Simplifying a quartic equation I have the following function to simplify and solve but there definitely is something wrong with my method as the initial conditions do not work with my final result so if anyone could pinpoint what I'm doing wrong, I would really appreciate it. 
Solving: $\frac{(N-0.5)^4}{N^2(-N+1)^2}=Ae^t$ where A is a constant. 
Essentially to solve for N my tutor recommended using the substitution $k=N-\frac{1}{2}$.
$\Rightarrow \frac{k^4}{(k+\frac{1}{2})^2(-k+\frac{1}{2})^2}=Ae^t$
$\Rightarrow \frac{k^2}{(k+(\frac{1}{2})(-k+\frac{1}{2})}=\sqrt{Ae^t}$
$\Rightarrow k^2=(k+\frac{1}{2})(-k+\frac{1}{2})\sqrt{Ae^t}$
$\Rightarrow k^2=(\frac{1}{4}-k^2)\sqrt{Ae^t}$
$\Rightarrow (1-\sqrt{Ae^t})k^2-\frac{1}{4}\sqrt{Ae^t}=0$
Then solving this like a quadratic gave me: 
$k= \pm\frac{(\sqrt{Ae^t})^\frac{1}{4}}{2\sqrt(Ae^t)^\frac{1}{4}+1}$
Subbing back $N$ we get the following formula:
$N= \frac{1}{2}\pm\frac{(\sqrt{Ae^t})^\frac{1}{4}}{2\sqrt(Ae^t)^\frac{1}{4}+1}$
However, I was given the initial condition $N(0)=2$ which does not hold for either of my equations, so I have gone wrong somewhere but I'm not sure where personally. 
Any insight would be much appreciated. 
 A: Beginning with
\begin{equation}
\frac{k^4}{\left(k^2-\frac{1}{4}\right)^2}=Ae^t
\end{equation}
we get
\begin{equation}
\frac{k^2}{\left(k^2-\frac{1}{4}\right)}=\pm\sqrt{Ae^t}
\end{equation}
Solving for $k^2$ gives
\begin{equation}
k^2=\frac{\pm\sqrt{Ae^t}}{4\left(1\pm\sqrt{Ae^t}\right)}
\end{equation}
So
\begin{equation}
k=\pm\sqrt{\frac{\pm\sqrt{Ae^t}}{4\left(1\pm\sqrt{Ae^t}\right)}}
\end{equation}
giving
\begin{equation}
N=\frac{1}{2}+\sqrt{\frac{\pm\sqrt{Ae^t}}{4\left(1\pm\sqrt{Ae^t}\right)}}
\end{equation}
with the negative option being ruled out by the requirement that $N(0)=2$.
So when $t=0$ it must be the case that
\begin{eqnarray}
\frac{1}{2}+\sqrt{\frac{\pm\sqrt{A}}{4\left(1\pm\sqrt{A}\right)}}&=&2\\
\frac{\pm\sqrt{A}}{1\pm\sqrt{A}}&=&9\\
\frac{-\sqrt{A}}{1-\sqrt{A}}&=&9
\end{eqnarray}
Choosing the positive option would require $\sqrt{A}$ to be negative.
Thus $\sqrt{A}=\frac{9}{8}$ and $A=\frac{81}{64}$ which is verified when substituted into the original equation.
A: First of all is $A$ any constant ? Because if $A<0$ there is no solutions.
But the real problem comes from 
$$\frac{k^4}{(k+\frac{1}{2})^2(-k+\frac{1}{2})^2}=Ae^t \implies \frac{k^2}{(k+(\frac{1}{2})(-k+\frac{1}{2})}=\sqrt{Ae^t}.$$
You're essentially saying that if $a^2 = b^2$ then $a=b$, but that's simply not true. For example $(-2)^2 = 2^2$ but $-2 \neq 2$. Actually if one knows that $a^2 = b^2$ one can only deduce that $a=b$ or $a=-b$. I suspect that if you take this into account you'll arrive to the right answer (but I'm not sure as I have not done the computations myself).
