Solution of the differential equation dy dx + y ^2 x = x ^2 x is - Vidyadip

MCQ

Solution of the differential equation $\frac{{dy}}{{dx}} + y{\sec ^2}x = \tan x{\sec ^2}x$ is

✓
$y = \tan x - 1 + c{e^{ - \tan x}}$
B
${y^2} = \tan x - 1 + c{e^{\tan x}}$
C
$y{e^{\tan x}} = \tan x - 1 + c$
D
$y{e^{ - \tan x}} = \tan x - 1 + c$

✓

Answer

Correct option: A.

$y = \tan x - 1 + c{e^{ - \tan x}}$

a
(a) $I.F.$ = ${e^{\int {{{\sec }^2}x\,dx} }} = {e^{\tan x}}$

Solution is $y{e^{\tan x}} = c + \int {\tan x{e^{\tan x}}{{\sec }^2}x\,dx} $

==> $y = c{e^{ - \tan x}} + \tan x - 1$.

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