# 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.

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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

$$f''(x) + g(x) f'(x)^2 = 0$$

$$f''(x) = - g(x) f'(x)^2$$

$$\frac{f''(x)}{f'(x)^2} = - g(x)$$

$$-d\left\{\frac{1}{f'(x)}\right\} = - g(x)dx$$

$$\frac{1}{f'(x)} = \int g(x) dx +C_0$$

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

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

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