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Find dy/dx for the equation x^3-2x^2y+3xy^2=38

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Differential equation.

 

asked Feb 28, 2014 in CALCULUS by andrew Scholar

1 Answer

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x^3-2x^2y+3xy^2 = 38

Apply derivative on each side.

Product rule d/dx(uv) = uv'+vu'

d/dx[x^3-2x^2y+3xy^2] = d/dx(38)

3x^2-2(x^2y'+y2x)+3(x2yy'+y^2) = 0

3x^2-2x^2y'-4xy+6xyy'+3y^2 = 0

3x^2-4xy+3y^2+y'(6xy-2x^2) = 0

y'(6xy-2x^2) = -3x^2+4xy-3y^2

y' = (4xy-3x^2-3y^2)/(6xy-2x^2)

answered Feb 28, 2014 by david Expert

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