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chikaa Group Title

how to find the general solution: dP/dt=kP cos(rt-v)^2 k, r and v are positive constants

  • one year ago
  • one year ago

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  1. myininaya Group Title
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    Separation of variables is needed.

    • one year ago
  2. myininaya Group Title
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    Put the problem in this form and the integrate both sides. \[g(p) dp=f(t) dt \]

    • one year ago
  3. myininaya Group Title
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    You can picture those k, r , and v's being like 1,2, and 3's. They are just constants. You need your P's together and your t's together. I don't care and you shouldn't care where this puts the constants.

    • one year ago
  4. myininaya Group Title
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    You can actually leave k on the side it is on or divide both sides by k. It doesn't matter.

    • one year ago
  5. chikaa Group Title
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    this is what I got: \[P=\exp(\frac{ k }{ 2r }\sin(2rt-2v)+\frac{ kt }{ 2 })\] However, this is what I should be getting according to the answer n my tutorial: \[P=\exp((\frac{ k }{ 2r }\sin(2rt-2v)+\frac{ kt }{ 2 }\]\[+\frac{ k }{ 2r }\sin(2v))\]

    • one year ago
  6. chikaa Group Title
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    I don't get where the last bit came from

    • one year ago
  7. myininaya Group Title
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    So did you set if up like I told you to?

    • one year ago
  8. myininaya Group Title
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    And then integrate?

    • one year ago
  9. myininaya Group Title
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    \[\frac{1}{p} dp=k \cos^2(rt-v) dt\]

    • one year ago
  10. myininaya Group Title
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    \[\ln|p|=k \int\limits_{}^{} \frac{1}{2}(\cos(2[rt-v]+1) dt\]

    • one year ago
  11. chikaa Group Title
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    yeah that's what I did

    • one year ago
  12. chikaa Group Title
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    i'm just not getting the last bit k/2r sin(2v)

    • one year ago
  13. chikaa Group Title
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    help needed urgently!!!

    • one year ago
  14. KenLJW Group Title
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    dP/dt=kP cos(rt-v)^2 dP/P=kcos(rt-v)^2dt ln(P)=(k/3r)cos(rt-v)^3 check with taking the Derivative with chain rule P=C1EXP[(k/3r)cos(rt-v)^3

    • one year ago
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