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.

  • $\begingroup$ Mathematica gives the answer instantly: $\left\{\frac{1}{2} \left(1-\sqrt{\frac{1-\sqrt{a} e^{t/2}}{a e^t-1}+1}\right),\frac{1}{2} \left(\sqrt{\frac{1-\sqrt{a} e^{t/2}}{a e^t-1}+1}+1\right),\frac{1}{2} \left(1-\sqrt{\frac{\sqrt{a} e^{t/2}+1}{a e^t-1}+1}\right),\frac{1}{2} \left(\sqrt{\frac{\sqrt{a} e^{t/2}+1}{a e^t-1}+1}+1\right)\right\}$ and $A=1$ $\endgroup$ – David G. Stork Feb 28 '17 at 19:51
  • $\begingroup$ What's $\,N(0)\,$ supposed to mean? $\endgroup$ – dxiv Feb 28 '17 at 19:54
  • $\begingroup$ if $N(0) = 2,$ you are going to have $k > (1/2)$ so need to change the step of taking square roots $\endgroup$ – Will Jagy Feb 28 '17 at 19:55
  • $\begingroup$ @dxiv $N(t)=2$ so when $t=0$ my equation should equal $2$, apologies. $\endgroup$ – Evan Feb 28 '17 at 19:56
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    $\begingroup$ @dxiv pretty sure this is solving a first order ODE where $N = N(t).$ The main error is that $N(0) = 2$ requires $\sqrt {(N-1)^2} = N-1$ $\endgroup$ – Will Jagy Feb 28 '17 at 19:58

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}


\begin{equation} k=\pm\sqrt{\frac{\pm\sqrt{Ae^t}}{4\left(1\pm\sqrt{Ae^t}\right)}} \end{equation}


\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.


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).

  • $\begingroup$ the initial condition demands $\sqrt {(-k + \frac{1}{2})^2} = k - \frac{1}{2} $ for the duration of this solution $N$ $\endgroup$ – Will Jagy Feb 28 '17 at 20:04
  • $\begingroup$ Well you're confirming what I'm saying then, from $a^2 = b^2$ he deduced $a=b$ while as you point out it was $a=-b$. Anyway if he was asked to find all the solutions and only then use the initial condition to find a particular solution then he must consider both cases. $\endgroup$ – Errol.Y Feb 28 '17 at 20:10

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