Can we solve the following ODE

Question

I would like to understand the solution to the following Ode, can we solve that? This there any idea that we can analysis something on that?

$\frac{d^2}{dx^2}u(x)+\sinh(u)=0$.

Thanks.

Answer 1

$$u'u''+u'\sinh u=0$$ integrates as $$u'^2+2\cosh u=c.$$

Then

$$\int\frac{du}{\sqrt{c-2\cosh u}}=x+c'.$$

This integral can be solved by the change of variable $t=e^u$, leading to the inverse of a quadratic polynomial, the antiderivative of which is an arc tangent or an argument of hyperbolic tangent.

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Update:

The square root was missing. This makes the integral non elementary.

Answer 2

Writing your equation in the form $$\frac{d^2u(x)}{dx^2}=-\sinh(u(x))$$ then we multiply by $$\frac{du(x)}{dx}$$ and integrate $$\int\frac{du(x)}{dx}\cdot \frac{d^2u(x)}{dx^2}dx=\int-\frac{du(x)}{dx}\sinh(u(x))dx$$ This is $$\frac{1}{2}\left(\frac{du(x)}{dx}\right)^2=-\cosh(u(x))+C$$ I hope you can finish now.