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Messages - Chang Peng (Eddie) Liu

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16
Quiz 2 / 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.

17
Quiz 1 / Re: Q1 problem 1 (L5101)
« on: September 25, 2014, 12:40:57 PM »
Hi Prof. Ivrii,

It took me close to an hour to come up with that in MSW because I'm not used to typing out equations; so by the time I finished, Roro already posted it! But thank you!

18
Quiz 1 / Re: Q1 problem 1 (L5101)
« on: September 25, 2014, 01:04:58 AM »
I'm having trouble typing out equations with proper format in this forum, so I did it in MSW and screenshot the work.. Apologies in advanced!

19
Quiz 1 / Re: Q1 problem 2 (Night sections)
« on: September 24, 2014, 11:04:15 PM »
Firstly, I'd like to apologize for the really ugly notations and equations.. there is the solution! I rewrote it--V.I.

Find the solution of the given initial value problem.
\begin{gather}
y' - 2y = e^{2t},\label{A}\\
y(0) = 2.\label{B}
\end{gather}

First, we need to find the integrating factor, which is $I = e^{\int -2\,dt}= e^{-2t}$

so we multiple the entire equation by I, thus, giving us
$e^{-2t} y' - 2e^{-2t} y = 1$.
We see that the left side of the equation can be rewritten as $[e^{-2t}y]'$
and we see that the right side of the equation is in fact 1

therefore: $[e^{-2t} y]' = 1$

we now take the integral of both sides, giving us: $e^{-2t} y = t + C$   where $C$ is a constant. To find  $C$, we substitute $y = 2$ when $t$ = 0 (this information is given in the question). We find out that $C = 2$.

rearranging the formula, we come to the solution as
\begin{equation*}
y = (t+2)  e^{2t}.
\end{equation*}

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