Upon re-examining this,

Would it be:

int((e^(2t^3/3)*P)') = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))

=> (e^(2t^3/3)*P) = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))

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- September 19th 2007, 07:41 PM #1

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## Dif EQ

Another Dif EQ, yay.

I have to solve the following Dif EQ, and give the interval for which the largest general solution is defined:

dP/dt + 2tP = P + 4t - 2

So I don't think I can do separation of variables, so I tried integrating factors...

P(x) = 2t

M(x) = e^(int(2t)dt) = e^(2t^3/3)

So multiply M(x) to each term:

e^(2t^3/3)*dP/dt + e^(2t^3/3)*2tP = e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3)

So this looks very messy now...

Again, it's the part after this that I get stuck.

- September 19th 2007, 08:00 PM #2

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- September 19th 2007, 08:03 PM #3

- September 19th 2007, 08:08 PM #4

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- September 19th 2007, 08:22 PM #5
you had

now what do you think is?

(in this case, using "P" gets confusing, because that's the variable we're working with in the first place, but oh well)

Suppose it was just

dP/dt + 2tP = 0. Would P(x) be 2t then?

- September 19th 2007, 08:36 PM #6

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2t - 1

Okay, so if P(x) = 2t - 1

That means M(x) = e^(int(2t - 1)dt) = e^(t^2 - t)

Multiply M(x) by each term:

e^(t^2 - t)*P' + e^(t^2 - t)*P = e^(t^2 - t)*4t - 2*e^(t^2 - t)

Okay:

int[e^(t^2 - t)*P)'] = int(e^(t^2 - t)*4t - 2*e^(t^2 - t))

=> e^(t^2 - t)*P = {Since this can't be integrated, I think I've done something wrong }

- September 19th 2007, 08:44 PM #7

- September 19th 2007, 09:34 PM #8

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Ah, tricky algebra manipulation.

Okay, so from where you left off:

dP/dt * 1/(2-P) = 2t - 1

=> dP/(2-P) = (2t-1)dt

Take integral of both sides:

-ln|P-2| = t^2 - t + C

Now the second part of the question asked to find the largest interval over which the GENERAL sol'n is defined. This looks defined everywhere, except for when the ln is 0 or negative, but since we have the absolute signs, we don't have to worry about it being neg. Therefore, it's defined (-infinity, 2) or (2, infinity). How am I supposed to know which it's in though.

- September 19th 2007, 09:52 PM #9

- September 19th 2007, 09:58 PM #10

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- September 19th 2007, 10:02 PM #11