Question

(2xy-y²+2x)dx-(2xy-x²-2y)dy=0

Answer 1

$$(2xy-y^2+2x)dx+(-2xy+x^2+2y)dy$$

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

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

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

We have the following system of differential equations to find the function $u(x, y)$

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

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

By integrating the first equation with respect to $x,$ we obtain

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

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

Substituting this expression for $u(x,y)$ into the second equation gives us:

$$x^2-2xy+\varphi'(y)=-2xy+x^2+2y$$

$$\varphi'(y)=2y$$

By integrating the last equation, we find the unknown function $\varphi(y)$

$$\varphi(y)=\int 2y dy=y^2-C$$

So that the general solution of the exact differential equation is given by

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