Determine if the given differential equation is separable I tried to solve this equation $dy/dx-\sin(x+y)=0$ but I don't know if is separable or not.
I did alternate form $dy/dx = (\sin x\cos y)+(\cos x\sin y)$.
Then i used trigonometry identities and algebra to simplify the equation.
$$(1/\sin x)dy/dx=\cos y+((\cos x\sin y)/\sin x)$$
$$(1/\sin y)(dy/\sin xdx)=\cos y+\cot x$$
$$dy/\sin x\sin ydx=(\cos y/\sin y)+\cot x$$
$$dy/\sin x\sin ydx=\cot y+\cot x$$
$$dy/dx=(\cot y+\cot x)(\sin x\sin y)$$
$$dy/dx=\cos y\sin x+\cos x\sin y$$
Then i suppose the "solution" should be integral of
$dy/(\cos y\sin y)=(\sin x+\cos x)dx$ ??
Please help me. 
 A: $$
\frac{dy}{dx} = \sin(x+y)
$$
let $v = y+x$ we find
$$
v' - 1 = \sin v
$$
thus we know that it is separable, and approachable by using $\tan$ substitution. 
A: Note that your last manipulation (before incorrectly "separating" the variables) is exactly what you obtained after expanding $$\sin(x+y) = \sin x \cos y + \cos x \sin y$$
As I said, you cannot separate the variables as you did. The equation is not separable into the form $f(y)\,dy = g(x) \,dx$.
What you can do is put $u = x+y$. $$\frac{du}{dx} = 1 + \frac{dy}{dx} = 1+\sin u$$ $$\frac{du}{dx} = 1 + \sin u \iff\frac{du}{1+\sin u} = dx$$
A: Here are the steps
$$ \frac{d}{dx}[y]- \sin(x+y) =0$$
$$ \frac{d}{dx}[y]= \sin(x+y) $$
Let $u=x+y$, then
$$ \frac{d}{dx}[u-x]=\sin(u) $$
$$ \frac{d}{dx}[u]-\frac{d}{dx}[x]= \sin(u) $$
$$ \frac{d}{dx}[u]-1= \sin(u) $$
$$ \frac{d}{dx}[u]= \sin(u) +1$$
$$ \frac{1}{\sin(u)+1}du= dx$$
$$ \int\frac{1}{\sin(u)+1}du= \int dx$$
$$ \int\frac{1}{\sin(u)+1}du= x+C$$
Hint: Let $t=\tan\left(\frac{u}{2}\right)$.
I'll leave the rest to for you to solve. 
