## Elementary Differential Equations and Boundary Value Problems 9th Edition

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
$$(2 x+3)+(2 y-2) y^{\prime}=0 \quad \quad (*)$$ 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