Question

The general solution of $\frac{dy}{dx}+ytanx=secx$ is
(a) y secx=tanx+C
(b) y tanx=secx+C
(c) tanx=y tanx+C
(d) x secx=tany+C

Answer 1

Given differential equation is
$\frac{dy}{dx}+ytanx=secx$
which is a linear differential equation
Here, P= tanx, Q secx
$\therefore IF=e^{\int tanxdx}=e^{log|secx|}=secx$
The general solution is
$y.secx=\int secx.secx+C\Rightarrow y.secx=\int se{c}^{2}xdx+C\Rightarrow y.secx=tanx+C$