Gabriela Duarte

2022-01-27

Solving a differential equation with trigonometric functions
How could I approach
$y-A\cdot \mathrm{sin}\left(\frac{{d}^{2}y}{{dx}^{2}}\right)=0$

helsinka04

Expert

Hint:
Let s=y′.
Then,
$y=A\mathrm{sin}\frac{ds}{dx}=A\mathrm{sin}\frac{ds}{dy}\frac{dy}{dx}=A\mathrm{sin}s\frac{ds}{dy}$
Rearranging yields
$sds=dy{\mathrm{sin}}^{-1}\frac{y}{A}$
You should finally obtain
$\sqrt{2}x+{C}_{2}=\int dy{\left(y{\mathrm{sin}}^{-1}\frac{y}{A}+A\sqrt{1-\frac{{y}^{2}}{{A}^{2}}}+{C}_{1}\right)}^{-\frac{1}{2}}$
The integral can be rewritten into
$A\int \frac{\mathrm{cos}g}{\sqrt{{C}_{1}+A\left(g\mathrm{sin}g+\mathrm{cos}g\right)}}dg$
by letting $y=A\mathrm{sin}g$
I do not expect the integral is elementary

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