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My turn to ask for maths help
Topic Started: Mar 6 2008, 12:33 PM (310 Views)
***musical princess***
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HOLY CARP!!!
I was given a homogenous equation in which i am trying to put y' in a general form of y; i have subbed in xv for y and done loads of rearranging to get it into a separable form. I know i have done it right up to this bit, but now that i am trying to integrate, i'm stumped.

I've got...

Int.[(1-v)/(1+v^2)]dv = Int.[dx/x]

I've got the right hand side. It's just... ln|x|+A


... but how the fook do i go about integrating the left hand side.... [(1-v)/(1+v^2)]dv ???


Anyone...???

x
x Caroline x
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Renauda
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HOLY CARP!!!
have you considered 2% or 1% skim rather than homogenized?
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***musical princess***
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That didn't work out too well either

( :rolleyes: :lol: )

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x Caroline x
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Klaus
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Can you post the full exercise?

I have no idea what "a homogenous equation in which i am trying to put y' in a general form of y" means.

Is all you want to know the integral of (1-v)/(1+v^2) ?
Trifonov Fleisher Klaus Sokolov Zimmerman
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***musical princess***
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Klaus
Mar 6 2008, 08:42 PM
Is all you want to know the integral of (1-v)/(1+v^2) ?

Yep, i've done the rest. All i need to know is how i intergrate that so i can finish the question

x
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***musical princess***
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The question i'm doing is:-

Find the general solution of the equation (x-y)y'=(x+y).

x
x Caroline x
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Klaus
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Is it possible that your derivation of (1-v)/(1+v^2) was wrong and what you really need is (1-v)/(1-v^2) ? Because the latter would be quite easy to solve, whereas I'd guess that the former has no analytical solution but only a numerical one.
Trifonov Fleisher Klaus Sokolov Zimmerman
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Klaus
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***musical princess***
Mar 6 2008, 10:45 PM
The question i'm doing is:-

Find the general solution of the equation (x-y)y'=(x+y).

x

What is y'?

Some kind of derivation? But of which function, and at which coordinate?
Trifonov Fleisher Klaus Sokolov Zimmerman
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***musical princess***
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Nope, it's definitely an addition sign. Believe me, i checked for that very thing about 30 times lol

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Klaus
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I have still no idea what the goal of this exercise is.

Here is a solution:

x=0, y=0, y'=0. Equation holds.

This is probably not what the teacher wants, but your (or the teacher's) description is very underspecified.
Trifonov Fleisher Klaus Sokolov Zimmerman
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***musical princess***
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Klaus
Mar 6 2008, 08:48 PM
***musical princess***
Mar 6 2008, 10:45 PM
The question i'm doing is:-

Find the general solution of the equation (x-y)y'=(x+y).

x

What is y'?

Some kind of derivation? But of which function, and at which coordinate?

y' (or y-prime) is just a short hand version of dy/dx.

It's the differential of the function of y with respect to x.


What the question is asking is for me to find what the original function is basically. So they want it written as y(x)=...

The way to do it is to substitute y for xv(x).

So the original question becomes...

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

Which when you substitute becomes...

v+xv'=(x+xv)/(x-xv)

Which, when you take out a factor of x, separate the variables and rearrange, becomes...

(1-v)/(1+v^2)dv=dx/x

x
x Caroline x
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sarah_blueparrot
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That's easy. The answer's 15.
Death is simply a shedding of the physical body like the butterfly shedding its cocoon. It is a transition to a higher state of consciousness where you continue to perceive, to understand, to laugh, and to be able to grow.

- Dr. Elizabeth Kubler-Ross
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***musical princess***
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sarah_blueparrot
Mar 6 2008, 09:14 PM
That's easy. The answer's 15.

or 42 ;)

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JBryan
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Have you tried integration by parts? I am too busy right now to try and work out whether it would work on this but I know it has been useful to me in the past on problems like this.
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There is a line we cross when we go from "I will believe it when I see it" to "I will see it when I believe it".


Henry II: I marvel at you after all these years. Still like a democratic drawbridge: going down for everybody.

Eleanor: At my age there's not much traffic anymore.

From The Lion in Winter.
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***musical princess***
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Oh you legend JB, i think that might actually work... i never thought about that what with it being a quotient but i'll just set dv as (1+v^2)^-1.

I'll give it a try...

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***musical princess***
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It didn't work.... The sub integral just getting a bigger and bigger degree of v so it's was a constant product integration :(

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Klaus
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Int.[(1-v)/(1+v^2)]dv = arctan(v) - 1/2 log(v^2+1)
Trifonov Fleisher Klaus Sokolov Zimmerman
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***musical princess***
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Oooh... how did you do that?

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Klaus
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Int.[(1-v)/(1+v^2)]dv = Int.[1/(1+v^2)]dv - Int.[v/(1+v^2)]dv = arctan(v) - 1/2 log(1+v^2)
Trifonov Fleisher Klaus Sokolov Zimmerman
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Klaus
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What kind of course is this, MP? High school maths? Or something else?
Trifonov Fleisher Klaus Sokolov Zimmerman
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JBryan
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I am the grey one
I was entertaining multiplying by (1 - v)/(1 - v) but I didn't really have the time to work on it. I have a life, you know. :D
"Any man who would make an X rated movie should be forced to take his daughter to see it". - John Wayne


There is a line we cross when we go from "I will believe it when I see it" to "I will see it when I believe it".


Henry II: I marvel at you after all these years. Still like a democratic drawbridge: going down for everybody.

Eleanor: At my age there's not much traffic anymore.

From The Lion in Winter.
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***musical princess***
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Klaus
Mar 6 2008, 11:07 PM
What kind of course is this, MP? High school maths? Or something else?

Degree (Bsc) - uni

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x Caroline x
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