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alyannahere

  • one year ago

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.

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

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

  3. PeterPan
    • one year ago
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    @alyannahere ?

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

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

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

  7. PeterPan
    • one year ago
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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

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

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

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

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

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

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

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

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

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

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

  18. alyannahere
    • one year ago
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    then ?

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

  20. alyannahere
    • one year ago
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    t+C ?

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

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

  23. alyannahere
    • one year ago
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    :o

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

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

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

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

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

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

  30. alyannahere
    • one year ago
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    yup i get it

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

  32. PeterPan
    • one year ago
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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 :)

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

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

  35. PeterPan
    • one year ago
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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. :)

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

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