Question

The general solution of the differential equation $\frac{dy}{dx}=ytanx-y^{2}secx$ is

• A. tan x = (c + sec x)y
• B. sec y = (c + tan y )x
• C. sec x = (c + tan x)y
• D. None of these

Answer 1

The correct option is C sec x = (c + tan x)y

We have $\frac{dy}{dx}=ytanx-y^{2}secx\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}tanx=-secx$
Putting $\frac{1}{y}=v\Rightarrow \frac{-1}{y^2}\frac{dy}{dx}=\frac{dv}{dx}$, we obtain
$\frac{dv}{dx}+tanx.v=secx$which is linear
$I.F=e^{\int tanxdx}=e^{logsecx}=secx$
$\therefore$ The solution is
$vsecx=\int se{c}^{2}xdx+c\Rightarrow \frac{1}{y}secx=tanx+c$
$\Rightarrow secx=y(c+tanx)$