#### Answer

The differential equation is exact and its solutions are given implicitly by
$$
x^{2}+3x+y^{2} -2y=c
$$
where $c$ is an arbitrary constant

#### Work Step by Step

$$
(2 x+3)+(2 y-2) y^{\prime}=0 \quad \quad (*)
$$
this equation can be written in the differential form
$$
M(x,y) d x+N(x,y) d y=0
$$
Therefore
$$
(2 x+3) d x+(2 y-2) d y=0
$$
By calculating $M_{y}$ and $N_{x}$ , we find that
$$
M_{y}(x, y)=0=N_{x}(x, y)
$$
so the given equation is exact. Thus there is a $ψ(x, y)$ such that
$$
\begin{aligned} \psi_{x}(x, y) &=(2 x+3)
\\ \psi_{y}(x, y) &=(2 y-2) \end{aligned}
$$
Integrating the first of these equations with respect to $x$ , we obtain
$$
\psi(x, y)=x^{2}+3x+h(y) \quad \quad (**)
$$
Setting $ψ_{y} = N$ gives
$$
\psi_{y}(x, y)=h^{\prime}(y)=2y-2
$$
Thus $h^{\prime}(y)=2y-2$ and $h(y)=y^{2} -2y$ The constant of integration can be omitted since any solution of the preceding differential equation is satisfactory.
Now substituting for h(y) in Eq. (**) gives
$$
\psi(x, y)=x^{2}+3x+y^{2} -2y
$$
Hence solutions of Eq. (*) are given implicitly by
$$
x^{2}+3x+y^{2} -2y=c
$$
where $c$ is an arbitrary constant