Werity that $(x)=\frac{4}{1-c{e}^x},$ where $c$ is an arbitrary constant, is a one parameter family of solutions to $\frac{dy}{dx}=\frac{y^2-4y}{4}$ Graph the solution curves corresponding to $c=0,-1,-2$ using the same coordinate axes.

Find $\frac{d}{dx}.$

$\frac{d}{dx}=\frac{4c{e}^x}{(1-c{e}^x{)}^2}$

$($Simplify your answer.$)$

Substitute $(x)$ for $y$ and the derivative for $\frac{dy}{dx}.$

$\frac{dy}{dx}=\frac{y^2-4y}{4}$

$=\frac{(\frac{4}{1-c{x}^x}{)}^2-4(\frac{4}{1-c{x}^x})}{4}$

$($Simplify your answer.$)$