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Mathematics Forum
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| Author | Message / Information |
| jharring Quote | Reply | | r' = r - r^2 posted on: 9/30/2006 11:59:33 PM Can somebody help me out with this. I'm probably being stupid but i can't even think of how to start this. You have r'(t) = r(t) - r(t)*r(t). I know r(t) = e^t/(e^t+c) but I can't think of the steps to go through to solve this. Thanks! |
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herbert1
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r' = r - r^2
replied on: 10/12/2006 4:00:54 PM dr/dx =r(1-r) dx/dr = 1/(r*(1-r)) dx = dr/(r*(1-r)) = dr*(1/r + 1/(1-r)) therefore x= ln(r) -ln(1-r) => x=ln(r/(1-r)) exp(x)= r/(1-r) therefore r=exp(x)/(1+exp(x)) if the result is beeing checked it should agree with the primary equation |
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Grogler
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r' = r - r^2
replied on: 11/8/2006 8:52:51 AM Why??? dx = dr/(r*(1-r)) = dr*(1/r + 1/(1-r)) therefore x= ln(r) - ln(1-r) There must be: x+C=ln|r|-ln|1-r|, isn't it? |
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herbert
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r' = r - r^2
replied on: 11/8/2006 1:21:45 PM dear Mr. Grogler, please note that my expression => x = ln(r/(1-r)) is substantially the same as your version of the same namely x = ln(r) - ln(1-r) but it perfectly legitimate to raise my expression ( or yours ) into exp() giving exp(x) = r/(1-r) thus removing the logarithm completely. Further : multiplication gives (1-r)*exp(x) = r or exp(x) - r*exp(x) =r or exp(x)=r+r*exp(x) or exp(x)=r*(1+exp(x)) or r(x)= exp(x)/(1+exp(x)) thus I got the final form of the solution which I think suites the basic differential equation according to its form r´= r - r^2 Of course there might be some constant lacking or the possibility that your form is more convenient in special cases. Besides - I was right with my guess that the decomposition into fractions lacks a minus throughout. But this does not afect the solution gravely.... Mathematiccs is simple only calculating is difficult.... Basically both our forms of solution are therefore equivalent With kind regards |
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Grogler
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r' = r - r^2
replied on: 11/8/2006 2:41:12 PM To herbert. I agree with you... |
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