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haganmc Group Title

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

  • one year ago
  • one year ago

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  1. lgbasallote Group Title
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    have you tried Bernoulli's?

    • one year ago
  2. Algebraic! Group Title
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    dx/dt = x+x^3 divide by x+x^3 multiply by dt

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

    • one year ago
  4. ironmanjimbo Group Title
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    Multiplying by dt is impossible! Rates of change are as is.

    • one year ago
  5. Algebraic! Group Title
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    lol

    • one year ago
  6. haganmc Group Title
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    you get \[\frac{ 1 }{ x+x^3 }dx=dt\]

    • one year ago
  7. haganmc Group Title
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    but how do you integrate this??

    • one year ago
  8. ironmanjimbo Group Title
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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

    • one year ago
  9. Algebraic! Group Title
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    @haganmc ... looks good to me.

    • one year ago
  10. Algebraic! Group Title
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    oh that wasn't your answer?

    • one year ago
  11. Algebraic! Group Title
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    book's answer?

    • one year ago
  12. haganmc Group Title
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    no now You have to integrate both sides.. i am trying to get x by itself

    • one year ago
  13. haganmc Group Title
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    but how do you do partial fraction decomposition?

    • one year ago
  14. Algebraic! Group Title
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    use partial frac. expansion on 1/(x+x^3)

    • one year ago
  15. Algebraic! Group Title
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    A / x + Bx /(x^2+1)

    • one year ago
  16. haganmc Group Title
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    thats where i messed up... i didnt put an x after the B

    • one year ago
  17. ironmanjimbo Group Title
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    Algebraic, YES, nice going!

    • one year ago
  18. Algebraic! Group Title
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    _Multiplying by dt is impossible! Rates of change are as is._

    • one year ago
  19. ironmanjimbo Group Title
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    That is correct. What you are actually doing is separation of variables, It is an important distinction!!!

    • one year ago
  20. Algebraic! Group Title
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    go learn the chain rule

    • one year ago
  21. ironmanjimbo Group Title
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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.

    • one year ago
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is replying to Can someone tell me what button the professor is hitting...

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