# What's my equation?

## Problem

The solution is:

$$X(t) = K \exp\left(\log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)\right)$$

Can you find a nice differential equation which this solution satisfies?

The solution is:

$$P(t) = \frac{a\exp(bt)}{a-1+exp(bt)}$$

Can you find a nice differential equation which this solution satisfies?

There is a branch of mathematics concerned with solving so-called 'inverse-problems'. In an inverse problem you begin with a solution, or some partial solution, and attempt to construct the equations or theories which might give rise to it. The first solution in this problem is called the Gomperz function and is used to model the size of tumors. Perhaps you might discover the uses for the second solution?

## Student Solutions

Part one:

We rewrite the equation as:

\begin{eqnarray*} {X(t)\over K} &=&
\exp\left(\log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)\right) \\
\log\left(\frac{X(t)}{K}\right) &=&
\log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)\\ \frac{
\frac{\mathrm{d}X(t)}{\mathrm{d}t} }{X(t)} &=&
\log\left(\frac{X(0)}{K}\right)(-\alpha)\exp(-\alpha t))\\
{\mathrm{d}X(t)\over \mathrm{d}t} &=&
\alpha\,X(t)\log\left(\frac{K}{X(t)}\right) \end{eqnarray*}

which is our required differential equation.

Part two:

Again we rewrite our equation as:

\begin{eqnarray*} P(t) &=& \frac{a\exp(bt)}{a-1+exp(bt)}
\\ P(t)(a-1)+P(t)\exp\left(bt\right) &=&
a\exp\left(bt\right) \\ \frac{P(t)(a-1)}{a-P(t)} &=&
\exp\left(bt\right)
\\ \log\left(P(t)(a-1)\right)-\log\left(a-P(t)\right)
&=& bt \\ \Biggr(\frac{(a-1) \mathrm{d}P}{(a-1)P(t)} +
\frac{\mathrm{d}P}{a-P(t)}\Biggr) &=& b\mathrm{d}t \\
\Biggr(\frac{1}{P(t)} + \frac{1}{a-P(t)}\Biggr) \mathrm{d}P
&=& b\mathrm{d}t \\ \frac{\mathrm{d}P}{\mathrm{d}t}
&=& \frac{b}{a} P(t)(a-P(t))

\end{eqnarray*}