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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.

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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 $f$.

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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).

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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}

$f(x) = \int{ \left(\int g(x) dx +C_0\right)^{-1}}dx+C_1$