Solution of the differential equation y' = y x - 2 x, is - Vidyadip

MCQ

Solution of the differential equation $y' = y\tan x - 2\sin x,$ is

A
$y = \tan x\, + 2c\,\cos x$
B
$y = \tan x\, + c\,\cos x$
C
$y = \tan x\, - 2c\,\cos x$
✓
None of these

✓

Answer

Correct option: D.

None of these

d
(d) $y' = y\tan x - 2\sin x$ ==> $\frac{{dy}}{{dx}} - y\tan x = - 2\sin x$
$I.F.$ $ = {e^{ - \int {\tan x\,dx} }} = {e^{\log \cos x}} = \cos x$
$\therefore$ $y\cos x = \int {( - 2\sin x)(\cos x)dx} + c$
==> $y\cos x = - \int {\sin 2x\,dx} + c$
==> $2y\cos x = \cos 2x + c$.

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