Solving a differential equation with trigonometric functions

Question

How could I approach $$\displaystyle y - A\cdot \sin\biggl(\frac{d^2y}{dx^2}\biggr) = 0\;?$$

Answer 1

Hint:

Let $s=y’$.

Then, $$y=A\sin\frac{ds}{dx}=A\sin\frac{ds}{dy}\frac{dy}{dx}=A\sin s\frac{ds}{dy}$$

Rearranging yields $$sds=dy\sin^{-1}\frac yA$$

You should finally obtain $$\sqrt2 x+C_2=\int dy\left(y\sin^{-1}\frac yA+A\sqrt{1-\frac{y^2}{A^2}}+C_1\right)^{-1/2}$$

The integral can be rewritten into $$A\int\frac{\cos g}{\sqrt{C_1+A(g\sin g+\cos g)}}dg$$ by letting $y=A\sin g$.

I do not expect the integral is elementary.