Answer to Question #200549 in Differential Equations for Chasey

Question

(xy²+ x -2y +3) dx ( x²y -2x -2y) dy =0

Answer 1

Let us solve the differential equation

$(xy^2+ x -2y +3) dx +( x^2y -2x -2y) dy =0$

Taking into account that $\frac{\partial (xy^2+ x -2y +3)}{\partial y}=2xy-2=\frac{\partial (x^2y -2x -2y)}{\partial x},$

we conclude that the equation is exact, and consequently there exists the function $u=u(x,y)$ such that

$\frac{\partial u}{\partial x}=xy^2+ x -2y +3,\ \frac{\partial u}{\partial y}=x^2y -2x -2y.$

Therefore, $u=\frac{1}2x^2y^2+\frac{1}2x^2-2xy+3x+c(y),$ and hence

$\frac{\partial u}{\partial y}=x^2y -2x +c'(y)=x^2y -2x -2y.$

It follows that $c'(y)=-2y,$ and thus $c(y)=-y^2+C.$

We conclude that thegeneral solution of the differential equation

$(xy^2+ x -2y +3) dx +( x^2y -2x -2y) dy =0$ is

$\frac{1}2x^2y^2+\frac{1}2x^2-2xy+3x-y^2=C.$