How solve this differential equation?

Question

This is the differential equations:

$$ y'+ a[sin(x)]+ b[cos(x)]/y =0 $$

where $a,b \in \mathbb{R}$$and$ $b>0$

How could i solve? Thank you in advance for any idea or proposed solution.

Answer 1

$y'+a\sin x+\dfrac{b\cos x}{y}=0$

$y\dfrac{dy}{dx}=-ay\sin x-b\cos x$

This belongs to an Abel equation of the second kind.

Let $t=a\cos x$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=-a(\sin x)\dfrac{dy}{dt}$

$\therefore-ay(\sin x)\dfrac{dy}{dt}=-ay\sin x-b\cos x$

$y\dfrac{dy}{dt}=y+\dfrac{b}{a\tan x}$

$y\dfrac{dy}{dt}-y=\dfrac{bt}{a\sqrt{a^2-t^2}}$

This belongs to an Abel equation of the second kind in the canonical form.

Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf