Question

The solution of $\cos y+(x\sin y-1)\frac{dy}{dx}=0$ is

• A. $x\sec y=\tan y+C$
• B. $\tan y-\sec y=Cx$
• C. $\tan y+\sec y=Cx$
• D. $x\sec y+\tan y=C$

Answer 1

Answer: (A)

Given differential equation can be rewritten as
$\cos y\frac{dy}{dx}+x\sin y-1=0$
$\Rightarrow \frac{dx}{dy}+(\tan y)x=\sec y$
It is a linear differential equation of the form
$\frac{dx}{dy}+Px=Q$
$\therefore IF=e^{\int Pdy}=e^{\int \tan ydy}$
$=e^{\log \sec y}=\sec y$
$\therefore$ Solution is
$x\sec y=\int {\sec }^{2}ydy$
$\Rightarrow x\sec y=\tan y+C$