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MathSofiya

Solve using variation of parameters \[y''-2y'+y=e^{2x}\]

  • one year ago
  • one year ago

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  1. TuringTest
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    complimentary....

    • one year ago
  2. MathSofiya
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    meaning?

    • one year ago
  3. MathSofiya
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    \[y_c=c_1e^{-x}+c_2xe^{-x}\]

    • one year ago
  4. TuringTest
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    right...

    • one year ago
  5. TuringTest
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    now we need the wronskian, do you know about that?

    • one year ago
  6. MathSofiya
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    nope

    • one year ago
  7. TuringTest
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    it is the determinant of a matrix

    • one year ago
  8. TuringTest
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    for\[y_c=c_1y_1+c_2y_2\]the Wronskian is\[W(y_1,y_2)=\left|\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right|\]

    • one year ago
  9. TuringTest
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    that's a determinant in case you forgot, so find that determinant

    • one year ago
  10. MathSofiya
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    \[\left|\begin{matrix}e^{-x}&xe^{-x}\\-e^{-x}&-e^{-x}(x-1)\end{matrix}\right|\] =\[e^{-x}(e^{-x}(x-1))-xe^{-x}e^{-x}\]

    • one year ago
  11. MathSofiya
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    completely wrong?

    • one year ago
  12. MathSofiya
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    oh it's plus

    • one year ago
  13. TuringTest
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    yep you caught it :)

    • one year ago
  14. MathSofiya
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    \[e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]

    • one year ago
  15. MathSofiya
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    that's it?

    • one year ago
  16. TuringTest
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    you dropped a minus

    • one year ago
  17. MathSofiya
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    where?

    • one year ago
  18. MathSofiya
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    ohhhh

    • one year ago
  19. MathSofiya
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    \[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]

    • one year ago
  20. TuringTest
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    right

    • one year ago
  21. TuringTest
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    simplify and be happy :)

    • one year ago
  22. MathSofiya
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    does it equal\[ e^{-x}\]? \[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}\] \[{-x}(e^{-x}(x-1))+xe^{-x}=1\]

    • one year ago
  23. MathSofiya
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    \[(e^{-x}(x-1))+xe^{-x}=1\]

    • one year ago
  24. MathSofiya
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    typo

    • one year ago
  25. TuringTest
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    no, why did you set it equal to e^-x, we can't set it equal to anything yet...

    • one year ago
  26. MathSofiya
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    oh my bad haha

    • one year ago
  27. MathSofiya
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    so how should I simplify it then?

    • one year ago
  28. TuringTest
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    \[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}\]now try again while I try to eat ;)

    • one year ago
  29. MathSofiya
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    -(x-1)=1-x 1-x=1-x

    • one year ago
  30. MathSofiya
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    no no no ...wait a second

    • one year ago
  31. MathSofiya
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    now i get 2-x=2-x

    • one year ago
  32. MathSofiya
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    nope still wrong

    • one year ago
  33. MathSofiya
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    x=1

    • one year ago
  34. MathSofiya
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    stop laughing

    • one year ago
  35. MathSofiya
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    Let's try that again \[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}\]

    • one year ago
  36. MathSofiya
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    they're just equal to each other

    • one year ago
  37. MathSofiya
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    \[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]

    • one year ago
  38. MathSofiya
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    \[-e^{-2x}(x-1)+xe^{-2x}\]

    • one year ago
  39. MathSofiya
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    \[e^{-2x}((1-x)+x)\]

    • one year ago
  40. MathSofiya
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    \[e^{-2x}\]

    • one year ago
  41. MathSofiya
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    yes?

    • one year ago
  42. TuringTest
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    yes, sorry, I was afk

    • one year ago
  43. MathSofiya
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    LOL ok

    • one year ago
  44. TuringTest
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    do you know the formula for the particular solution for variation of parameters?

    • one year ago
  45. MathSofiya
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    • one year ago
  46. MathSofiya
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    something like that?

    • one year ago
  47. TuringTest
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    almost unidentifiable to me in that form, but correct I'm sure I have the more direct formula memorized http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx (the green box with the formula that says "variation of parameters)

    • one year ago
  48. MathSofiya
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    is the first one a y''? \[y''+q(t)y'+r(t)y=g(t)\]

    • one year ago
  49. TuringTest
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    yeah, it looks screwy I know

    • one year ago
  50. MathSofiya
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    do we know what q(t) and r(t) are? are they the coefficients of y' and y? Probably not it looks like

    • one year ago
  51. TuringTest
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    yes and in this case they are the coefficients, which in this case are constants the formula is just pointing out that they may not be

    • one year ago
  52. MathSofiya
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    oh I see, soo... \[y''+q(t)y'+r(t)y=g(t)\] \[y''-2y'+y=e^{2x}\] so now I integrate

    • one year ago
  53. TuringTest
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    right, using the formula... r(t) and r(t) don't even come into the formula for the particular solution in variation of parameters

    • one year ago
  54. TuringTest
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    q(t) and r(t)... they determine the complimentary only, which as you can see *is* part of the formula for the particular

    • one year ago
  55. MathSofiya
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    \[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{-x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx\]

    • one year ago
  56. TuringTest
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    the first integrand is wrong I think

    • one year ago
  57. TuringTest
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    I think you put the wrong g(x)

    • one year ago
  58. MathSofiya
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    \[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx\]

    • one year ago
  59. TuringTest
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    oh wait, wait, the denominator in each integrand is the wronskian, which we found to be e^(-2x) I think you are confusing notation with the parentheses

    • one year ago
  60. MathSofiya
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    Oh lol so the denominator is just e^{-2x}

    • one year ago
  61. TuringTest
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    right

    • one year ago
  62. TuringTest
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    notice that makes our integrals quite tolerable :)

    • one year ago
  63. MathSofiya
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    \[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}}dx\] That's a lot easier to integrate

    • one year ago
  64. TuringTest
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    oh yeah :) especially after a little simplification

    • one year ago
  65. MathSofiya
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    \[-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]

    • one year ago
  66. MathSofiya
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    so what was the point of doing variation of parameters

    • one year ago
  67. TuringTest
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    well now you have a particular solution

    • one year ago
  68. TuringTest
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    y(x)=yc+yp you got you now so you're good to go

    • one year ago
  69. TuringTest
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    you got yp*

    • one year ago
  70. TuringTest
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    I did not check your integral btw

    • one year ago
  71. MathSofiya
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    that's ok I did wolfram

    • one year ago
  72. TuringTest
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    right-o, I'd do the same right now :)

    • one year ago
  73. MathSofiya
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    oh I see \[c_1e^{-x}+c_2xe^{-x}-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]

    • one year ago
  74. TuringTest
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    yep, and of course if this is an IVP we now proceed to find c1 and c2 otherwise we're done having found the general solution

    • one year ago
  75. MathSofiya
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    good, wow, that was a lot THanks Turing!

    • one year ago
  76. TuringTest
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    Always a pleasure Sofiya :D

    • one year ago
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