Question

The solution of the differential equation $\frac{d y}{d x}=1-\cos (y-x) \cot (y-x)$ is

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

Answer 1

Answer: (D)

We have,
$ \frac{d y}{d x}=1-\cos (y-x) \cot (y-x) $
Put $ y -x=v$
$ \Rightarrow \, \frac{d y}{d x}=1+\frac{d v}{d x} $
$\therefore \, 1+ \frac{d v}{d x}=1-\cos v \cot v$
$ \Rightarrow \,\frac{d v}{d x}=-\frac{\cos ^{2} v}{\sin v} $
$\Rightarrow \, -\int \frac{\sin v}{\cos ^{2} v} d v=\int d x$
$ \Rightarrow -\int \sec v \tan v d v=d x $
$\Rightarrow \,-\sec v=x+c$
$ \Rightarrow \,x+\sec (y-x)=c$