### It's Only a Minus Sign

Solve these differential equations to see how a minus sign can change the answer

### Differential Equation Matcher

Match the descriptions of physical processes to these differential equations.

### Taking Trigonometry Series-ly

Look at the advanced way of viewing sin and cos through their power series.

# What's My Equation?

##### Age 16 to 18 Challenge Level:

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*}