Discussion Forum

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

Answer:

Explanation:

We have   $\frac{dy}{dx}=1- \cos (y-x) \cot (y-x) $

Put  $y-x = v \Rightarrow  \frac{dy}{dx}=1+\frac{dv}{dx}$

 $\therefore$    $1+ \frac{dv}{dx}=1- \cos v \cot v $  $\Rightarrow$ $\frac{dv}{dx}= -\frac{\cos ^{2} v}{\sin v}$

 $\Rightarrow$     $-\int\frac{\sin v}{\cos^{2} v}dv=\int dx\Rightarrow -\int \sec v \tan v dv=dx$

 $\Rightarrow -\sec v=x+c\Rightarrow x+\sec(y-x)=c$