Question

(𝑥𝑦²+ 𝑦 − 𝑥)𝑑𝑥 + 𝑥(𝑥𝑦 + 1)𝑑𝑦 = 0

Answer 1

$$\dfrac{\partial Q}{\partial x}=2xy+1=\dfrac{\partial P}{\partial y}$$

The system of two differential equations that define the function $u(x, y)$ is

$$\begin{cases} \dfrac{\partial u}{\partial x}=xy^2+y-x \\ \\ \dfrac{\partial u}{\partial y}=x^2y+x \end{cases}$$

Integrate the first equation over the variable $x$

$$u=\int(xy^2+y-x)dx+\varphi(y)$$

$$=\dfrac{x^2y^2}{2}+xy-\dfrac{x^2}{2}+\varphi(y)$$

Differentiate with respect to $y$

$$\dfrac{\partial u}{\partial y}=x^2y+x+\varphi'(y) =x^2y+x$$

$$\varphi'(y) =0$$

$$\varphi(y)=-\dfrac{C}{2}$$

Then

$$u=\dfrac{x^2y^2}{2}+xy-\dfrac{x^2}{2}-\dfrac{C}{2}$$

The general solution of the exact differential equation is given by

$$x^2y^2+2xy-x^2=C$$