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
$\frac{dL}{dt} = 2(1-L)$, where $0 \le L \le 1$
(A) If one of the students knows one-half of the list at time $t=0$ 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:
$\frac{dL}{dt} = 2(1-0.5) = 1$
Student B:
$\frac{dL}{dt} = 2(1-0) = 2$
Therefore, student B is learning at a faster rate than student A

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

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

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

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

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

No, student B will NOT catch up.

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

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

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

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

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

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

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

Now try plugging in the initial conditions.