Question

(x²+2xy-7x)dx+(x²+2y²-3y)dy=0

Answer 1

$$M(x, y)=x^2+2xy-7x$$

$$\dfrac{\partial M}{\partial y}=2x$$

$$N(x, y)=x^2+2y^2-3y$$

$$\dfrac{\partial N}{\partial x}=2x$$

$$\dfrac{\partial M}{\partial y}=2x=\dfrac{\partial N}{\partial x}$$

The differential equation $(x^2+2xy-7x)dx+(x^2+2y^2-3y)dy=0$  is an exact equation.

$$\begin{cases} \dfrac{\partial u}{\partial x}=M(x, y) \\ \\ \dfrac{\partial u}{\partial y}=N(x, y) \end{cases}$$

$$\begin{cases} \dfrac{\partial u}{\partial x}=x^2+2xy-7x \\ \\ \dfrac{\partial u}{\partial y}=x^2+2y^2-3y \end{cases}$$

$$u(x, y)=\int(x^2+2xy-7x)dx+\varphi(y)$$

$$=\dfrac{1}{3}x^3+x^2y-\dfrac{7}{2}x^2+\varphi(y)$$

$$\dfrac{\partial u}{\partial y}=x^2+\varphi'(y)=x^2+2y^2-3y$$

$$\varphi'(y)=2y^2-3y$$

$$\varphi(y)=\int(2y^2-3y)dy=\dfrac{2}{3}y^3-\dfrac{3}{2}y^2-C$$

$$u=\dfrac{1}{3}x^3+x^2y-\dfrac{7}{2}x^2+\dfrac{2}{3}y^3-\dfrac{3}{2}y^2-C$$

$$\dfrac{1}{3}x^3+x^2y-\dfrac{7}{2}x^2+\dfrac{2}{3}y^3-\dfrac{3}{2}y^2=C$$