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alyannahere

hi ! can you help me solve this ? Find the general or particular solution for this differential equation 1. dy/dt=2y 2. dy/dx = -0.5y; y(0) = 100 i realized that it's all y so i don't really know how to go about that.

  • one year ago
  • one year ago

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  1. PeterPan
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    First one :) \[\huge \frac{dy}{dt}=2y\] Let's multiply both sides by dt, and divide both sides by 2y

    • one year ago
  2. PeterPan
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    \[\huge \frac{dy}{2y}=dt\] Now integrate both sides :)

    • one year ago
  3. PeterPan
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    @alyannahere ?

    • one year ago
  4. alyannahere
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    solving it... :)

    • one year ago
  5. alyannahere
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    not really sure how.. ?

    • one year ago
  6. alyannahere
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    i'm really sorry.. i don't know how to go about this problem @PeterPan

    • one year ago
  7. PeterPan
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    Well, maybe if it's tweaked a bit, you'll find it simpler... \[\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt\] Sorry about the other one, it's a typo o.O

    • one year ago
  8. alyannahere
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    so that is 1/2 1/y = C ?

    • one year ago
  9. PeterPan
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    No... remember \[\huge \int\limits \frac{dx}{x}=\ln(x) + C\]

    • one year ago
  10. alyannahere
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    so it's 1/2 ln y = C ?

    • one year ago
  11. alyannahere
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    my friend said the answer is y=C e ^2t

    • one year ago
  12. alyannahere
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    i don't know how he got it

    • one year ago
  13. PeterPan
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    Hang on, just integrate for now, and tell me what you got :)\[\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt\]

    • one year ago
  14. alyannahere
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    1/2 ln y = C ?

    • one year ago
  15. PeterPan
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    Well, isn't it though \[\huge \int\limits dx = x + C\]

    • one year ago
  16. alyannahere
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    how did it become dx ? i'm so sorry i just really don't understand this lesson

    • one year ago
  17. PeterPan
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    This is just a reminder :) The integral of 1 is just x (plus a constant)

    • one year ago
  18. alyannahere
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    then ?

    • one year ago
  19. PeterPan
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    Then... What's \[\huge \int\limits dt\]

    • one year ago
  20. alyannahere
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    t+C ?

    • one year ago
  21. PeterPan
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    Very good :) So now, the result of \[\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt\] is...?

    • one year ago
  22. alyannahere
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    1/2 ln y = t + C ?

    • one year ago
  23. alyannahere
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    :o

    • one year ago
  24. alyannahere
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    wait question, why is did you put 2y in the denominator in the first place?

    • one year ago
  25. PeterPan
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    Well, because we want to put everything involving y with the dy and everything involving t with the dt

    • one year ago
  26. alyannahere
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    that makes sense... but was my answer correct ?

    • one year ago
  27. PeterPan
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    Yeah :) But we're not yet done, hang on

    • one year ago
  28. PeterPan
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    let's \[\huge \frac{1}{2}\ln(y)=t+C\]playing with it a bit... \[\huge \ln(y) = 2t + 2C\]

    • one year ago
  29. PeterPan
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    But 2C is just another constant, so we can disregard the coefficient ln(y) = 2t + C Getting it so far?

    • one year ago
  30. alyannahere
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    yup i get it

    • one year ago
  31. alyannahere
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    oh then you put e to both sides to get sid of the ln ?

    • one year ago
  32. PeterPan
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    That's right :) and e^C is again, just another constant, so just replace it with C And you're done :)

    • one year ago
  33. alyannahere
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    how come it's addition and not multiplication ?

    • one year ago
  34. alyannahere
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    y=C e ^2t ? that's the answer..

    • one year ago
  35. PeterPan
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    Actually, it's \[\huge y = e^{2t + C}\] right? By laws of exponents, this is just \[\huge y = e^Ce^{2t}\] And once again, I said e^C is just another constant anyway, so you can just replace it with C. :)

    • one year ago
  36. alyannahere
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    ohhh !!! i see... thank you sooo much !!!!!!!!!

    • one year ago
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