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# P3 differential help watch

1. ok problem is with:

find solution curves to following equation:

dy/dx = y + y^2 , through (0,1)

and suggest restrictions which should be placed on the values of x.
it probably is very easy, im just not getting it,

(this is from ocr book ex.9B q.3)

cheerz.
2. 1/[y(y + 1)] dy/dx = 1
1/y - 1/(y + 1) dy/dx = 1
ln(y/(y + 1)) = x + ln(1/2) . . . using the initial condition
y/(y + 1) = (1/2)e^x
(y + 1)/y = 2 e^(-x)
1/y = 2 e^(-x) - 1
y = 1 / (2 e^(-x) - 1)

This solution is valid when (2 e^(-x) - 1) > 0. Ie, it is valid when x < ln(2).

Does the question really ask for solution curves?
3. (Original post by Jonny W)

y = 1 / (2 e^(-x) - 1)

This solution is valid when (2 e^(-x) - 1) > 0. Ie, it is valid when x < ln(2).

Does the question really ask for solution curves?
cheerz, thanx a lot but yeah, but the book is a ***** and this stuff hasnt come up in the last 5 years of exams, so i;ll leave that for now.

when solution is valid, why cant (2 e^(-x) - 1) < 0, surely as long as it doesnt equal 0 it is valid
4. (Original post by KerChing)
cheerz, thanx a lot but yeah, but the book is a ***** and this stuff hasnt come up in the last 5 years of exams, so i;ll leave that for now.

when solution is valid, why cant (2 e^(-x) - 1) < 0, surely as long as it doesnt equal 0 it is valid
It has something to do with the fact that e^x is asymtopic.
5. (Original post by Bhaal85)
It has something to do with the fact that e^x is asymtopic.
cheerz, still confused, but i;ll just hope that it wont come up on wednesday
6. (Original post by KerChing)
cheerz, still confused, but i;ll just hope that it wont come up on wednesday
I doubt it will come up, your on the OCR board like me too, have you done any past papers? I can do everything apart from setting up differential equations and some vectors.
7. (Original post by Bhaal85)
I doubt it will come up, your on the OCR board like me too, have you done any past papers? I can do everything apart from setting up differential equations and some vectors.
yeah differential equations,especially the wordy ones are annoying, especially because the P4 stuff is actually easy because there's a rule you can learn there.

but you cant apply it to p3, so it can be annoying, vectors should hopefully be easy asking point of intersection or angle, hopefully nothing difficult
8. (Original post by KerChing)
why cant (2 e^(-x) - 1) < 0, surely as long as it doesnt equal 0 it is valid
The given data doesn't determine y(x) for x > ln(2). Loosely speaking, the system explodes as we pass x = ln(2), and the initial condition no longer influences it. If the problem is to find a function y: R \ {ln(2)} -> R such that

dy(x)/dx = y(x) + y(x)^2 . . . for all x =/= ln(2),
y(0) = 1

then there are infinitely many solutions. (Here R is the set of all real numbers and R \ {ln(2)} is the set of all real numbers except ln(2).) One solution, as you said, is given by y(x) = 1 / (2 e^(-x) - 1) for x =/= ln(2). But there are others:

y(x) = 1 / (2 e^(-x) - 1) for x < ln(2),
y(x) = 2004 + 1 / (2 e^(-x) - 1) for x > ln(2)

works just as well.

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Updated: June 5, 2004
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