Question

\[\frac{dy}{dx}\cos\left( x - y \right) = 1\]

Answer 1

We have, 
\[\frac{dy}{dx}\cos\left( x - y \right) = 1\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{\cos\left( x - y \right)}\]

Putting x - y = v

\[ \Rightarrow 1 - \frac{dy}{dx} = \frac{dv}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = 1 - \frac{dv}{dx}\]

\[ \therefore 1 - \frac{dv}{dx} = \frac{1}{\cos v}\]

\[ \Rightarrow \frac{dv}{dx} = 1 - \frac{1}{\cos v}\]

\[ \Rightarrow \frac{dv}{dx} = \frac{\cos v - 1}{\cos v}\]

\[ \Rightarrow \frac{\cos v}{\cos v - 1}dv = dx\]

Integrating both sides, we get

\[\int\frac{\cos v}{\cos v - 1}dv = \int dx\]

\[ \Rightarrow - \int\frac{\cos v\left( 1 + \cos v \right)}{1 - \cos^2 v}dv = \int dx\]

\[ \Rightarrow - \int\frac{\cos v\left( 1 + \cos v \right)}{\sin^2 v}dv = \int dx\]

\[ \Rightarrow - \int\left( \cot v\ cosec\ v + \cot^2 v \right)dv = \int dx\]

\[ \Rightarrow - \int\left( \cot v\ cosec\ v + {cosec}^2 v - 1 \right)dv = \int dx\]

\[ \Rightarrow - \left( - cosec\ v - \cot v - v \right) = x + C\]

\[ \Rightarrow cosec \left( x - y \right) + \cot \left( x - y \right) + x - y = x + C\]

\[ \Rightarrow cosec \left( x - y \right) + \cot \left( x - y \right) - y = C\]

\[ \Rightarrow \frac{1 + \cos \left( x - y \right)}{\sin \left( x - y \right)} - y = C\]

\[ \Rightarrow \cot\left( \frac{x - y}{2} \right) = y + C\]