The general
- (a) y $\sec x = \tan x + C$ ✓ Correct
- (b) $y\tan x = \sec x + C$
- (c) $\tan x = y\tan x + C$
- (d) $x\sec x = \tan y + C$
The general
of $\frac{{dy}}{{dx}} + y\tan x = \sec x$ is
Solution
Given differential equation is
$\frac{{dy}}{{dx}} + y\tan x = \sec x$
which is a linear differential equation Here,
$P = \tan x,Q = \sec x$
$\therefore {\rm{IF}} = {e^{\int {\tan } xdx}} = {e^{\log |\sec x|}} = \sec x$
The general Solution is
$y \cdot \sec x = \int {\sec } x \cdot \sec x + C$
$\Rightarrow$ $y \cdot \sec x = \int {{{\sec }^2}} xdx + C$
$\Rightarrow$ $y \cdot \sec x = \tan x + C$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Differential Equations. Curated by Sachin Sharma. Free for all students.