haganmc
how to solve this diff equation?? dx/dt−x^3=x
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lgbasallote
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have you tried Bernoulli's?
Algebraic!
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dx/dt = x+x^3
divide by x+x^3
multiply by dt
haganmc
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the final answer should be
\[x=\pm \sqrt{C ^{2t}/(1-Ce ^{2t})}\]
ironmanjimbo
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Multiplying by dt is impossible! Rates of change are as is.
Algebraic!
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lol
haganmc
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you get
\[\frac{ 1 }{ x+x^3 }dx=dt\]
haganmc
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but how do you integrate this??
ironmanjimbo
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haganmc is right on, what he did is separation of variables which is not the same as multiplying by a component of a rate of change. Well done. Now go with factoring and partial fraction decomposition and you are on your way to a solution
Algebraic!
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@haganmc ... looks good to me.
Algebraic!
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oh that wasn't your answer?
Algebraic!
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book's answer?
haganmc
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no now You have to integrate both sides.. i am trying to get x by itself
haganmc
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but how do you do partial fraction decomposition?
Algebraic!
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use partial frac. expansion on 1/(x+x^3)
Algebraic!
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A / x + Bx /(x^2+1)
haganmc
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thats where i messed up... i didnt put an x after the B
ironmanjimbo
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Algebraic, YES, nice going!
Algebraic!
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_Multiplying by dt is impossible! Rates of change are as is._
ironmanjimbo
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That is correct. What you are actually doing is separation of variables, It is an important distinction!!!
Algebraic!
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go learn the chain rule
ironmanjimbo
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I don't mind if you do not want to agree with me, but I will simply suggest that you look into what I'm pointing out regarding how the rate of change is separated in such equations. Simply put, it LOOKS like dt was multiplied, but it was not.