- Describe as best you can at this point in your own words what a differential equation is.
- Following earnings example from the previous chapter, if the number of employees in a company is growing at a rate of [latex]0.05[/latex] times the number of employees, what is a differential equation that describes this situation?
[latex]E'(t) = 0.05 E(t)[/latex].ans
- Verify the function [latex]f(x) = e^x - x - 1[/latex] solves the differential equation:
[latex]f'(x) = f(x) + x[/latex]
We see
\begin{align*}
f'(x) & = f(x) + x \\
e^x – 1 & = (e^x – x – 1) + x \\
e^x – 1 & = e^x – 1
\end{align*}
as desired.ans
- Verify the function [latex]f(x) = 2 \sqrt{x}[/latex] satisfies the differential equation:
[latex]f'(x) = \frac{2}{f(x)}.[/latex]
We see
\begin{align*}
f'(x) & = \frac{2}{f(x)} \\
x^{-1/2} & = \frac{2}{2 \sqrt{x}} \\
\frac{1}{\sqrt{x}} & = \frac{1}{\sqrt{x}}
\end{align*}
as desired.ans
- For each differential equation, find [latex]f'(t)[/latex] for the given value of [latex]t[/latex], or state there is not enough information.
- Suppose [latex]f'(t) = 3 f(t) + 5[/latex] and [latex]f(3) = -1[/latex]. Find [latex]f'(3)[/latex].
[latex]2[/latex]ans
- Suppose [latex]f'(t) = t + f(t)[/latex], and [latex]f(7) = 1[/latex]. Find [latex]f'(7)[/latex].
[latex]8[/latex]ans
- Suppose [latex]f'(t) = \frac{1}{ \sqrt{f(t)} }[/latex] and [latex]f(0) = 9[/latex]. Find [latex]f'(0)[/latex].
[latex]\frac{1}{3}[/latex]ans
- Suppose [latex]f'(t) = e^{-f(t)}[/latex] and [latex]f(0) = 1[/latex]. Find [latex]f'(1)[/latex].
Not enough information.ans
- Suppose [latex]f'(t) = 3 f(t) + 5[/latex] and [latex]f(3) = -1[/latex]. Find [latex]f'(3)[/latex].
- For each relationship between the value of a function and its derivative, write down a differential equation. For example, if I said “a function is growing at a rate equal to seven times the value of the function” you’d write down [latex]f'(t) = 7 f(t)[/latex].
- A function is growing at a rate equal to twice the function value.
[latex]f'(t) = 2 f(t)[/latex]ans
- A function is growing at a rate equal to the square root of the function value.
[latex]f'(t) = \sqrt{f(t)}[/latex]ans
- A function is growing at a rate equal to [latex]t[/latex] times the function value.
[latex]f'(t) = t f(t)[/latex]ans
- A function is accelerating at a rate equal to the sum of the function value and how quickly the function is growing.
[latex]f''(t) = f'(t) + f(t)[/latex].ans
- A function is growing at a rate equal to twice the function value.
- Verify that the given solution to each differential equation is correct.
- Differential equation [latex]f'(t) = f(t) + 3[/latex], solution [latex]f(t) = 3 e^t - 3[/latex].
\begin{align*}
f'(t) & = f(t) + 3 \\
\frac{d}{dt}(3e^t – 3) & = (3e^t – 3) + 3 \\
3e^t & = 3e^t
\end{align*}ans
- Differential equation [latex]f'(t) = 4\sqrt{f(t)}[/latex], solution [latex]f(t) = 4 t^2[/latex].
\begin{align*}
f'(t) & = 4 \sqrt{f(t)} \\
\frac{d}{dt}(4 t^2) & = 4 \sqrt{4 t^2} \\
8t & = 4(2t) \\
8t & = 8t.
\end{align*}ans
- Differential equation [latex]f'(t) = (f(t))^2[/latex], solution [latex]f(t) = -t^{-1}[/latex].
\begin{align*}
f'(t) & = (f(t))^2 \\
\frac{d}{dt} (-t^{-1}) & = (-t^{-1})^2 \\
t^{-2} & = t^{-2}
\end{align*}ans
- Differential equation [latex]f'(t) = e^{-f(t)}[/latex], solution [latex]f(t) = \ln(t)[/latex].
\begin{align*}
f'(t) & = e^{-f(t)} \\
\frac{d}{dt} \ln(t) & = e^{-\ln(t)} \\
\frac{1}{t} & = \frac{1}{e^{\ln(t)}} \\
\frac{1}{t} & = \frac{1}{t}
\end{align*}ans
- Differential equation [latex]f'(t) = f(t) + 3[/latex], solution [latex]f(t) = 3 e^t - 3[/latex].
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Informal Calculus Copyright © by Tyler Seacrest is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.