Question

Solve the differential equation: $\tan y\frac{dy}{dx}+\tan x=\cos y{\cos }^{2}x.$

• A. $\sec y\sec x=\sin x-c.$
• B. $\sec y\sec x=-\sin x+c.$
• C. $\sec y\sec x=\sin x+c.$
• D. None of these.

Answer 1

Answer: (A), (C)

The correct options are
A $\sec y\sec x=\sin x-c.$
C $\sec y\sec x=\sin x+c.$

$\tan y\frac{dy}{dx}+\tan x=\cos y{\cos }^{2}x$

$\Rightarrow \sec y\tan y\frac{dy}{dx}+\sec y\tan x={\cos }^{2}x$
Put $\sec y=v\Rightarrow \sec y\tan ydy=dv$
$\therefore \frac{dv}{dx}+v\tan x={\cos }^{2}x$ ...(1)
Here $P=\tan x\Rightarrow \int Pdx=\int \tan xdx=\log \sec x$
$\therefore I.F.=e^{\log \sec x}=\sec x$
Multiplying (1) by $I.F.$ we get
$\sec x\frac{dv}{dx}+v\tan x\sec x=\cos x$
Integrating both sides we get
$\sec xv=\int \cos xdx+c\Rightarrow \sec y\sec x=\sin x+c$

Constant can also be negative hence option (A) is also true