Question #338b6

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Answer 1 · 2017-08-15T18:11:49

$\frac{dy}{dx} = \frac{-y^3cos(x) + sin(x - 2y)}{3y^2sin(x) + 2sin(x - 2y)}$

$y^3sin(x) + cos(x - 2y) = 5$

Take derivative of both sides:

$3y^2 \frac{dy}{dx}sin(x) + y^3cos(x) - sin(x - 2y)(1 - 2\frac{dy}{dx}) = 0$

Solve for dy/dx:

$3y^2 \frac{dy}{dx}sin(x) + y^3cos(x) - sin(x - 2y) + 2sin(x - 2y)\frac{dy}{dx} = 0$

$3y^2\frac{dy}{dx}sin(x)+ 2sin(x-2y)\frac{dy}{dx} = -y^3cos(x)+ sin(x-2y)$

$\frac{dy}{dx}(3y^2sin(x) + 2sin(x - 2y)) = -y^3cos(x) + sin(x - 2y)$

$\frac{dy}{dx} = \frac{-y^3cos(x) + sin(x - 2y)}{3y^2sin(x) + 2sin(x - 2y)}$