Innocent-looking ODE

Question

I am struggling to show that the solution of

$$y'' (x) + b \exp(y) = 0$$

with initial conditions $y(0) = y_0$ and $y'(0) = 0$ is

$$y (x) - y_0 - \ln \cosh(x) \sqrt{2 b \exp(y_0)}$$

I have used the substitution $p = y'$ and got as far as getting

$$\frac{dy}{\sqrt{1 - \exp(y)}} = \pm \sqrt{2b} + \text{Constant}$$

Intergating and getting the constast seems to have me stumped. Can anyone halp, please.

This is evidently an example from Jaeger & Starfield - Applied Math book from years ago

Sorry guys, I am retired and struggle with Latex etc ! Regarding the last comment re' equality, this is what you have to do ,surely ? This solution will differ depending on the the choice of the constant of integration from the solution of pdp = -bexp(y)dy, surely

Answer 1

$$\frac{dy}{dx}=-2\tanh(x)\sqrt{\frac12be^{y_0}}$$ and

$$\frac{d^2y}{dx^2}=-2\text{ sech}^2(x)\sqrt{\frac12be^{y_0}}.$$

Then

$$be^y=b\text{ sech}^{2\sqrt{\frac12be^{y_0}}}(x)$$ doesn't seem correct. Review the parenthesing.

Answer 2

The equation you should get using $p=y'$ is $y' = \pm\sqrt{2b(e^{y_0} - e^y)}$. Ain't gonna elaborate it further.