general solution of the FODE $\frac{1}{\sin(x)} \frac{dy}{dx} = y\sec(x) -2$ What's the general solution of the first order differential equation ? $$ \begin{align*} \frac{1}{\sin(x)} \frac{dy}{dx} = y\sec(x)-2, \quad 0<x< \frac{\pi}{2} \end{align*} $$ my solution is as follows but i think it's wrong $$ \frac{dy}{dx} = y\tan(x) -2\sin(x) \\\ \frac{dy}{dx} - y\tan(x) = -2\sin(x) \\\ e^{-\int \tan(x)dx} = \cos(x)\\\ y\,cos(x) = \int(-2\cos(x)\sin(x))dx\\\ y =\cos(x) + \frac{C}{\cos(x)} $$

It seems correct to me...

$$y'-y\tan(x)=-2\sin(x)$$ $$y'-y\frac {\sin(x)}{\cos(x)}=-2\sin(x)$$ $$\frac {y'\cos(x)-y\sin(x)}{\cos(x)}=-2\sin(x)$$ $$({y}{\cos(x)})'=-\sin(2x)$$ Integrating .. $${y(x)}{\cos(x)}=-\int \sin(2x) dx$$ $$y(x)=\frac 1 {\cos(x)}(\frac {\cos(2x)}2+K)$$

The same answer as yours.....with $\cos(2x)=2cos^2(x)-1$

$$\boxed {y(x)=\frac C {\cos(x)}+\cos(x)}$$