How to represent the given information correctly to solve for a particular solution to a differential equation? Lupita's lawn is left unattended so that an infestation of weeds begins to take over. The rate of growth of the weeds is proportional to the area of lawn not yet invaded by weeds.
a) If $W \space m^2$ is the area of lawn taken over by weeds after t weeks, and the lawn has a total area of $A \space m^2$, write the differential equation that models the situation.
b) The area taken over by weeds grows from one quarter to one half the total area of the lawn in $T$ weeks. If $t = 0$ when $W = \frac{1}{4}A$, solve the differential euqation in part a).
First I write down the differential equation, which is: $\frac{dW}{dt} = k(A - W)$. I proceed and find the general solution for this differential equation which is: $W = A - Be^{-kt}$ for that I let $e^c = B$
.
I continue this with the information I know to interpret; that is when $t = 0$,  $W = \frac{1}{4}A$.
This gives:
$\frac{1}{4}A = A - Be^{-k(0)}$
$\frac{3}{4}A = B$. Substituting this back to the equation:
$W = A - \frac{3}{4}Ae^{-kt}$
Now this is where I am stuck. I am unsure how to represent "The area taken over by weeds grows from one quarter to one half the total area of the lawn in $T$ weeks" mathematically. I have the proportional constant $k$ left to solve.
Do I have to do something like:$ \int _{\frac{1}{4}A}^{\frac{1}{2}A}\:\frac{1}{A-W}dW\:=\:\int _0^T\:k\:dt$ How can I solve for $k$, hence the equation?
FYI the answer is:
$W\:=\:A\:-\frac{3}{4}Ae^{ln\left(\frac{2}{3}\right)t}$
Thanks.
 A: 
Lupita's lawn is left unattended so that an infestation of weeds begins to take over. The rate of growth of the weeds is proportional to the area of lawn not yet invaded by weeds.


(a) If $W \space m^2$ is the area of lawn taken over by weeds after $\color{red}{t_0}$ weeks, and the lawn has a total area of $A \space m^2$, write the differential equation that models the situation.


(b) The area taken over by weeds grows from one quarter to one half the total area of the lawn in $T$ weeks. If $t = 0$ when $W = \frac{1}{4}A$, solve the differential equation in part (a).

Let $x\equiv x(t)$ be the area covered by weeds at any time $t$, then
\begin{align*}
\frac {dx}{dt}&=k(A-x)\\
x&=A-B\exp\left({-kt}\right)\\
W&=A-B\exp\left({-kt_0}\right)\\
x&=A-(A-W)\exp\left({-k(t-t_0)}\right)\\
\frac{A-x}{A-W}&=\exp\left({-k(t-t_0)}\right)\\
\frac{A-\frac A4}{A-W}&=\exp\left({-k(t_1-t_0)}\right)\tag{1}\\
\frac{A-\frac A2}{A-W}&=\exp\left({-k(t_2-t_0)}\right)\tag{2}\\
\frac32&=\exp\left({k(t_2-t_1)}\right)&(\because\text{Dividing }(1)\text{ by }(2))\\
\frac32&=\exp\left({kT}\right)\\
\frac{\ln\left(\frac32\right)}T&=k\\
\Rightarrow x&=A-(A-W)\exp\left[{-\ln\left(\frac32\right)\left(\frac{t-t_0}T\right)}\right]\\
\Rightarrow \frac A4&=A-(A-W)\exp\left[{\frac{t_0}T\ln\left(\frac32\right)}\right]\\
\frac{3A}{4(A-W)}&=\exp\left[{\frac{t_0}T\ln\left(\frac32\right)}\right]\\
\ln\left(\frac{3A}{4(A-W)}\right)&=\frac{t_0}T\ln\left(\frac32\right)\\
\Rightarrow x&=A-(A-W)\exp\left[{-\ln\left(\frac32\right)\left(\dfrac{t-\frac{T\ln\left(\frac{3A}{4(A-W)}\right)}{\ln\left(\frac32\right)}}T\right)}\right]\\
x&=A-(A-W)\exp\left[\ln\left(\frac{3A}{4(A-W)}\right)-\ln\left(\frac32\right)\frac tT\right]\\
x&=A-(A-W)\exp\left[\ln\left(\frac{3A}{4(A-W)}\right)\right]\exp\left[-\ln\left(\frac32\right)\frac tT\right]\\
x&=A-\frac{3A}4\exp\left[\ln\left(\frac23\right)\frac tT\right]&\left(\because -\ln\left(\frac ab\right)=\ln\left(\frac ba\right)\right)\\
x&=A-\frac{3A}4\left(\frac32\right)^{-\dfrac tT}&(\because \exp\left(\ln x\right)=x)\\
\end{align*}
A: These are first $T$ weeks (since the initial condition is $W_0 = \frac{1}{4} A$), you have:
$$ \frac{1}{4} A = A(1-\frac{3}{4} e^{k*0}) $$
and
$$ \frac{1}{2} A = A (1-\frac{3}{4} e^{k*T}) $$
Dividing the second one with the first one:
$$\begin{align}
 2 &= \frac{(1-\frac{3}{4} e^{kT})}{(1-\frac{3}{4} e^{k*0})} \\
1 &= 2 - 2 \frac{3}{4} e^{kT} \\
\frac{1}{2} &= \frac{3}{4} e^{kT} \\
\ln(\frac{2}{3}) &= kT \\
k &= \frac{\ln(\frac{2}{3})}{T}
\end{align} $$
