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anonymous

  • 5 years ago

hi i need help with differentiation equation

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  1. anonymous
    • 5 years ago
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    post yur question

  2. anonymous
    • 5 years ago
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    Solve the seperable differential equation for u

  3. anonymous
    • 5 years ago
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    du/dt =e^(4u-16t)

  4. anonymous
    • 5 years ago
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    Use the following initial condition: u(0) = 0 U = ?

  5. anonymous
    • 5 years ago
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    u*

  6. anonymous
    • 5 years ago
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    first lets re-write the equation:\[e^{4u-16t} = e^{4u}*e^{-16t} = {e^{4u} \over e^{-16t}} \]

  7. anonymous
    • 5 years ago
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    can you solve from there?

  8. anonymous
    • 5 years ago
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    this is calculus btw

  9. anonymous
    • 5 years ago
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    ops typo, the last one should be => e^(16t) no negative cause it is as a fraction form

  10. anonymous
    • 5 years ago
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    i did do that wat u rewritten

  11. anonymous
    • 5 years ago
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    \[{du \over dt} = {e^{4u} \over e^{16t} } \]

  12. anonymous
    • 5 years ago
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    is this calculus 1 ?

  13. anonymous
    • 5 years ago
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    calculus 2 and i did do e^4u/e^16t how to proceed ?

  14. anonymous
    • 5 years ago
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    e(-4u)du=e(-16t)dt solve it for each variable: (-1/4)* e(-4u)=(-1/16)*e^(-16t) + cont simplify

  15. anonymous
    • 5 years ago
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    i found the constant to be -3/16

  16. anonymous
    • 5 years ago
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    good?

  17. anonymous
    • 5 years ago
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    ok. So, now all we do is separate 't' terms and 'u' tems\[{1\over e^{4u}} du= {1 \over e^{16t}}dt\]

  18. anonymous
    • 5 years ago
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    yes but i found constant to be -3/16.... what to do next to find u?

  19. anonymous
    • 5 years ago
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    is that constant correct?

  20. anonymous
    • 5 years ago
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    mathmind, what happens to (-1/4) & (-1/16) coefficients (you have to get it after integration)

  21. anonymous
    • 5 years ago
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    thats wat u get when u substitute both u and t with 0

  22. anonymous
    • 5 years ago
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    initial condition u(o)=0, what about condition for t?

  23. anonymous
    • 5 years ago
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    wasnt mentioned in my question

  24. anonymous
    • 5 years ago
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    i found c = 0

  25. anonymous
    • 5 years ago
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    agree with MathMind.. c=0

  26. anonymous
    • 5 years ago
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    c = 0 u = 4t

  27. anonymous
    • 5 years ago
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    is that answer?

  28. anonymous
    • 5 years ago
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    yup

  29. anonymous
    • 5 years ago
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    it doesnt work, my homework dont accept that

  30. anonymous
    • 5 years ago
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    (-1/4)e^(-4u)=(-1/16)e^(-16t) +C -4u=C*(-16t) if u(0)=0, co C=0

  31. anonymous
    • 5 years ago
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    HINT: To determine the constant of integration after you integrated both sides, DO NOT take natural logs, but rather just set u = 0 and t = 0 with u and t both still in the exponents. After determining the constant, then you need to take logs on both sides to solve for u. This is what the hint that my questioned offered.

  32. anonymous
    • 5 years ago
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    same thing.. c will be equal 0.. that's just a shortcut.. kinda

  33. anonymous
    • 5 years ago
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    you didn't tell this before... if u=0 & t=0 C=-1/4 +1/16 = -3/16

  34. anonymous
    • 5 years ago
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    'oh wait nvm.. inik is right

  35. anonymous
    • 5 years ago
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    so C is not zero? inik do u agree with mathmind?

  36. anonymous
    • 5 years ago
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    can u figure out the u now?

  37. anonymous
    • 5 years ago
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    comes from: (-1/4)e^(-4u)=(-1/16)e^(-16t) +C (-1/4)*1= (-1/16)*1 +C we did it!

  38. anonymous
    • 5 years ago
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    so u = ?

  39. anonymous
    • 5 years ago
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    just put value of C=-3/16 in: (-1/4)e^(-4u)=(-1/16)e^(-16t) +C simplify & take ln to both sides

  40. anonymous
    • 5 years ago
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    I'm doing it now...

  41. anonymous
    • 5 years ago
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    how do i take log of both sides am not good with logs

  42. anonymous
    • 5 years ago
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    that is where i am confused at

  43. anonymous
    • 5 years ago
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    give me sec

  44. anonymous
    • 5 years ago
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    ok

  45. anonymous
    • 5 years ago
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    you take ln to take away the e^\[\ln (e^x) = x\]

  46. anonymous
    • 5 years ago
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    it only takes away e^?

  47. anonymous
    • 5 years ago
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    wat if it is -e^

  48. anonymous
    • 5 years ago
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    \[{-1 \over 4} e^{-4u}={-1 \over 16}e^{-16t} +C\] taking the Ln of everything so we can solve for u:\[{-1 \over 4}\ln (e^{-4u}) = {-1 \over 16} \ln(e^{-16t}) + \ln C\]\[{-1\over4} (-4u) = {-1 \over 16} (-16t) + \ln C\]

  49. anonymous
    • 5 years ago
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    now do i plug in 0 for t?

  50. anonymous
    • 5 years ago
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    let's try: e^(-4u)=1/4 * e^(-16t) +3/4 -4u=ln[1/4*(e^(-16t) +3)] u=-1/4 *[ln(e^(-16t)+3) - ln4] u=ln4/4 -1/4* ln(e^(-16t) +3)

  51. anonymous
    • 5 years ago
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    t= variable unless you been asked to put t=o, you can't

  52. anonymous
    • 5 years ago
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    u(t)=... u(0) = 0

  53. anonymous
    • 5 years ago
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    not for general solution

  54. anonymous
    • 5 years ago
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    what you mean?

  55. anonymous
    • 5 years ago
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    it was not for you... sorry. you posted your response the same time as me...

  56. anonymous
    • 5 years ago
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    oh haha alriught

  57. anonymous
    • 5 years ago
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    I mean that the answer should be in form u(t)=...one thing if u(o) means t=0 - another. your response would be correct :)

  58. anonymous
    • 5 years ago
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    have to go. Thank you MathMind & jophil - was fun!

  59. anonymous
    • 5 years ago
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    np :D thanks for joining

  60. anonymous
    • 5 years ago
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    i have to go now baii

  61. anonymous
    • 5 years ago
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    ty to u 2 mathmind cya

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