In problems, 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form ${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{G}\left(\mathrm{ax}+\mathrm{by}\right)$.

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

If the right-hand side of the equation $\frac{dy}{dx}=f\left(x,y\right)$can be expressed as a function of the ratio $\frac{y}{x}$alone, then we say the equation is homogeneous.

Equations of the form$\frac{dy}{dx}=G\left(ax+by\right)$

When the right-hand side of the equation $\frac{dy}{dx}=f\left(x,y\right)$can be expressed as a function of the combination $ax+by$, where a and b are constants, that is,$\frac{dy}{dx}=G\left(ax+by\right)$then the substitution$z=ax+by$ transforms the equation into a separable one.

A first-order equation that can be written in the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)y^n$, where P(x) and Q(x) are continuous on an interval (a, b) and n is a real number, is called a Bernoulli equation.

We have used various substitutions for y to transform the original equation into a new equation that we could solve. In some cases, we must transform both x and y into new variables, say u and v. This is the situation for equations with linear coefficients-that is, equations of the form

.$\left(a_{1}x+b_{1}y+c_1\right)dx+\left(a_{2}x+b_{2}y+c_2\right)dy=0$

Given, $\cos \left(x+y\right)dy=\sin \left(x+y\right)dx$.

By evaluating

$\begin{array}{rcl}\cos \left(x+y\right)dy& =& \sin \left(x+y\right)dx\\ \frac{dy}{dx}& =& \frac{\sin \left(x+y\right)}{\cos \left(x+y\right)}\\ & =& \tan \left(x+y\right)\end{array}$

It seems that the given equation is $y^{\text{'}}=G\left(ax+by\right)$

Therefore, the given equation is the form of$y^{\text{'}}=G\left(ax+by\right)$