Question

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

Answer 1

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