Question

Solve the following differential equations:
${\cos }^{2}x\frac{dy}{dx}+y=\tan x$

Answer 1

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

$\frac{1}{y^2}\frac{dy}{dx}-\frac{\tan x}{y}=\sec x$

Let $z=\frac{1}{y}$

$\frac{dz}{dx}=\frac{-1}{y^2}\frac{dy}{dx}$

$\frac{-dz}{dx}=\frac{1}{y^2}\frac{dy}{dx}$

$\frac{-dz}{dx}-z\tan x=\sec x$

$\frac{dz}{dx}+z\tan x=-\sec x$

Integrating factor (I.F)= $e^{\int f\left(x\right)dx}=e^{\int \tan xdx}$

$=e^{\log \sec x}=secx$

$Z(I.F)=\int Q\left(x\right).(I.F)dx+c$

$z\sec x=\int -secx.\sec xdx+c$

$z\sec x=\int -{\sec }^{2}xdx+c$

$z\sec x=-\tan x+c$

$\frac{1}{y}\sec x=-\tan x+c$

$y=\frac{\sec x}{-\tan x+c}$