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haganmc

  • 2 years ago

how to solve this diff equation?? dx/dt−x^3=x

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  1. lgbasallote
    • 2 years ago
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    have you tried Bernoulli's?

  2. Algebraic!
    • 2 years ago
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    dx/dt = x+x^3 divide by x+x^3 multiply by dt

  3. haganmc
    • 2 years ago
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    the final answer should be \[x=\pm \sqrt{C ^{2t}/(1-Ce ^{2t})}\]

  4. ironmanjimbo
    • 2 years ago
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    Multiplying by dt is impossible! Rates of change are as is.

  5. Algebraic!
    • 2 years ago
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    lol

  6. haganmc
    • 2 years ago
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    you get \[\frac{ 1 }{ x+x^3 }dx=dt\]

  7. haganmc
    • 2 years ago
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    but how do you integrate this??

  8. ironmanjimbo
    • 2 years ago
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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

  9. Algebraic!
    • 2 years ago
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    @haganmc ... looks good to me.

  10. Algebraic!
    • 2 years ago
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    oh that wasn't your answer?

  11. Algebraic!
    • 2 years ago
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    book's answer?

  12. haganmc
    • 2 years ago
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    no now You have to integrate both sides.. i am trying to get x by itself

  13. haganmc
    • 2 years ago
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    but how do you do partial fraction decomposition?

  14. Algebraic!
    • 2 years ago
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    use partial frac. expansion on 1/(x+x^3)

  15. Algebraic!
    • 2 years ago
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    A / x + Bx /(x^2+1)

  16. haganmc
    • 2 years ago
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    thats where i messed up... i didnt put an x after the B

  17. ironmanjimbo
    • 2 years ago
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    Algebraic, YES, nice going!

  18. Algebraic!
    • 2 years ago
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    _Multiplying by dt is impossible! Rates of change are as is._

  19. ironmanjimbo
    • 2 years ago
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    That is correct. What you are actually doing is separation of variables, It is an important distinction!!!

  20. Algebraic!
    • 2 years ago
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    go learn the chain rule

  21. ironmanjimbo
    • 2 years ago
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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.

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