In problem $\mathbf{1}\mathbf{-}\mathbf{6}$, determine whether the differential equation is separable $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{0}$.

A first-order ordinary differential equation $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{f}\left(x,y\right)$is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions$\mathit{g}\left(x\right)$ that is a function of $\mathit{x}$alone and $\mathit{h}\left(y\right)$that is a function of $\mathit{y}$alone.

Mathematically, the equation is $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{f}\left(x,y\right)$ separable, when $\mathit{f}\left(xy\right)\mathbf{=}\mathit{g}\left(x\right)\mathit{h}\left(y\right)$

The given equation is

$\frac{d}{dx}-\sin \left(x+y)=0......(1\right)$

Write the given equation as follows:

role="math" localid="1654761564667" $\frac{d}{dx}=\sin \left(x+y\right)......\left(2\right)$

Now, use the trigonometrical to identity $\sin \left(A+B\right)=\sin A\cos B=\cos A\sin B$. It results,

$\sin \left(x+y\right)=\sin x\cos y+\cos x\sin y$

In this case, equation (2) becomes,

$\frac{dy}{dx}=\sin x\cos y+\cos x\sin y........\left(3\right)$

The function in the right – hand side of equation (3) is

$f\left(x,y\right)=\sin x\cos y+\cos x\sin y$

This function can’t be written as a product of two functions$g\left(x\right)$ and $h\left(y\right)$.

Therefore, the given differential equation is not separable.