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chikaa

  • 2 years ago

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

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  1. myininaya
    • 2 years ago
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    Separation of variables is needed.

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

  3. myininaya
    • 2 years ago
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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.

  4. myininaya
    • 2 years ago
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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.

  5. chikaa
    • 2 years ago
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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: \[P=\exp((\frac{ k }{ 2r }\sin(2rt-2v)+\frac{ kt }{ 2 }\]\[+\frac{ k }{ 2r }\sin(2v))\]

  6. chikaa
    • 2 years ago
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    I don't get where the last bit came from

  7. myininaya
    • 2 years ago
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    So did you set if up like I told you to?

  8. myininaya
    • 2 years ago
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    And then integrate?

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

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

  11. chikaa
    • 2 years ago
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    yeah that's what I did

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

  13. chikaa
    • 2 years ago
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    help needed urgently!!!

  14. KenLJW
    • 2 years ago
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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

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