$\cos^2 x \frac{dy}{dx}$ + y = tan x $\frac{dy}{dx} + \sec^2 x.y$ = tan x . $\sec^2$ x It is a linear differential equation of the first order. Comparing it with $\frac{dy}{dx}$ + Py = Q, we get P = $\sec^2$ x and Q = tan x. $\sec^2$ x Integration factor = $e^{\int P\,dx}$ = $e^{\int \sec^2 x\,dx}$ = $e^{\tan x}$ The solution of the given linear differential equation is given as: y $e^{\tan x}$ = ∫ tan x . $\sec^2$ x . $e^{\tan x}$ dx + C Let tan x = t ⇒ $\sec^2$ x dx = dt $y e^t$ = ∫ t . $e^t$ . dt + C $y e^t$ = $t\cdot e^t$ - ∫ 1.$e^t$ dt + C $y e^t$ = $t\cdot e^t - e^t$ + C $y e^t$ = $e^t$ (t - 1) + C $y e^{\tan x}$ = $e^{\tan x}$ (tan x - 1) + C y = tan x - 1 + $C e^{-\tan x}$