How do you find #(dy)/(dx)# given #y(x+1)-y^2=x#?

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Answer 1 · 2016-10-30T06:54:20

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

Distribute y:

#yx + y - y^2 = x#

Differentiate:

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

Move y to the right side:

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

Factor out #dy/dx#

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

Divide both sides by the leading coefficient:

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