Question

$2\frac{dy}{dx}-y\sec x=y^3\tan x$.

Answer 1

$2\frac{dy}{dx}-y\sec x=y^3\tan x$.

$2\frac{dy}{dx}=y^3\tan x+y\sec x$

$2\frac{dy}{dx}=y(y^2\tan x+\sec x)$

$2\frac{dy}{y}=(y^2\tan x+\sec x)dx$

$2\log y=y^2{\sec }^{2}x+\ln |\sec x+\tan x|+C$

$\frac{2}{y^2}\cdot \log y={\sec }^{2}x+\ln |\sec x+\tan x|$

$\frac{2}{y^2}\cdot \log y=\log e^{{\sec }^{2}x}+\ln |\sec x+\tan x|$

$\frac{-2}{y^2}=-1+(\pi +x)\cot (\frac{\pi }{4}+\frac{x}{2})$