The Bernouli's equation
$\frac{d y}{d x}-y \tan x=\frac{\sin x \cos ^2 x}{y^2}$ can be transformed to
Answer & explanation
Correct answer: option 3
We have,
$y^2 \frac{d y}{d x}-y^3 \tan x=\sin x \cos ^2 x$
Putting $y^3=z$ and $y^2 \frac{d y}{d x}=\frac{1}{3} \frac{d z}{d x}$, we get
$\frac{1}{3} \frac{d z}{d x}-z \tan x=\sin x \cos ^2 x$
$\Rightarrow \frac{d z}{d x}-3 z \tan x =3 \sin x \cos ^2 x$