Solve the following differential equation: cos² x (dy)/(dx) + y = tan x, given that y(0) = 1…

The given D.E is $\frac{dy}{dx} + y\sec^2 x = \tan x\sec^2 x$ $\therefore$ I.F $= e^{\int\sec^2 x\,dx} = e^{\tan x}$ So, the solution of the D.E is given by $y\cdot e^{\tan x} = \int\tan x\cdot\sec^2 x\cdot e^{\tan x}\,dx + c$ Let $\tan x = t$, so that $\sec^2 x\,dx = dt$ $y\cdot e^{\tan x} = \int t\cdot e^t\,dt + c = te^t - e^t + c$ $y e^{\tan x} = \tan x\cdot e^{\tan x} - e^{\tan x} + c$ Given $x = 0$ then $y = 1$, $c = 2$ $\therefore$ Required Solution: $y\cdot e^{\tan x} = \tan x\,e^{\tan x} - e^{\tan x} + 2$ $y = \tan x - 1 + 2e^{-\tan x}$ $\therefore y\left(\frac{\pi}{4}\right) = \frac{2}{e}$