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lgbasalloteBest ResponseYou've already chosen the best response.0
have you tried Bernoulli's?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
dx/dt = x+x^3 divide by x+x^3 multiply by dt
 one year ago

haganmcBest ResponseYou've already chosen the best response.0
the final answer should be \[x=\pm \sqrt{C ^{2t}/(1Ce ^{2t})}\]
 one year ago

ironmanjimboBest ResponseYou've already chosen the best response.0
Multiplying by dt is impossible! Rates of change are as is.
 one year ago

haganmcBest ResponseYou've already chosen the best response.0
you get \[\frac{ 1 }{ x+x^3 }dx=dt\]
 one year ago

haganmcBest ResponseYou've already chosen the best response.0
but how do you integrate this??
 one year ago

ironmanjimboBest ResponseYou've already chosen the best response.0
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
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
@haganmc ... looks good to me.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
oh that wasn't your answer?
 one year ago

haganmcBest ResponseYou've already chosen the best response.0
no now You have to integrate both sides.. i am trying to get x by itself
 one year ago

haganmcBest ResponseYou've already chosen the best response.0
but how do you do partial fraction decomposition?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
use partial frac. expansion on 1/(x+x^3)
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
A / x + Bx /(x^2+1)
 one year ago

haganmcBest ResponseYou've already chosen the best response.0
thats where i messed up... i didnt put an x after the B
 one year ago

ironmanjimboBest ResponseYou've already chosen the best response.0
Algebraic, YES, nice going!
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
_Multiplying by dt is impossible! Rates of change are as is._
 one year ago

ironmanjimboBest ResponseYou've already chosen the best response.0
That is correct. What you are actually doing is separation of variables, It is an important distinction!!!
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
go learn the chain rule
 one year ago

ironmanjimboBest ResponseYou've already chosen the best response.0
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.
 one year ago
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