Step By Step Calculus » 16.1 - Exponential Growth/Decay

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Now, motivated by practical applications, we are interested in finding solutions to
\displaystyle \frac{d}{dt}F(t)=k F(t) \quad \displaystyle \frac{d}{dt}F(t)=k F(t) \quad for \displaystyle t\geq 0 \quad \displaystyle t\geq 0 \quad and \displaystyle F(0)=C, \displaystyle F(0)=C,      \displaystyle \displaystyle
i.e., f(t)f(t) is replaced by kF(t)kF(t). Since the left and right hand sides involve the same function, this is not just an antiderivative but rather Equation (1) is known as a (linear ordinary) differential equation.
Herein, we show that F(t)=Ce^{kt}F(t)=Ce^{kt} is the unique solution to (1).
To determine kk and CC, the general equations to use are:
\displaystyle k\displaystyle k
\displaystyle =\displaystyle =
\displaystyle \displaystyle \frac{\ln \left(\frac{f(t_2)}{f(t_1)}\right) }{t_2-t_1}\qquad\qquad C=\displaystyle \frac{f(t_1)}{e^{kt_1}}\displaystyle \displaystyle \frac{\ln \left(\frac{f(t_2)}{f(t_1)}\right) }{t_2-t_1}\qquad\qquad C=\displaystyle \frac{f(t_1)}{e^{kt_1}}
\displaystyle \displaystyle
Finally, we show that the n^{th}n^{th} derivative, F^{(n)}(t)F^{(n)}(t), of F(t)F(t) also satisfies an equation like (1) but with CC replaced with C \cdot k^{n}C \cdot k^{n}:
\displaystyle \frac{d}{dt}F^{(n)}(t)=kF^{(n)}(t) \quad \displaystyle \frac{d}{dt}F^{(n)}(t)=kF^{(n)}(t) \quad with\displaystyle \quad F^{(n)}(0)=Ck^{n}. \displaystyle \quad F^{(n)}(0)=Ck^{n}.