Question

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

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

Answer 1

Answer: (A)

Given differential equation can be rewritten as
$ \cos y \frac{d y}{d x}+x \sin y-1=0 $
$\Rightarrow \frac{d x}{d y}+(\tan y) x=\sec y$
It is a linear differential equation of the form
$\frac{d x}{d y}+P x =Q $
$\therefore I F =e^{\int \,P \,d y}=e^{\int \,\tan \,y \,d y}$
$=e^{\log \sec y}=\sec y$
$\therefore $ Solution is
$ x \sec y=\int \sec ^{2}\, y\, d y$
$\Rightarrow x \sec y=\tan\, y+C$