Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

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 $f$.

share|improve this answer

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).

share|improve this answer

The general approach would be

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

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

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

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

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

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.