Given \(\dfrac{dy}{dx}=2xy-y\) and \(y(0)=3\), which of the following is equal to \(y\) ?

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$$ \frac{dy}{dx}=2xy-y $$ $$ \frac{dy}{dx}=y(2x-1) $$ $$ \frac{1}{y}   dx = 2x-1   dx $$ $$ \ln|y|=x^2-x+C $$ $$ y=e^{x^2-x+C} $$ $$ y=Ae^{x^2-x} $$

Using the initial condition:

$$ 3=Ae^{(0)^2-(0)} $$ $$ 3=A $$ $$ \boxed{y=3e^{x^2-x}} $$