- A student has 3 hours to cram for an examination and during this time wishes to memorize a set of 60 facts. According to psychologists, the rate at which a person can memorize a set of facts is proportional to the number of facts remaining to be memorized. Thus, if a student memorizes y facts in t minutes, then
$$ \frac{dy}{dt} = k (60 - y) $$ where k is a positive constant and and y < 60 for all values t $ \ge $ 0
Assume that zero facts are memorized initially and the student memorizes 15 facts in the first 20 minutes . a. Express y as a function of t. b. How many facts will the student memorize in 1 hour? c. How many facts will the student memorize in 3 hours? d. How many facts will the student memorize in the long run? e. How long will the student memorize 40 facts?
My main problem is a.
I used the differential equation $ y(t) = k (y(t) - A) $ and I got $ y(t) = 60 -60e^{-kt} $.
However, it took few trial and error methods to do this. But is there more detailed solution, especially by integrating $ \frac{dy}{dt} = k (60 - y) $?
You are provided that,
Also, $\frac{dy}{dt} = k(60-y)$ Integrating using the limits as $y(0)= 0$ and taking $y(t) = y$ at some $t$,
$ \begin{align} &\int_0^y\frac{dy}{60-y} = \int_0^t kdt \\&\Rightarrow -\ln(\frac{60-y}{60}) = kt \\&\Rightarrow \ln(1- \frac{y}{60}) = -kt\\ &\Rightarrow \boxed{y\equiv y(t)= 60 - 60^{-kt}}\end{align}$
Now using the fact $y(20) = 15$, you can get $k$