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July 29, 2014

July 29, 2014

Posted by **lana** on Friday, June 26, 2009 at 12:45pm.

x^2y + y^2x = 3

- Math -
**Count Iblis**, Friday, June 26, 2009 at 12:59pmd[x^2y + y^2x] = 0 -->

2xydx + x^2dy + 2xydy + y^2dx = 0 ---->

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

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

- Math -
**Reiny**, Friday, June 26, 2009 at 1:01pmuse the product rule for each term on the left side

x^2(dy/dx) + y(2x)(dx/dx) + y^2(dx/dx) + x(2y)(dy/dx) = 0

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

dy/dx is a common factor, so ...

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

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

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