MCQ
Solution of $\cos x\frac{{dy}}{{dx}} + y\sin x = 1$ is
  • A
    $y\sec x\tan x = c$
  • $y\sec x= \tan x  c$
  • C
    $y\tan x = \sec x + c$
  • D
    $y\tan x = \sec x\tan x + c$

Answer

Correct option: B.
$y\sec x= \tan x  c$
b
(b) Given equation can be written as $\frac{{dy}}{{dx}} + y\tan x = \sec x$

$\therefore $ $I.F.$ $ = {e^{\int_{}^{} {\tan xdx} }} = {e^{\log \sec x}} = \sec x$

Hence solution is $y\sec x = \int_{}^{} {{{\sec }^2}x + c} = \tan x + c$.

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