Answer to Question #300869 in Differential Equations for Cross

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

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