The solution of the differential equation (d y/d x)=y tan x-2 sin x is

Question

The solution of the differential equation (d y/d x)=y tan x-2 sin x is (A)  y= cos x+c sec x (B) y= cos x+c (C) y= sin x+c (D) y= operatornamecosec x+c

Tardigrade Admin · Accepted Answer

(d y/d x)=y tan x-2 sin x (d y/d x)-y tan x=-2 sin x I.F. =e-∫ tan x d x=| cos x| y cos x=( cos 2 x/2)+k ⇒ y=( cos 2 x/2 cos x)+k sec x ⇒ y= cos x+c sec x

Q. The solution of the differential equation $\frac{d y}{d x} = y tan x - 2 sin x$ is

$y = cos x + c sec x$

$y = cos x + c$

$y = sin x + c$

$y = cosec x + c$

$\frac{d y}{d x} = y tan x - 2 sin x$
$\frac{d y}{d x} - y tan x = - 2 sin x$
I.F. $= e^{- \int t a n x d x} = ∣ cos x ∣$
$y cos x = \frac{c o s 2 x}{2} + k$
$\Rightarrow y = \frac{c o s 2 x}{2 c o s x} + k sec x \Rightarrow y = cos x + c sec x$

Q. The solution of the differential equation dxdy​=ytanx−2sinx is

y=cosx+csecx

y=cosx+c

y=sinx+c

y=cosecx+c