Question

\[\frac{dy}{dx} = \tan\left( x + y \right)\]

Answer 1

We have,
\[\frac{dy}{dx} = \tan\left( x + y \right)\]
\[\frac{dy}{dx} = \frac{\sin\left( x + y \right)}{\cos\left( x + y \right)}\]
Let x + y = v
\[ \therefore 1 + \frac{dy}{dx} = \frac{dv}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]
\[ \therefore \frac{dv}{dx} - 1 = \frac{\sin v}{\cos v}\]
\[ \Rightarrow \frac{dv}{dx} = \frac{\sin v}{\cos v} + 1\]
\[ \Rightarrow \frac{dv}{dx} = \frac{\sin v + \cos v}{\cos v}\]
\[ \Rightarrow \frac{\cos v}{\sin v + \cos v}dv = dx\]
Integrating both sides, we get
\[\int\frac{\cos v}{\sin v + \cos v}dv = \int dx\]
\[ \Rightarrow \frac{1}{2}\int\frac{\left( \sin v + \cos v \right) + \left( \cos v - \sin v \right)}{\sin v + \cos v}dv = \int dx\]
\[ \Rightarrow \frac{1}{2}\int dv + \frac{1}{2}\int\frac{\cos v - \sin v}{\sin v + \cos v}dv = \int dx\]
\[ \Rightarrow \frac{1}{2}v + \frac{1}{2}\int\frac{\cos v - \sin v}{\sin v + \cos v}dv = x\]
\[\text{ Putting }\sin v + \cos v = t\]
\[ \Rightarrow \left( \cos v - \sin v \right)dv = dt\]
\[ \therefore \frac{1}{2}v + \frac{1}{2}\int\frac{dt}{t} = x\]
\[ \Rightarrow \frac{1}{2}v + \frac{1}{2}\log \left| t \right| = x + C\]
\[ \Rightarrow \frac{1}{2}\left( x + y \right) + \frac{1}{2}\log \left| \sin \left( x + y \right) + \cos \left( x + y \right) \right| = x + C\]
\[ \Rightarrow \frac{1}{2}\left( y - x \right) + \frac{1}{2}\log \left| \sin \left( x + y \right) + \cos \left( x + y \right) \right| = C\]
\[ \Rightarrow \left( y - x \right) + \log \left| \sin \left( x + y \right) + \cos \left( x + y \right) \right| = 2C\]
\[ \Rightarrow y - x + \log \left| \sin \left( x + y \right) + \cos \left( x + y \right) \right| = K ...........\left(\text{where, }K = 2C \right)\]