Re: help with 1st order ODEs

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Posted by Soroban on September 18, 2002 at 14:12:23:

In Reply to: help with 1st order ODEs posted by Matt on September 18, 2002 at 12:55:20:

Hello, Matt!

You said you "just enrolled", so I assume you haven't heard of
"Undetermined Coefficients", "Method of Operators", "Variation
of Parameters" ~ good stuff ~ to be shown later.

I assume you can use "I" (Integrating Factors), so here's my solution.

y' - y = 2te^2t, where y(0) = 1

In this problem: I = e^[INT(-1)dt] = e^(-t)

Mutliply through by e^(-t):
e^(-t)*y' - e^(-t)*y = 2te^t

The left side is the differential of e^(-t)*y:
d[e^(-t)*y] = 2te^t dx

Integrating on the left, we get: e^(-t)*y

Integrating on the right, we do it "by parts":
2te^t - 2e^t + C

We have: e^(-t)*y = 2te^t - 2e^t + C

or: y = 2te^2t - 2e^2t + Ce^t

From y(0) = 1, we have: 1 = 0 - 2 + C, or C = 3

Therefore: y = 2te^2t - 2e^2t + 3e^t

or: y = 2(t - 1)e^2t + 3e^t

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