Author Topic: Quiz 2 Problem 1 (night sections)  (Read 799 times)

Chang Peng (Eddie) Liu

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Quiz 2 Problem 1 (night sections)
« on: October 01, 2014, 10:39:30 PM »
3.2 #17

If the Wronskian $W$ of $f$ and $g$ is $3e^{4t}$ , and if $f (t) = e^{2t}$ , find $g(t)$.


Can anyone type solution? - V.I.
« Last Edit: October 02, 2014, 03:57:50 AM by Victor Ivrii »

Roro Sihui Yap

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Re: Quiz 2 Problem 1 (night sections)
« Reply #1 on: October 02, 2014, 11:16:18 AM »
Consider Wronskian  $W = \det \begin{bmatrix} f(t) \ \ \ g(t) \\f’(t) \ \ g’(t)\end{bmatrix}$
\begin{equation}
3e^{4t} = f(t)g’(t) - g(t)f’(t)
\end{equation}
We know $f(t) = e^{2t}$. Therefore $f’(t) = 2e^{2t}$. Substitute the terms in
\begin{equation} 3e^{4t} = (e^{2t})g’(t) - (2e^{2t})g(t) \end{equation}
Divide all terms by $(e^{2t})$
\begin{gather} 3e^{2t} = g’(t) - 2g(t) \\
g’(t) - 2g(t) = 3e^{2t} \end{gather}
We need to find an integrating factor $\mu (t) = e^{\int -2\,dt}= e^{-2t}$ . Multiply all terms by $e^{-2t}$
\begin{gather} g’(t)e^{-2t} - 2g(t)e^{-2t} = 3 \\
 [g(t)e^{-2t}]' = 3 \\
g(t)e^{-2t} = 3t + c \\
 g(t) = 3te^{2t} + ce^{2t} \end{gather}
Nice—I made minor improvements. V.I.
« Last Edit: October 02, 2014, 11:47:39 AM by Victor Ivrii »