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

An integrating factor is $\mu(x)= e^{3x}$ So The differential equation : $$\left(3 x^{2} y+2 x y+y^{3}\right) d x+\left(x^{2}+y^{2}\right) d y=0$$ becomes exact when multiplied by the integrating factor $\mu(x)= e^{3x}$ and its solutions are given implicitly by $$e^{3 x} (3yx^{2}+y^{3}) =c$$ where $c$ is an arbitrary constant
$$\left(3 x^{2} y+2 x y+y^{3}\right) d x+\left(x^{2}+y^{2}\right) d y=0 \quad \quad (i)$$ Comparing this Equation with the differential form: $$M(x,y) d x+N(x,y) d y=0$$ we observe that \begin{aligned}M(x,y) &=\left(3 x^{2} y+2 x y+y^{3}\right) \\ N(x,y) &=\left(x^{2}+y^{2}\right) \end{aligned} By calculating $M_{y}$ and $N_{x}$ , we find that \begin{aligned} M_{y}(x, y) &= 3 x^{2}+2 x +3 y^{2} \\ N_{x}(x, y) & = 2x \end{aligned} we find that $$\frac{M_{y}(x, y)-N_{x}(x, y)}{N(x, y)}=\frac{\left(3 x^{2}+2 x +3 y^{2} \right) -2x }{\left(x^{2}+y^{2}\right) }=3.$$ Thus there is an integrating factor $μ$ that is a function of $x$ only, and it satisfies the differential equation $$\frac{d \mu}{d x}=3 \mu .$$ Hence $$\frac{d \mu}{ \mu}=3 dx$$ Integrating, we obtain $$\mu=e^{3x}$$ Multiplying Eq. (i) by this integrating factor, we obtain $$\left(3 x^{2} y+2 x y+y^{3}\right) e^{3x} d x+\left(x^{2}+y^{2}\right)e^{3x} d y=0$$ The latter equation is exact, since $$M_{y}(x, y)=\left(3 x^{2} +2 x+3y^{2 }\right) e^{3x} =N_{x}(x, y).$$ Thus there is a $\psi (x, y)$ such that \begin{aligned} \psi_{x}(x, y) &=M(x,y) =\left(3 x^{2} y+2 x y+y^{3} \right) e^{3x} , \\ \psi_{y}(x, y) &=N(x,y) =\left(x^{2}+y^{2} \right) e^{3x} \end{aligned} \quad \quad (ii) Integrating the first of these equations with respect to $x$ , we obtain $$\begin{split} \psi(x, y) & =\int { \left(3 x^{2} y+2 x y+y^{3} \right) e^{3x} } dx \\ & = 3y \int { x^{2} e^{3x}dx}+2y \int {x e^{3x}dx}+y^{3} \int { e^{3x}dx } +h(y) \\ & = 3y (\frac{1}{3} e^{3 x} x^{2}-\frac{2}{27}\left(3 e^{3 x} x-e^{3 x}\right)) +2y (\frac{1}{9}\left(e^{3 x} \cdot 3 x-e^{3 x}\right)) +y^{3} (\frac{1}{3} e^{3 x}) +h(y) \\ & = e^{3 x} (yx^{2}+\frac{1}{3}y^{3})+h(y) \quad \quad \quad\quad\quad (iii) \end{split}$$ where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (ii), we compute $\psi_{y}(x, y)$ from Eq. (iii) and set it equal to $N$, obtaining $$\psi_{y}(x, y)= e^{3 x} (x^{2}+y^{2}) +h^{\prime}(y)= \left(x^{2}+y^{2} \right) e^{3x}$$ Thus $h^{\prime}(y)=0$ and $h(y)=0$ , the constant of integration can be omitted since any solution of the preceding differential equation is satisfactory. Now substituting for $h(y)$ in Eq. (iii) gives $$\psi(x, y)= e^{3 x} (yx^{2}+\frac{1}{3}y^{3})$$ Hence solutions of Eq. (i) are given implicitly by $$\psi(x, y) = e^{3 x} (yx^{2}+\frac{1}{3}y^{3}) =c_{1}$$ where $c_{1}$ is an arbitrary constant. The solution can be rewritten as follows $$\psi(x, y) = e^{3 x} (3yx^{2}+y^{3}) =c$$ where $c=3c_{1}$ is an arbitrary constant