The solution of differential equation $(\sin x+\cos x)dy+(\cos x-\sin x)dx=0$ is :

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

We can write the given equation as: 

  $\mathrm{dy}+\frac{\cos \mathrm{x}-\sin \mathrm{x}}{\sin \mathrm{x}+\cos \mathrm{x}}\mathrm{dx}=0$

By integrating both sides, we get:

   $y+\int \frac{\cos x-\sin x}{\sin x+\cos x}dx=c ...\left(1\right)$

Put $\sin x+\cos x=t$

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

 Thus from (1) we have 

$$\begin{gathered} \Rightarrow \mathrm{y}+\int \frac{\mathrm{dt}}{\mathrm{t}}=\mathrm{c}\Rightarrow \mathrm{y}+\log \mathrm{t}=\mathrm{c} \\ \begin{array}{l}\Rightarrow \log \mathrm{t}=-\mathrm{y}+\mathrm{c}\Rightarrow \mathrm{t}={\mathrm{e}}^{-\mathrm{y}}\cdot {\mathrm{e}}^{\mathrm{c}}\\ \Rightarrow \sin \mathrm{x}+\cos \mathrm{x}=\frac{{\mathrm{e}}^{\mathrm{c}}}{{\mathrm{e}}^{\mathrm{y}}}\\ \Rightarrow {\mathrm{e}}^{\mathrm{y}}(\sin \mathrm{x}+\cos \mathrm{x})={\mathrm{e}}^{\mathrm{c}}=\mathrm{c}\end{array} \end{gathered}$$