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Date: Wed, 16 May 2007 22:32:32 -0700
From: "Young Rao" <young....@gmail.com>
To: Math54@googlegroups.com
Subject: Re: Fall 2006 Final #5
In-Reply-To: <1179378265.528601.92140@q75g2000hsh.googlegroups.com>
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so I just worked it out... I think that all solutions tend to zero as long
as a =/= 0

in the case where a < -1 or a >1, we will get a complex eigenvalue, with the
real part always = -1.
if -1 < a < 0 or 0 < a < 1, then you will add a real number to -1, that
number will always be < -1, so the eigenvalue will stay negative, netting
two negative eigenvalues. The only case left to consider is a = 0, and in
that case we will get -2 and 0 as eigenvalues, which does not satisfy the
conditions.

On 5/16/07, shervinat...@gmail.com <shervinat...@gmail.com> wrote:
>
>
> For part A, we got that lambda = (-2 +- [4 - 4(alpha)^2]^1/2 ) / 2,
> which implies that alpha > 1 and alpha < -1 gives complex eigenvalues.
>
> For part b, we get that lamba = -alpha, which implies that any
> positive alpha gives a negative lambda, which is what we want for the
> ODE to tend to 0 as t goes to infinity. However, we also realized that
> if alpha < -1, we have a complex eigenvalue, where the real part of
> the eigenvalue is -1, which is what we want for the ODE to tend to
> zero, right?
>
> My question is - would the solution to part b simply be the interval
> from zero to infinity, or negative infinity to -1 union zero to
> infinity?
>
>
> >
>

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so I just worked it out... I think that all solutions tend to zero as long as a =/= 0<br><br>in the case where a &lt; -1 or a &gt;1, we will get a complex eigenvalue, with the real part always = -1. <br>if -1 &lt; a &lt; 0 or 0 &lt; a &lt; 1, then you will add a real number to -1, that number will always be &lt; -1, so the eigenvalue will stay negative, netting two negative eigenvalues. The only case left to consider is a = 0, and in that case we will get -2 and 0 as eigenvalues, which does not satisfy the conditions.
<br><br><div><span class="gmail_quote">On 5/16/07, <b class="gmail_sendername"><a href="mailto:shervinat...@gmail.com">shervinat...@gmail.com</a></b> &lt;<a href="mailto:shervinat...@gmail.com">shervinat...@gmail.com</a>&gt; wrote:
</span><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"><br>For part A, we got that lambda = (-2 +- [4 - 4(alpha)^2]^1/2 ) / 2,<br>which implies that alpha &gt; 1 and alpha &lt; -1 gives complex eigenvalues.
<br><br>For part b, we get that lamba = -alpha, which implies that any<br>positive alpha gives a negative lambda, which is what we want for the<br>ODE to tend to 0 as t goes to infinity. However, we also realized that<br>
if alpha &lt; -1, we have a complex eigenvalue, where the real part of<br>the eigenvalue is -1, which is what we want for the ODE to tend to<br>zero, right?<br><br>My question is - would the solution to part b simply be the interval
<br>from zero to infinity, or negative infinity to -1 union zero to<br>infinity?<br><br><br></div><br>

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