6359
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2011-06-09T09:13:16
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<a href="http://nrich.maths.org/5874&part=">Warm-up problem</a>
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<a href="http://nrich.maths.org/6569&part=">Try this next</a>
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<a href="https://nrich.maths.org/discus/messages/27/27.html?1307565044">Ask NRICH</a>
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<a href="http://en.wikipedia.org/wiki/Inverse_problem">Read all about it</a>
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<a href="http://nrich.maths.org/7062&part=solution">Last week's solution</a>
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<div>
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<span style="font-weight: bold;">The solution is:</span>
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$$X(t) = K \exp\left(\log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)\right)$$<br />
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<span style="font-style: italic;">Can you find a nice differential equation which this solution satisfies?</span>
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<span style="font-weight: bold;">The solution is:</span>
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$$P(t) = \frac{a\exp(bt)}{a-1+exp(bt)}$$<br />
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<span style="font-style: italic;">Can you find a nice differential equation which this solution satisfies?</span>
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<span style="font-style: italic;">Did you know ... ?</span>
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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 <a href="http://en.wikipedia.org/wiki/Gompertz_function">Gomperz function</a> and is used to model the size
of tumors. Perhaps you might discover the uses for the second solution?</div>
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<p>Part one:<br></br>
<br></br>
We rewrite the equation as:<br></br>
<br></br>
\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*} <comment>
$${X(t)\over K} = exp\left(\log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)\right)$$ <br /><br />
$$\log\left(\frac{X(t)}{K}\right) = \log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)$$ <br />
<br />
$$\frac{ \frac{\mathrm{d}X(t)}{\mathrm{d}t} }{X(t)} = log\left(\frac{X(0)}{K}\right)(-\alpha)\exp(-\alpha t)$$<br /><br />
$${\mathrm{d}X(t)\over \mathrm{d}t} = \alpha\,X(t)\log\left(\frac{K}{X(t)}\right) $$
</comment></p>
<p>which is our required differential equation.<br></br>
<br></br>
Part two:<br></br>
<br></br>
Again we rewrite our equation as:<br></br>
<br></br>
\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))<br></br>
\end{eqnarray*}<br></br>
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<p>Part one:<br></br>
<br></br>
We rewrite the equation as:<br></br>
<br></br>
\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*} <comment>
$${X(t)\over K} = exp\left(\log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)\right)$$ <br /><br />
$$\log\left(\frac{X(t)}{K}\right) = \log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)$$ <br />
<br />
$$\frac{ \frac{\mathrm{d}X(t)}{\mathrm{d}t} }{X(t)} = log\left(\frac{X(0)}{K}\right)(-\alpha)\exp(-\alpha t)$$<br /><br />
$${\mathrm{d}X(t)\over \mathrm{d}t} = \alpha\,X(t)\log\left(\frac{K}{X(t)}\right) $$
</comment></p>
<p>which is our required differential equation.<br></br>
<br></br>
Part two:<br></br>
<br></br>
Again we rewrite our equation as:<br></br>
<br></br>
\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))<br></br>
\end{eqnarray*}<br></br>
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Weekly challenge 42: What's my equation?
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