Question

Find:
$\frac{dy}{dx}=sin(x+y)+cos(x+y)$

Answer 1

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

Substitute $x+y=v$

$\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$

The given differential equation becomes

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

$\Rightarrow \frac{dv}{dx}-1=\sin v+\cos v$

$\Rightarrow \frac{dv}{dx}=\sin v+\cos v+1$

$\Rightarrow \frac{dv}{1+\sin v+\cos v}=dx$

$\Rightarrow \frac{dv}{1+\frac{2\tan \frac{v}{2}}{1+{\tan }^2\frac{v}{2}}+\frac{1-{\tan }^2\frac{v}{2}}{1+{\tan }^2\frac{v}{2}}}=dx$

$\Rightarrow \frac{1+{\tan }^2\frac{v}{2}}{2(1+\tan \frac{v}{2})}dv=dx$

$\Rightarrow \frac{{\sec }^2\frac{v}{2}}{2(1+\tan \frac{v}{2})}dv=dx$

$\Rightarrow \log \left|2(1+\tan \frac{v}{2})\right|+c=x$

Re-substituting $x+y=v$ we get

$\Rightarrow x=\log \left|2(1+\tan \frac{x+y}{2})\right|+c$ where $c$ is the constant of integration.