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MathSofiya

  • 3 years ago

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

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  1. TuringTest
    • 3 years ago
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    complimentary....

  2. MathSofiya
    • 3 years ago
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    meaning?

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

  4. TuringTest
    • 3 years ago
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    right...

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

  6. MathSofiya
    • 3 years ago
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    nope

  7. TuringTest
    • 3 years ago
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    it is the determinant of a matrix

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

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

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

  11. MathSofiya
    • 3 years ago
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    completely wrong?

  12. MathSofiya
    • 3 years ago
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    oh it's plus

  13. TuringTest
    • 3 years ago
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    yep you caught it :)

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

  15. MathSofiya
    • 3 years ago
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    that's it?

  16. TuringTest
    • 3 years ago
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    you dropped a minus

  17. MathSofiya
    • 3 years ago
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    where?

  18. MathSofiya
    • 3 years ago
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    ohhhh

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

  20. TuringTest
    • 3 years ago
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    right

  21. TuringTest
    • 3 years ago
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    simplify and be happy :)

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

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

  24. MathSofiya
    • 3 years ago
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    typo

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

  26. MathSofiya
    • 3 years ago
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    oh my bad haha

  27. MathSofiya
    • 3 years ago
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    so how should I simplify it then?

  28. TuringTest
    • 3 years ago
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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 ;)

  29. MathSofiya
    • 3 years ago
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    -(x-1)=1-x 1-x=1-x

  30. MathSofiya
    • 3 years ago
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    no no no ...wait a second

  31. MathSofiya
    • 3 years ago
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    now i get 2-x=2-x

  32. MathSofiya
    • 3 years ago
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    nope still wrong

  33. MathSofiya
    • 3 years ago
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    x=1

  34. MathSofiya
    • 3 years ago
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    stop laughing

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

  36. MathSofiya
    • 3 years ago
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    they're just equal to each other

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

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

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

  40. MathSofiya
    • 3 years ago
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    \[e^{-2x}\]

  41. MathSofiya
    • 3 years ago
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    yes?

  42. TuringTest
    • 3 years ago
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    yes, sorry, I was afk

  43. MathSofiya
    • 3 years ago
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    LOL ok

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

  45. MathSofiya
    • 3 years ago
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  46. MathSofiya
    • 3 years ago
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    something like that?

  47. TuringTest
    • 3 years ago
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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)

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

  49. TuringTest
    • 3 years ago
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    yeah, it looks screwy I know

  50. MathSofiya
    • 3 years ago
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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

  51. TuringTest
    • 3 years ago
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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

  52. MathSofiya
    • 3 years ago
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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

  53. TuringTest
    • 3 years ago
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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

  54. TuringTest
    • 3 years ago
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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

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

  56. TuringTest
    • 3 years ago
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    the first integrand is wrong I think

  57. TuringTest
    • 3 years ago
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    I think you put the wrong g(x)

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

  59. TuringTest
    • 3 years ago
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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

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

  61. TuringTest
    • 3 years ago
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    right

  62. TuringTest
    • 3 years ago
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    notice that makes our integrals quite tolerable :)

  63. MathSofiya
    • 3 years ago
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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

  64. TuringTest
    • 3 years ago
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    oh yeah :) especially after a little simplification

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

  66. MathSofiya
    • 3 years ago
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    so what was the point of doing variation of parameters

  67. TuringTest
    • 3 years ago
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    well now you have a particular solution

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

  69. TuringTest
    • 3 years ago
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    you got yp*

  70. TuringTest
    • 3 years ago
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    I did not check your integral btw

  71. MathSofiya
    • 3 years ago
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    that's ok I did wolfram

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

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

  74. TuringTest
    • 3 years ago
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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

  75. MathSofiya
    • 3 years ago
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    good, wow, that was a lot THanks Turing!

  76. TuringTest
    • 3 years ago
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    Always a pleasure Sofiya :D

  77. Mendicant_Bias
    • one year ago
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    Looking back over this, I don't understand how you guys got a negative double root to the characteristic equation. I got m_1 = m_2 = +1. Plugging into the normal formula for repeated roots, getting e^x and xe^x, not e^-x. Am I missing something? @ganeshie8

  78. Mendicant_Bias
    • one year ago
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    @jim_thompson5910 @agent0smith

  79. Mendicant_Bias
    • one year ago
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    @Zarkon @terenzreignz Is anybody else familiar with ODE's seeing this, or am I missing something?

  80. Mendicant_Bias
    • one year ago
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    @zepdrix Anybody?

  81. zepdrix
    • one year ago
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    @Mendicant_Bias For the homogeneous solution? Yes the characteristic equation appears to be giving +1 as a repeated root, as you indicated

  82. zepdrix
    • one year ago
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    So yah, that was a mistake early in the thread :C

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