Answer to Question #296627 in Differential Equations for Tobias Felix

Question

determine the general/particular solution for each equation using the applicable solution to equations of order one (separable, homogenous,linear,exact)

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

Answer 1

Let us determine the general solution of the differential equation

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

Taking into account that

$\frac{\partial( 3x(xy-2))}{\partial y} =3x^2=\frac{\partial (x^3-2y)}{\partial x},$

we conclude that this equation is exact, and hence there is a function $u=u(x,y)$ such that

$\frac{\partial u}{\partial x}=3x(xy-2)=3x^2y-6x,\ \ \frac{\partial u}{\partial y}=x^3-2y.$

It follows that $u=x^3y-3x^2+c(y),$ and thus

$\frac{\partial u}{\partial y}=x^3+c'(y)=x^3-2y.$

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

We conclude that the general solution is of the differential equation

$x^3y-3x^2-y^2=C.$