Author Topic: TUT0102 Quiz3  (Read 308 times)

Changhao Jiang

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TUT0102 Quiz3
« on: October 11, 2019, 02:04:50 PM »
\documentclass{article}
\usepackage[utf8]{inputenc}


\usepackage{natbib}
\usepackage{graphicx}
\usepackage{amsmath}
\begin{document}


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

Solution: We calculate the Wronskian of f and g:

$W =
\begin{vmatrix}
f(t) & g(t) \\
f'(t) & g'(t) \\
\end{vmatrix}
=
\begin{vmatrix}
e^{2t} & g(t) \\
2e^{2t} & g'(t) \\
\end{vmatrix} $
= $e^{2t}g'(t)-2e^{2t}g(t)=3e^{4t}$ \newline
$e^{2t}g'(t)-2e^{2t}g(t)=3e^{4t}$ \newline
Divided by $e^{2t}$ on both sides, we get $g'(t)-2g(t)=3e^{3t}$ \newline
$\mu(t)$ = $e^{\int-2dt}$ = $e^{-2t}$ \newline
multiply $e^{-2t}$ on both sides, $e^{-2t}g'(t)-2e^{-2t}g(t)=3$ \newline
$(e^{-2t}g(t))'=3$ \newline
Integral on both sides, $e^{-2t}g(t)=3t+C$ \newline
g(t) = $3te^2t+Ce^{2t}$

Coollight

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Re: TUT0102 Quiz3
« Reply #1 on: October 11, 2019, 02:13:44 PM »
Hi, I believe the final answer should be

\begin{align*}
\ g(t)= 3te^{2t} + ce^{2t} \\
\end{align*}

 :)