Question #21718

Question

807 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-20T10:13:28

#dy/dx=(2xy^3-4xy^3 -y)/(x-3x^2y^2 +6x^2y^2)#

#xy=y^3x^2−2x^2y^3#

#x.dy/dx+y.1=[(y^3. 2x)+(x^2 . 3y^2 dy/dx)]−[(2x^2 . 3y^2 dy/dx)+(2y^3 . 2x)]#

#x.dy/dx+y=[2xy^3+3x^2y^2 dy/dx]−[6x^2y^2 dy/dx+4xy^3 ]#

#x.dy/dx+y=2xy^3+3x^2y^2 dy/dx−6x^2y^2 dy/dx-4xy^3 #

#x.dy/dx-3x^2y^2 dy/dx +6x^2y^2 dy/dx=2xy^3-4xy^3 -y#

#dy/dx(x-3x^2y^2 +6x^2y^2)=2xy^3-4xy^3 -y#

#dy/dx=(2xy^3-4xy^3 -y)/(x-3x^2y^2 +6x^2y^2)#

#dy/dx=-(2xy^3+y)/(x+3x^2y^2)=-(y(2xy^2+1))/(x(3xy^2+1))#