Question

(sin x + cos x) dy + (cos x − sin x) dx = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(\sin x+\cos x\right)dy+\left(\cos x-\sin x\right)dx=0 \\ \Rightarrow dy=-\left(\frac{\cos x-\sin x}{\sin x+\cos x}\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=-\int \left(\frac{\cos x-\sin x}{\sin x+\cos x}\right)dx \\ \Rightarrow y=-\int \left(\frac{\cos x-\sin x}{\sin x+\cos x}\right)dx \\ \mathrm{Putting}\sin x+\cos x=t \\ \Rightarrow \left(\cos x-\sin x\right)dx=dt \\ \therefore y=-\int \frac{dt}{t} \\ \Rightarrow y=-\log \left|t\right|+C \\ \Rightarrow y=-\log \left|\sin x+\cos x\right|+C \\ \Rightarrow y+\log \left|\sin x+\cos x\right|=C \\ \mathrm{Hence},y+\log \left|\sin x+\cos x\right|=C\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}. \end{gathered}$$