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I have this particular question which i'm having trouble with... could someone help me?
"Solve the following initial value problem
(1+x)y' + y^2 - 1 = 0, y(0)=0"
Thanks in advance!
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This differential equation is of the variables-separable type. You can separate the variables and integrate.
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If you don't mind checking my answer:
arctanh(y) = ln (x+1)
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Thats what Ive got as well.
Since
you can also write the answer as
Last edited by JaneFairfax (2007-03-10 12:01:05)
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Cheers!
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I have another similar question this time, but it's proving too be a little bit more difficult:
y' + tan(x)y + x = 0, y(0)=2
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This is an equation of the form
I believe theres a standard method for solving it. (You know, the one involving particular integrals and complementary functions?)
EDIT: Er, I found the formula to use:
http://eqworld.ipmnet.ru/en/solutions/ode/ode0103.pdf
Last edited by JaneFairfax (2007-03-11 04:14:14)
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I'm not sure if that applies to this question
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Why not? The solution is
where C is a constant to be determined. You just have to find a way to integrate xsecx so good luck with that one.
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I have a very strong suspicion that you got one of the signs in your equation wrong. Instead of
Im very sure it should be
with a minus sign for the tanx. Then instead of xsecx you would be integrating xcosx which is really a piece of cake. The solution to the second equation above satisfying the given initial condition would then be
Frankly I dont see how you can integrate xsecx without making so much of a mess that the question becomes not worth answering at all.
Last edited by JaneFairfax (2007-03-11 05:43:13)
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I agree with you as well but the question seems to be what i have typed out
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Well, then youll have to find a way to integrate xsecx. If you integrate by parts, youll have to integrate the integral of secx, which is ln|secx+tanx|. Im not sure how you do it, really.
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Mathematica gives:
Therefore, you probably wrote the question wrong. QED.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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