve the following differential equation:
$$
f''(x)+\frac{(n-1)(f'(x))^2}{\sinh(x)}=0
$$
under the boundary conditions $f(1)=1$ and $\lim_{x\to\infty}f(x)=0$.
More generally, how to solve
$$
f''(x)+g(x)(f'(x))^2=0
$$
for some known function $g(x)$ for the same boundary conditions.
 A: Let $u=f'$. Then $-\frac{u'}{u^2} = g$ and so $(\frac{1}{u})'= g$. Integrate both sides to find $u$ and then integrate once again to find $f$.
A: To be more explicit (now that some time has passed), this particular equation can be restated as $(\frac{1}{u})'= \frac{(n-1)}{\sinh(x)}$, so $\frac{1}{f'(x)}=(n-1)(\ln(\tanh(x/2))+A)$
However, Wolfram Alpha indicates that the A=0 case has 1/f'(x) approaching 0 in the limit, which means that f'(x) will not approach 0 for any finite value of A, so it seems like f(x) will never approach a finite limit at infinity (Wolfram Alpha was also not able to find an explicit formula for f(x), even in the A=0 case, so the boundary-value problem is difficult to solve by brute force).
A: The general approach would be
$$f''(x) + g(x) f'(x)^2 = 0$$
$$f''(x) = - g(x) f'(x)^2 $$
$$\frac{f''(x)}{f'(x)^2} = - g(x) $$
$$-d\left\{\frac{1}{f'(x)}\right\} = - g(x)dx $$
$$\frac{1}{f'(x)} = \int g(x) dx +C_0$$
$$f'(x) = \left(\int g(x) dx +C_0\right)^{-1}$$
$$f(x) = \int{ \left(\int g(x) dx +C_0\right)^{-1}}dx+C_1$$
