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**MT / Re: MT problem 3**

« **on:**December 05, 2014, 11:57:12 PM »

can someone please tell me what section of the textbook this question was from?

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can someone please tell me what section of the textbook this question was from?

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I just showed the steps for finding the second eigenvector..

After that, you would proceed to plug it into the standard general equation for a repeated eigenvalue question.

After that, you would proceed to plug it into the standard general equation for a repeated eigenvalue question.

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This is what I did... Please feel free to correct me if I'm wrong! :p

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Solve the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing $t$.

\begin{equation*}

y''+ 4y'+ 4y = 0, \qquad y(-1) = 2,\quad y'(-1) = 1.

\end{equation*}

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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.

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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!

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!

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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!

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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*}

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 $[e^{-2t}y]'$

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

rearranging the formula, we come to the solution as

\begin{equation*}

y = (t+2) e^{2t}.

\end{equation*}

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