Suppose a student has 3 hours to cram for an exam and during this time wishes to memorize 60 facts. The formula for facts memorized per hour is $\frac{dy}{dt}=k(60-y)$ where $y$ is facts memorized, $t$ is minutes and $k$ is a positive constant. It is assumed that $y<60$ for all $t \geq0$.
If the student memorizes $15$ facts in $20$ minutes, how many facts will the student memorize in one hour.
So take the given formula and moving $dt$ over gives $$dy=k(60-y)dt$$ now this is where I'm stuck because I don't know how to get the $y$ on the right over the the left. If I distribute the $k$ and the divide by $y$ I'm left with $$\frac{dy}{y}=60k-k$$ but I don't believe that this is the way to go about it.
This is a differential equation, where the goal is to find $y(t).$ You can solve this one by separation of variables: \begin{align*} dy&=k(60-y)\,dt\\ \frac{dy}{y-60}&=-k\,dt\\ \ln|y-60|&=-kt+C\\ y-60&=Ce^{-kt}\\ y&=60+Ce^{-kt}. \end{align*} To find $C,$ you must plug in the information you know: $y(20)=15,$ to get \begin{align*} 15&=60+Ce^{-k20}\\ -45&=Ce^{-k20}\\ C&=-45e^{k20}. \end{align*} This makes sense, as we must have $C<0$ to force $y<60.$ So we can write $$y=60-45e^{k20}e^{-kt}=60-45e^{k(20-t)}. $$ Can you finish?