How do you find $x^2y- xy^3 = 2-2x^3$ implicitly?

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Answer 1 · 2015-08-07T11:51:32

$dy/dx=(y^3-6x^2-2xy)/(x^2-3xy^2)$

Differentiate term by term (using the product rule):

$(d(x^2y))/dx=2xy+x^2dy/dx$

$(d(xy^3))/dx=y^3+3xy^2dy/dx$

$(d(2))/dx=0$

$(d(2x^3))/dx=6x^2$

This gives:
$2xy+x^2dy/dx-(y^3+3xy^2dy/dx)=-6x^2$

Rearranging terms to make $dy/dx$ the subject we have:

$(x^2-3xy^2)dy/dx=y^3-6x^2-2xy$

=>

$dy/dx=(y^3-6x^2-2xy)/(x^2-3xy^2)$

How do you find x^2y- xy^3 = 2-2x^3x^2y- xy^3 = 2-2x^3 implicitly?