Solve $\frac {dy}{dx} + y\tan x=y^3 \sec x$ Solve 
$$\dfrac {dy}{dx}+y\tan x= y^3 \sec x$$
My Attempt:
$$\dfrac {dy}{dx} + y\tan x= y^3 \sec x$$
Dividing both sides $y^3$, 
$$y^{-3} \dfrac {dy}{dx} + y^{-2} \tan x= \sec x$$
Put $y^{-2} = z$
$$(-2)y^{-2} \dfrac {dy}{dx} = \dfrac {dz}{dx}$$
$$y^{-3} \dfrac {dy}{dx} = \dfrac {-1}{2} \dfrac {dz}{dx}$$
Then,
$$-\dfrac {1}{2} \dfrac {dz}{dx} + z.\tan x= \sec x$$
$$-\dfrac {dz}{dx} + (2\tan x) z =2\sec x$$

$$\frac{y'}{y^3}+\frac{\tan x}{y^2}=\frac{1}{\cos x}$$
$$ z'-2z\tan x=-\frac{2}{\cos x}$$
The integrating factor would be now:
$$e^{-2\int \tan x}dx=e^{\ln(\cos^2 x)+C}$$
So multiply both sides by $\cos^2 x$
$$(z\cos^2 x)'=-2\cos x\Rightarrow z\cos^2 x=-2\sin x+C$$
$$y=\sqrt{\frac{\cos^2 x}{-2\sin x+C}}$$
