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lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0have you tried Bernoulli's?

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

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

ironmanjimbo
 2 years ago
Best ResponseYou've already chosen the best response.0Multiplying by dt is impossible! Rates of change are as is.

haganmc
 2 years ago
Best ResponseYou've already chosen the best response.0you get \[\frac{ 1 }{ x+x^3 }dx=dt\]

haganmc
 2 years ago
Best ResponseYou've already chosen the best response.0but how do you integrate this??

ironmanjimbo
 2 years ago
Best ResponseYou've already chosen the best response.0haganmc 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!
 2 years ago
Best ResponseYou've already chosen the best response.1@haganmc ... looks good to me.

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

haganmc
 2 years ago
Best ResponseYou've already chosen the best response.0no now You have to integrate both sides.. i am trying to get x by itself

haganmc
 2 years ago
Best ResponseYou've already chosen the best response.0but how do you do partial fraction decomposition?

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

haganmc
 2 years ago
Best ResponseYou've already chosen the best response.0thats where i messed up... i didnt put an x after the B

ironmanjimbo
 2 years ago
Best ResponseYou've already chosen the best response.0Algebraic, YES, nice going!

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

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

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

ironmanjimbo
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.
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