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