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anonymous

  • 5 years ago

Find the solution to the differential equation, subject to the given initial condition 6*dp/dt=p^4, p(0)=9. it seems so simple but cant get the answer right please Help!

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  1. anonymous
    • 5 years ago
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    This is a separable differential equation. You just need to rearrange:\[6p^{-4}dp=dt\]integrate and solve for your constant.

  2. anonymous
    • 5 years ago
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    Is this how you started?

  3. anonymous
    • 5 years ago
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    Anyway, integrate both sides:\[6\frac{p^{-3}}{-3}=t+c \rightarrow -2p^{-3}=t+c \rightarrow p^{-3}=c-\frac{t}{2}\]

  4. anonymous
    • 5 years ago
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    So\[p(0)=9 \rightarrow 9^{-3}=c-\frac{0}{2}=c\]

  5. anonymous
    • 5 years ago
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    yes and i have to write it out as p(t)= but i keep getting it wrong so far i havep^3=(t+C)/-2

  6. anonymous
    • 5 years ago
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    I get\[p(t)=\sqrt[3]{\frac{2}{c-t}}\]

  7. anonymous
    • 5 years ago
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    \[p(0)=9 \rightarrow 9=\sqrt[3]{\frac{2}{c}} \rightarrow c=\frac{2}{9^3}\]

  8. anonymous
    • 5 years ago
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    yep thats what I've been plugging into wiley plus to no avail

  9. anonymous
    • 5 years ago
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    What exactly have you been plugging in?

  10. dumbcow
    • 5 years ago
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    i get a constant of -2/243 -> p(t) = cubed root (-486/(243t - 2))

  11. anonymous
    • 5 years ago
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    cubed root of (2/729-t)

  12. anonymous
    • 5 years ago
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    Maybe you need to use \[c=-\frac{2}{9^3}\]for a solution\[p(t)=\sqrt[3]{\frac{-2}{t+c}}\]

  13. anonymous
    • 5 years ago
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    \[p(t)=9\frac{\sqrt[3]{-2}}{\sqrt[3]{9^3t-2}}\]

  14. anonymous
    • 5 years ago
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    These are just different ways of saying the same thing. These online submission things are such a crapshoot.

  15. anonymous
    • 5 years ago
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    i have new values now 3 *du/dt= u^3, u(0)=6 which then i get once integrated u^-2= 2/3t+C

  16. anonymous
    • 5 years ago
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    substituting gives me c=1/12 so u(t)=square root (2/3t+1/12)?

  17. anonymous
    • 5 years ago
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    you are correct a crapshoot it is

  18. anonymous
    • 5 years ago
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    \[u= \frac{6}{\sqrt{1-24t}}\]

  19. anonymous
    • 5 years ago
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    positive root only satisfies the boundary condition.

  20. anonymous
    • 5 years ago
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    I had c=-1/24

  21. anonymous
    • 5 years ago
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    once you integrated did you get 3/2 u^2=t+C

  22. anonymous
    • 5 years ago
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    and this checks out with wolfram alpha

  23. anonymous
    • 5 years ago
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    \[-\frac{3}{2}u^{-2}=t+c\]after integration.

  24. anonymous
    • 5 years ago
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    aha i see what you did, nice job! Thank you for the help!

  25. anonymous
    • 5 years ago
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    np

  26. anonymous
    • 5 years ago
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    So, does that mean your other question is defunct now?

  27. anonymous
    • 5 years ago
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    well i have an understanding of the problem but all of my attepmts are used

  28. anonymous
    • 5 years ago
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    but thank you

  29. anonymous
    • 5 years ago
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    you're welcome :)

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