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# differencial equations

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the exam in math is comming near, so i need to "brush up" a bit... example: condtion: P(2,2) xy' - y' = xy + y how does one solve this? i know that y' is dy/dx... as i know u have forst solve the "homogenoues" part of the equation ie: y' +- something = 0 equation can also be written as: y'(x-1) = y (x+1) do u have to move all y-params on left side or can u say that y'(x-1) = 0 ?? [edited by - original vesoljc on January 21, 2003 3:35:01 PM]

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You''re on the right track; you have to separate your variables before integrating.

This is a game development board, so you should not ask this kind of question here. Look at the Forum FAQ.

Cédric

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cant we say that i''m making a "math" based game?

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Homework questions are reluctantly permitted, as long as you follow the guidelines of the FAQ, e.g., the main thing is that you should not just ask for an answer. Show your work and ask for guidance not answers.

In any case, please do try to keep even proper homework questions to a minimum, .

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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this equation is homgenous

y''(x-1) = y (x+1)

y''/y = (x+1)/(x-1)

(1/y) * dy/dx = (x+1)/(x-1)

now integration (dx) will give you a solution.

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ga,

Please do not post answers to homework questions. The forum FAQ explains my policy on this.

Forum FAQ

Original vesoljc''s question did follow forum policy on homework, e.g., they showed enough of their work to prove they weren''t merely asking for an answer. And cedricl''s reply was appropriate, since it gave a hint at how to proceed. An explicit answer to the question is not appropriate.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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so can "we" solve this as a first order linear differential equation?
y'' = dy/dx
dy/dx + p(x)y = q(x)
y''(x-1) = y(x+1)
y''(x-1) - y(x+1) = 0 /x-1)
y'' - y(x+1)/(x-1) = 0
p(x) = (x+1)/(x-1)
q(x) = 0

first, the integrating factor:
u(x) = e^(INT[p(x)dx]) = e^(INT[(x+1)/(x-1)dx]) which is umm...
we say that t = (x-1)
dt = 1*dx
so, INT[t+2/t dt] -> INT[t/t dt] + INT[2/t dt]
INT[dt] + 2*INT[1/t dt] -> INT[dx] + 2*INT[1/t dt]
since INT[1/x dx] = log[x] ->
x + 2*log[x-1],
which gives us u(x) = e^(x+2*Log[x-1])
second,
y = (INT[u(x)q(x)dx] + C ) / u(x)
q(x) = 0 ?
y = (INT[u(x)*0* dx] + C) / u(X)

so y = (0 + C) /u(x) ?
correct?

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the second smile " " should not be a smile but a division

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I''ll remind folks to review original vesoljc''s most recent reply, offering suggestions but not answers if you find anything wrong.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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original vesoljc, your integration was executed nicely.

No need to go through the complications at the end, however. Once you've integrated both sides, you are given the starting conditions x and y, and thus enough information to find C. I don't know why you started putting it in to all this crazy notation with p(x), q(x), and u(x) flying all over the place, maybe it's just the way your Calculus professor has taught these integration techniques. I'm not saying it's a bad thing, but it can lead to confusion if used more often than necessary.

You already found the original function y, which was e[x + 2ln(x-1)] - as always, take the derivative of that to verify its correctness. First try to put it in a form that will yield the original, un-integrated equation. Anyway, in the final form y = e[x + 2ln(x-1)] + C, you can plug in the initial conditions to find C.

Also, smooth move substituting t for (x - 1) That was quite clever, I liked it. I was still looking for the pattern

[edited by - Zipster on January 27, 2003 2:33:25 AM]

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