solve when x=0 and y= 2
ok I have setup as shown`(is this ok?). is there some rule about F/F'? to make it easier to solve?
many thanks
$\displaystyle \displaystyle \begin{align*} (2y + 4)\,\frac{dy}{dx} &= y^2 + 4y \\ \left[\frac{2y + 4}{y(y + 4)}\right]\,\frac{dy}{dx} &= 1 \\ \int{\left[\frac{2y + 4}{y(y+4)}\right]\,\frac{dy}{dx}\,dx} &= \int{1\,dx} \\ \int{\frac{2y + 4}{y(y + 4)}\,dy} &= \int{1\,dx}\end{align*}$
Go from here, you will need to use a substitution or Partial Fractions on the LHS.