Find DY/DX Using implicit differentation Y=Y^3-XY?

Answer 1

y = y^3 - xy

Differentiating implicitly,

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

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

Move everything with a dy/dx to the left hand side.

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

Factor dy/dx out of these terms.

(dy/dx) [ 1 - 3y^2 + x ] = -y

Divide both sides to isolate dy/dx

dy/dx = -y / [1 - 3y^2 + x]

Answer 2

y = y^3 - xy

(using product rule)

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

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

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

therefore, dy/dx = -y / (1 - 3y^2 + x)

Answer 3

dY / dX = 3Y².(dY/dX) - Y - X.(dY/dX)
(3Y² - X - 1).dY/dX = Y
dY/dX = Y / (3Y² - X - 1)