Q. Solution of the differential equation $\frac{dy}{dx} + y\sec^{2}{x = \tan{x\sec^{2}x}}$ is
A) $y = \tan{x - 1 + ce^{- \tan x}}$
B) $y^{2} = \tan{x - 1 + ce^{\tan x}}$
C) $ye^{\tan x} = \tan{x - 1 + c}$
D) $ye^{{- tan}x} = \tan{x - 1 + c}$
Explanation
(a)
It is a linear differential equation of the form of $\frac{dy}{dx} + Py = Q$.
$\ \Rightarrow P = \sec^{2}{x,\ Q = \tan{x\sec^{2}x}}$
$\therefore IF = e^{\int_{}^{}{P\ dx}} = e^{\int_{}^{}{\sec^{2}{x\ dx}}} = e^{\tan x}$
Solution is ${ye}^{\tan x} = \int_{}^{}{\tan{{xe}^{\tan x}\sec^{2}{x\ dx + c}}}$
$\Rightarrow {ye}^{\tan x} = \tan{{x\ e}^{\tan x} - e^{\tan x}} + c$
$\Rightarrow y = \tan{x - 1 + ce^{- \tan x}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.