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anonymous

  • 5 years ago

(Cosx )dY +(sinx)Y=1 anyone who can solve it?

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  1. anonymous
    • 5 years ago
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    I first divide with cosx. So i get dY+ sinx/cosx Y =cosx..

  2. anonymous
    • 5 years ago
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    Then i Take U as... \[U=e ^{(sinx \div cosx)} ..... \] \[e ^{\int\limits_{}^{}(sinx \div cosx)} Y \int\limits_{}^{}e ^{\int\limits_{}^{}(sinx \div cosx)} \times Cos x\]

  3. anonymous
    • 5 years ago
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    \[e ^{\int\limits_{}^{}sinx \div cosx} = e ^{-\ln(cosx)} = e^{\ln(cosx)^{-1}} = cosx ^{-1}\]

  4. anonymous
    • 5 years ago
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    \[1\div \cos x \times Y =\int\limits_{}^{} (1\div cosx) \times cox x\] \[1 \div cosx \times Y =\int\limits 1\] \[1 \div cosx \times Y =x +c\] \[Y= xcos(x) + Ccos(x)\]

  5. anonymous
    • 5 years ago
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    Anyone no were i did the wrong?

  6. anonymous
    • 5 years ago
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    Yes

  7. anonymous
    • 5 years ago
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    Umm...I'll write out a solution, scan and attach. You're on the right track, though.

  8. anonymous
    • 5 years ago
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    Thank you!

  9. anonymous
    • 5 years ago
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    Can you wait a bit? I want to make some tea.

  10. anonymous
    • 5 years ago
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    Np!

  11. anonymous
    • 5 years ago
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    Scanning now...

  12. anonymous
    • 5 years ago
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    okeej! i dont see... anywrong.. hm.. but.. were is the wrong.

  13. anonymous
    • 5 years ago
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    1 Attachment
  14. anonymous
    • 5 years ago
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    If it's unfamiliar, it's because I just derive the integrating factor from first principles. It's just one of those differential equations where you use an integrating factor, as you have done.

  15. anonymous
    • 5 years ago
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    Okej, so i can t do like i have done?..

  16. anonymous
    • 5 years ago
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    I'm not too sure what you've done (on the face of it). Did you attempt to use an integrating factor, or are you looking at a book and seeing the form of your equation in there and following the book's description on how to solve it?

  17. anonymous
    • 5 years ago
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    Yes, the symbol U is the as the integrating factor

  18. anonymous
    • 5 years ago
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    brb

  19. anonymous
    • 5 years ago
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    ok

  20. anonymous
    • 5 years ago
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    What i did, was that used U =e^ (int (Sinx\cosx))

  21. anonymous
    • 5 years ago
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    or you could right int ( tan x) like you did.

  22. anonymous
    • 5 years ago
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    Yes, I saw.

  23. anonymous
    • 5 years ago
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    The method of integrating factors boils down to this: for a differential equation \[y'+p(x)y=q(x)\]then the solution y is found by solving\[\frac{d(\mu y)}{dx}=\mu q(x) \rightarrow \mu y= \int\limits_{}{}\mu q(x) dx + c\]where c is a constant and\[\mu = e^{\int\limits_{}{} p(x)dx}\]

  24. anonymous
    • 5 years ago
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    Then\[y=\frac{1}{\mu}\int\limits_{}{}\mu q(x) dx+\frac{c}{\mu}\]

  25. anonymous
    • 5 years ago
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    Were you jumping to that formula to work it out?

  26. anonymous
    • 5 years ago
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    Yes Exactly

  27. anonymous
    • 5 years ago
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    That was, what i was trying to do.

  28. anonymous
    • 5 years ago
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    \[y=\frac{1}{\sec x}\int\limits_{}{}\sec x \sec x dx+\frac{c}{\sec x}=\cos x \int\limits_{}{}\sec^2x dx+c \cos x\]\[=\cos x \tan x + c \cos x=\sin x + c \cos x\]

  29. anonymous
    • 5 years ago
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    okeey, i see ..

  30. anonymous
    • 5 years ago
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    i must have done wrong with the e^∫sinx÷cosx

  31. anonymous
    • 5 years ago
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    The integration factor

  32. anonymous
    • 5 years ago
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    What happens when you try it again?

  33. anonymous
    • 5 years ago
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    The forumla is not wrong? huh? I uses this formula, \[U \times Y =\int\limits_{}^{}U \times q(x)\]

  34. anonymous
    • 5 years ago
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    Where, DyDx + r(x)Y= q(x)

  35. anonymous
    • 5 years ago
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    The formula will give you the same answer AS LONG AS you integrate the right-hand side BEFORE dividing by the U on the left-hand side. If you don't do this, you won't get full use of the constant of integration. It forms part of the solution.

  36. anonymous
    • 5 years ago
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    For example

  37. anonymous
    • 5 years ago
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    Yes of course. that i now..

  38. anonymous
    • 5 years ago
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    imagine, you divide through by U now. Then you'd have\[Y=\frac{1}{\sec x}\int\limits_{}{}\sec x \sec x dx=\cos x \tan x +c = \sin x +c\]

  39. anonymous
    • 5 years ago
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    Okay

  40. anonymous
    • 5 years ago
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    What is your Integration factor ?

  41. anonymous
    • 5 years ago
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    sec x

  42. anonymous
    • 5 years ago
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    same as before

  43. anonymous
    • 5 years ago
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    Equals = 1/ cosx

  44. anonymous
    • 5 years ago
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    as my.

  45. anonymous
    • 5 years ago
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    You don't get the solution.

  46. anonymous
    • 5 years ago
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    \[U=\sec x = \frac{1}{\cos x}\]

  47. anonymous
    • 5 years ago
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    I found my wrong... I ve right... , at first when i divided with Cosx to get Dydx free... i, wrote dY+ sinx/cosx Y =cosx. it should be dY+ sinx/cosx Y =1\cosx.

  48. anonymous
    • 5 years ago
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    Yes

  49. anonymous
    • 5 years ago
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    ahhhh for god sake...

  50. anonymous
    • 5 years ago
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    Thank you sooooooooooooooooooooooooo much!!!!! appreciate the help

  51. anonymous
    • 5 years ago
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    Sorry, I saw that before - I thought you saw it too after the attachment was sent...lol, sorry.

  52. anonymous
    • 5 years ago
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    that's fine. become a fan! ;)

  53. anonymous
    • 5 years ago
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    Already one..

  54. anonymous
    • 5 years ago
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    nice

  55. anonymous
    • 5 years ago
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    ;)

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