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G.3.b.iii) Use the formula to come up with the exact initial value (starter) on y[0] that gives the break between the two families

Stay with the same differential equation:

y'[t] = f[t_,y_] = - t^2 + y;
y[0] = starter

y [t] = 2 - 2 * [ExponentialE]^t + starter * [ExponentialE]^t + 2t + t^2

Use this formula to come up with the exact initial value (starter) on y[0] that gives the break between the two families
----
Yeah, so how do find the exact initial value that'll give the break between a solution that'll graph upwards, and one that'll graph downwards? I'm pretty sure the answer should be 2, but I'm not sure...

Date: 2005-04-08 11:50 am (UTC)
From: [identity profile] p-trekkie.livejournal.com
Yeah it's definitely 2. If less than two the negative exponential will dominate as t-->inf, but if greater than two the positive exponential will dominate. t^2 and lower order terms can't hold a candle to the exponential as t--> inf, so you can basically ignore them. At least I think that's what they're asking.

Date: 2005-04-08 12:19 pm (UTC)
From: [identity profile] ibneko.livejournal.com
::nods:: probably... thanks. That makes more sense to me now... ::was looking at it some completely different way::

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