Calculus Small Changes, Big Ideas

Section 4.2 Exponential functions

• Population growth:

  \begin{equation*} \dv{y}{t}=ky \end{equation*}

  How can we figure out which function satisfies this relationship?

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• Introduce slope fields and Euler’s method

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• We find that the solution curves are always exponential curves. When \(k=1\text{,}\) we have \(\DS\dv{y}{t}=y\text{,}\) so we’ll call the solution for which \(y(0)=1\) the exponential function and label it \(y=\exp(t)\text{.}\)

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• By properties of derivatives, we know that the solution of \(\DS\dv{y}{t}=y\) with \(y(0)=c\) must be \(y=c\exp(t)\text{.}\) In particular, the solution of \(\DS\dv{y}{t}=y\) with \(y(0)=\exp(a)\) must be \(y=\exp(a)\exp(t)\text{.}\)

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  However, using the Chain Rule, we can also show that \(y=\exp(a+t)\) is a solution of \(\DS\dv{y}{t}=y\) with \(y(0)=\exp(a)\text{,}\) since

  \begin{equation*} \dv{t}\exp(a+t) = \exp(a+t)\cdot\dv{t}(a+t) = \exp(a+t)\cdot 1 = \exp(a+t)\text{.} \end{equation*}

  Since the solution to a linear differential equation is unique, we must have \(\exp(a+t)=\exp(a)\exp(t)\text{.}\) This is exactly the behavior we expect from exponents, since \(b^{x+y}=b^xb^y\) for any base \(b\gt 0\text{.}\)

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  It can be shown numerically that

  \begin{equation*} \exp(1)=2.718281828459045235\cdots=\ee\text{,} \end{equation*}

  so instead of \(\exp(t)\text{,}\) we can write \(\ee^t\text{,}\) and hence we have

  \begin{equation*} \ee^{a+t}=\ee^a\ee^t \end{equation*}

  as expected.

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• We can write this as an integral as well:

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  \begin{align*} \int \ee^x\dd{x} \amp = \ee^x+C \end{align*}
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• Overall, the solution to the original differential equation \(\DS\dv{y}{t}=ky\) is

  \begin{equation*} y=y_{0}\ee^{kt}\text{.} \end{equation*}

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• When \(k\gt 0\text{,}\) we get exponential growth.

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  When \(k\lt 0\text{,}\) we get exponential decay.

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