Question

The solution of differential equation $(sinx+cosx)dy$ + $(cosx-sinx)dx=0$ is

Answer 1

$(\sin x+\cos x)dy+(\cos x-\sin x)dx=0$

$(\sin x+\cos x)dy=-(\cos x-\sin x)dx$

$\frac{dy}{dx}=-\frac{(\cos x-\sin x)}{(\sin x+\cos x)}$

$dy=-\frac{(\cos x-\sin x)}{(\sin x+\cos x)}dx$

$\int dy=-\int \frac{\cos x-\sin x}{\sin x+\cos x}dx$

$\sin x+\cos x=t$

$(\cos x-\sin x)dx=dt$

$(\sin x-\cos x)dx=-dt$

$y=\int -\frac{dt}{t}+c$

$=-log|t|+c$

$y=-log|\sin x+\cos x|+c$.