1. Easy differential equation

Can someone help me with the problem below? I'm 99% positive i screwed up part B
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Suppose two students memorize lists according to the model
$\displaystyle \frac{dL}{dt} = 2(1-L)$, where $\displaystyle 0 \le L \le 1$
(A) If one of the students knows one-half of the list at time $\displaystyle t=0$ and the other knows none of the list, which student is learning most rapidly at this instant?

(B) Will the student who starts out knowing none of the list ever catch up to the student who starts out knowing one-half of the list?

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(A)

Student A:
$\displaystyle \frac{dL}{dt} = 2(1-0.5) = 1$
Student B:
$\displaystyle \frac{dL}{dt} = 2(1-0) = 2$
Therefore, student B is learning at a faster rate than student A

(B)
$\displaystyle \frac{dL}{dt} = 2(1-L)$

$\displaystyle \frac{dL}{1-L}= 2 dt$

$\displaystyle -ln |1-L| = 2t +c$

$\displaystyle L = e^{2t +c} = ce^{2t}$

Using the given information $\displaystyle L_A(0) = 0.5$; $\displaystyle L_B(0) = 0$:
$\displaystyle L_A = 0.5e^2t$
$\displaystyle L_A = 0$

No, student B will NOT catch up.

^I was going to set $\displaystyle L_A$ equal to $\displaystyle L_B$, but then I'll end up with trying to solve $\displaystyle ln(0)$, which is not possible, seeing as it approaches negative infinity...

$\displaystyle -\ln |1-L| = 2t +c$ to

$\displaystyle L = e^{2t +c} = ce^{2t}$

Instead, it should be this (I can't skip that many steps safely!):

$\displaystyle \ln|1-L|=-2t-c$

$\displaystyle 1-L=e^{-2t}e^{-c}$

$\displaystyle L=1-e^{-2t}e^{-c}.$

Now try plugging in the initial conditions.