- Verify that the given solution to each differential equation is correct, and solve for the free parameter.
- Differential equation [latex]f'(t) = f(t) + 3[/latex], solution [latex]f(t) = A e^t - 3[/latex], [latex]f(0) = 4[/latex].
\begin{align*}
f'(t) & = f(t) + 3 \\
\frac{d}{dt} (A e^t – 3) & = (A e^t – 3) + 3 \\
A e^t & = A e^t.
\end{align*}
If [latex]f(0) = 4[/latex], then [latex]A = 7[/latex].ans
- Differential equation [latex]f'(t) = 2 f(t) - 2[/latex], solution [latex]f(t) = A e^{2t} + 1[/latex], [latex]f(0) = 0[/latex].
\begin{align*}
f'(t) & = 2 f(t) – 2 \\
\frac{d}{dt} (A e^{2t} + 1) & = 2(A e^{2t} + 1) – 2 \\
2 A e^{2t} & = 2 A e^{2t} + 2 – 2 \\
2 A e^{2t} & = 2 A e^{2t}
\end{align*}
If [latex]f(0) = 0[/latex], then [latex]A = -1[/latex].ans
- Differential equation [latex]f'(x) = \frac{1}{f(x) + 1}[/latex], solution [latex]f(x) = \sqrt{A+2 x+1} - 1[/latex], [latex]f(0) = 4[/latex].
\begin{align*}
f'(x) &= \frac{1}{f(x) + 1} \\
\frac{d}{dx} (\sqrt{A+2 x+1} – 1) &= \frac{1}{(\sqrt{A+2 x+1} – 1) + 1} \\
\frac{d}{dx} (A + 2x + 1)^{1/2} & = \frac{1}{\sqrt{A + 2x + 1}} \\
\frac{1}{2} (A + 2x + 1)^{-1/2} \cdot 2 & = \frac{1}{\sqrt{A + 2x + 1}} \\
(A + 2x + 1)^{-1/2} & = \frac{1}{\sqrt{A + 2x + 1}} \\
\frac{1}{\sqrt{A + 2x + 1}} & = \frac{1}{\sqrt{A + 2x + 1}}
\end{align*}
If [latex]f(0) = 4[/latex], then [latex]A = 24[/latex].ans
- Differential equation [latex]f'(t) = (f(t))^2 + f(t)[/latex], solution [latex]f(t) = -\frac{Ae^{t}}{Ae^t - 1}[/latex], [latex]f(0) = 3[/latex].
\begin{align*}
f'(t) & = (f(t))^2 + f(t) \\
\frac{d}{dt} \left( -\frac{Ae^{t}}{Ae^t – 1} \right) & = \left(-\frac{Ae^{t}}{Ae^t – 1}\right)^2 + \left( -\frac{Ae^{t}}{Ae^t – 1} \right) \\
-\frac{(A e^t – 1)Ae^t – Ae^t(Ae^t)}{(A e^t – 1)^2} & = \frac{(A e^t)^2}{(A e^t -1 )^2} – \frac{Ae^t}{A e^t – 1} \\
-\frac{(A e^t)^2 – Ae^t – (Ae^t)^2}{(A e^t – 1)^2} & = \frac{(A e^t)^2}{(A e^t -1 )^2} – \frac{Ae^t}{A e^t – 1} \cdot \frac{(A e^t – 1)}{(A e^t – 1)} \\
-\frac{- Ae^t}{(A e^t – 1)^2} & = \frac{(A e^t)^2}{(A e^t -1 )^2} – \frac{(A e^t)^2 – Ae^t}{(A e^t – 1)^2} \\
\frac{Ae^t}{(A e^t – 1)^2} & = \frac{(A e^t)^2}{(A e^t -1 )^2} + \frac{-(A e^t)^2 + Ae^t}{(A e^t – 1)^2} \\
\frac{Ae^t}{(A e^t – 1)^2} & = \frac{(A e^t)^2 – (Ae^t)^2 + Ae^t}{(A e^t -1 )^2} \\
\frac{Ae^t}{(A e^t – 1)^2} & = \frac{Ae^t}{(A e^t -1 )^2}
\end{align*}
If [latex]f(0) = 3[/latex], then [latex]A = 1.5[/latex].ans
- Differential equation [latex]f'(t) = f(t) + 3[/latex], solution [latex]f(t) = A e^t - 3[/latex], [latex]f(0) = 4[/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.