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MathSofiya Group Title

Solve using variation of parameters \[y''-y'=e^x\] \[r^2-r=0\] \[r_1=r_2=1\] \[y_c=c_1e^x+c_2xe^x\] \[W(y_1,y_2)=0\] Is that possible?

  • one year ago
  • one year ago

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  1. MathSofiya Group Title
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    No I did that wrong, one moment

    • one year ago
  2. MathSofiya Group Title
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    no it equals \[e^{2x}\]

    • one year ago
  3. MathSofiya Group Title
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    \[y_p(x)=-e^xx^2+x^2e^x\]

    • one year ago
  4. TuringTest Group Title
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    wrong complimentary

    • one year ago
  5. TuringTest Group Title
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    \[r^2-r=r(r-1)=0\implies r=\{?,?\}\]

    • one year ago
  6. MathSofiya Group Title
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    \[r^2-r=0\]

    • one year ago
  7. MathSofiya Group Title
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    r=0 and r=1

    • one year ago
  8. TuringTest Group Title
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    so the complimentary is...?

    • one year ago
  9. MathSofiya Group Title
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    \[y_c=c_1+c_2e^x\]

    • one year ago
  10. TuringTest Group Title
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    right, now I get a bit dicey.... since you have e^x on the RHS as well I feel you may need to multiply this by x to keep it linearly independent

    • one year ago
  11. MathSofiya Group Title
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    sure \[y_c=c_1+c_2xe^x\]

    • one year ago
  12. TuringTest Group Title
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    \[y_c=c_1x+c_2xe^x\]

    • one year ago
  13. MathSofiya Group Title
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    why on both c_1 and c_2, how did come to that conclusion?

    • one year ago
  14. TuringTest Group Title
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    or do you do it to the particular instead? hm... well normally you do it to the whole RHS particular, but since you need the complimentary for the particular in VP I'm not sure...

    • one year ago
  15. MathSofiya Group Title
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    I'm not sure... DunnBros is about to kick me out, they close at 11pm

    • one year ago
  16. MathSofiya Group Title
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    I'll be back on in like 15 mins when I get home

    • one year ago
  17. TuringTest Group Title
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    fair enough, I'm about to call it a night as well see ya, I will investigate this :)

    • one year ago
  18. MathSofiya Group Title
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    Make that 7 mins, ha!

    • one year ago
  19. MathSofiya Group Title
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    \[W(y_1,y_2)=x^2e^x\]

    • one year ago
  20. MathSofiya Group Title
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    \[y_1=x\] \[y_2=xe^x\]

    • one year ago
  21. MathSofiya Group Title
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    \[y_p(x)=-x\int \frac{e^x}xdx+xe^x\int x^{-1}dx\]

    • one year ago
  22. wasiqss Group Title
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    well ik it

    • one year ago
  23. wasiqss Group Title
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    my twin :D

    • one year ago
  24. MathSofiya Group Title
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    hi twin

    • one year ago
  25. TuringTest Group Title
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    that first integral is not going to be so nice I don't think.... you know who knows exactly what we're missing @lalaly DE help please!!!!!

    • one year ago
  26. wasiqss Group Title
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    well i can do it

    • one year ago
  27. wasiqss Group Title
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    tell me equation for which i have to solve\y complimentary

    • one year ago
  28. MathSofiya Group Title
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    y''-y'=e^x

    • one year ago
  29. TuringTest Group Title
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    we need variation of parameters @wasiqss

    • one year ago
  30. wasiqss Group Title
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    yehh just tell me equation i ll take out solution. as y complimentary is solved i ll solve for y particular

    • one year ago
  31. TuringTest Group Title
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    Our problem is that the solutions to the complimentary are not independent from the particular, which you can tell since we got W=0 on our first try

    • one year ago
  32. TuringTest Group Title
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    ok @wasiqss let's see what you got :)

    • one year ago
  33. wasiqss Group Title
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    tell me equation man completely cause this seems messy

    • one year ago
  34. TuringTest Group Title
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    y''-y'=e^x

    • one year ago
  35. wasiqss Group Title
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    easy one lol

    • one year ago
  36. TuringTest Group Title
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    then do it, by all mean!

    • one year ago
  37. MathSofiya Group Title
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    http://www.wolframalpha.com/input/?i=integral+of+e^%28x%29%2Fx

    • one year ago
  38. wasiqss Group Title
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    y complimentary is y= c1 +c2e^-x

    • one year ago
  39. TuringTest Group Title
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    ^yes, answer not important, we need the process...

    • one year ago
  40. TuringTest Group Title
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    complimentary is y=c1+c2e^x

    • one year ago
  41. MathSofiya Group Title
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    i disagree with the above answer, shouldn't it be e^x, yeah what turing has

    • one year ago
  42. wasiqss Group Title
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    yes cause roots are zero and -1

    • one year ago
  43. TuringTest Group Title
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    no they are not

    • one year ago
  44. TuringTest Group Title
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    r^2-r=0 r(r-1)=0 r={0,1}

    • one year ago
  45. MathSofiya Group Title
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    do you think my W is right? x^2 e^x

    • one year ago
  46. TuringTest Group Title
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    that I doube @MathSofiya because we have e^x on both left and right our solutions are linearly dependet (that means we can't solve it yet) so maybe if we multiply the g(x) by x ? not sure, but worth a try

    • one year ago
  47. TuringTest Group Title
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    with W=e^2e^x I don't think that first integral is closed, check the wolf, I have not yet

    • one year ago
  48. TuringTest Group Title
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    w=x^2e^x *

    • one year ago
  49. MathSofiya Group Title
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    do we agree that \[y_c=c_1x+c_2xe^x\] or should it be \[y_c=c_1+c_2e^x\]

    • one year ago
  50. MathSofiya Group Title
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    yeah that's what I had

    • one year ago
  51. TuringTest Group Title
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    that I am not sure, but based on our attempt to use the second one failing I would now try the first (we are discovering this together right now :)

    • one year ago
  52. TuringTest Group Title
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    however if we are to try the first we need to multiply g(x) by x (I'm thinking....)

    • one year ago
  53. TuringTest Group Title
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    so let's try that

    • one year ago
  54. TuringTest Group Title
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    otherwise we get W=0 which is a failure

    • one year ago
  55. TuringTest Group Title
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    @lalaly I know you can help us, please where are you?

    • one year ago
  56. MathSofiya Group Title
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    \[y_c=c_1x+c_2xe^x\] \[\left|\begin{matrix}x&xe^x\\1&e^x(x-1)\end{matrix}\right|=x^2e^x\]

    • one year ago
  57. MathSofiya Group Title
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    can you explain the g(x) concept again and why it being e^x would be a problem because that cancels something out?...

    • one year ago
  58. TuringTest Group Title
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    no I think that is wrong because that leads to the particular having\[\int\frac{e^x}dxx\]

    • one year ago
  59. TuringTest Group Title
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    *\[\int\frac{e^x}xdx\]

    • one year ago
  60. TuringTest Group Title
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    here is the problem... have you taken linear algebra?

    • one year ago
  61. MathSofiya Group Title
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    no

    • one year ago
  62. MathSofiya Group Title
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    @lalaly is here

    • one year ago
  63. TuringTest Group Title
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    okay. then we have a set of vectors (in this case equations) if their determinant is zero we know they are linearly dependent

    • one year ago
  64. wasiqss Group Title
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    now listen

    • one year ago
  65. wasiqss Group Title
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    for y particular

    • one year ago
  66. TuringTest Group Title
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    that is a bad thing in DE's we need all of our solutions to be linearly independent, so the determinant of the solutions, (the wronskian) must not be zero, otherwise our solutions are linearly dependent

    • one year ago
  67. wasiqss Group Title
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    is it xe^x

    • one year ago
  68. wasiqss Group Title
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    so as the denominator becomes zero

    • one year ago
  69. TuringTest Group Title
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    and would you care to explain how/why ?

    • one year ago
  70. wasiqss Group Title
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    we multiply by x

    • one year ago
  71. wasiqss Group Title
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    cofficeint of e^1x

    • one year ago
  72. wasiqss Group Title
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    =1

    • one year ago
  73. wasiqss Group Title
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    [xe^]/[(-1)^2-1)]

    • one year ago
  74. wasiqss Group Title
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    HENCE we get zero in denominator

    • one year ago
  75. TuringTest Group Title
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    anyway... to keep the particular linearly independent from the complimentary we often need to multiply by x, so that is what I am thinking must happen somehow @lalaly will confirm if I am making any sense or not now

    • one year ago
  76. wasiqss Group Title
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    now we need to apply binomeal expansion

    • one year ago
  77. TuringTest Group Title
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    @wasiqss sorry to be so direct, but no we do not need any binomial expansion here please let's just have a listen to @lalaly

    • one year ago
  78. wasiqss Group Title
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    arghh i wish i was good at making you understand it

    • one year ago
  79. wasiqss Group Title
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    we need it!

    • one year ago
  80. TuringTest Group Title
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    you cannot have 0 in the denom, that shows you that you have already made a mistake !!!!

    • one year ago
  81. wasiqss Group Title
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    well to counter this zero!! we need to multiply x

    • one year ago
  82. TuringTest Group Title
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    ...as I was saying...

    • one year ago
  83. wasiqss Group Title
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    plz no offense but plz go revise DE s

    • one year ago
  84. TuringTest Group Title
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    ._.

    • one year ago
  85. wasiqss Group Title
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    you know what .. we can approach a DE in differnt ways and this one i will certainly approach by binomeal.. cause we studied this way for this case

    • one year ago
  86. TuringTest Group Title
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    yes, but I think it is way more work than is necessary now I just wanna hear @lalaly

    • one year ago
  87. wasiqss Group Title
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    kay

    • one year ago
  88. TuringTest Group Title
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    oh dear, we seem to have lost her I am going to try this op paper I guess

    • one year ago
  89. wasiqss Group Title
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    hahaha lana dear

    • one year ago
  90. wasiqss Group Title
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    \[[1+(D(D+1)-1)^-1[x^2e^x\]

    • one year ago
  91. wasiqss Group Title
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    we have to expand binomeally [1+(D(D+1)−1)−1

    • one year ago
  92. wasiqss Group Title
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    well i have a solution on page i wonder how to give you that

    • one year ago
  93. MathSofiya Group Title
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    take a picture with your cellphone and attach it

    • one year ago
  94. MathSofiya Group Title
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    or use \[\LaTeX\]

    • one year ago
  95. wasiqss Group Title
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    good idea twin can i do it tomorow because i would need to upload it too?

    • one year ago
  96. TuringTest Group Title
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    okay got it :)

    • one year ago
  97. wasiqss Group Title
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    latex is something not my type

    • one year ago
  98. wasiqss Group Title
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    turing my approach?

    • one year ago
  99. TuringTest Group Title
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    do everything normally, then multiply the particular by x at the end that is all

    • one year ago
  100. MathSofiya Group Title
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    get used to it, it's a very helpful Latex is a very useful tool

    • one year ago
  101. wasiqss Group Title
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    turing binomeal is essential in it!

    • one year ago
  102. TuringTest Group Title
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    oh yeah?, just watch...

    • one year ago
  103. TuringTest Group Title
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    actually no need to multiply by x at all\[y_c=c_1+c_2e^x\]\[W=e^x\]\[y_p=-\int e^{-x}+e^x\int dx=e^x+xe^x\]\[y_p=e^x+xe^x\]\[y=y_c+y_p=c_1+c_2e^x+e^x+xe^x=c_1+e^x(1+x+c_2)\]and yes, but is your solution in 5 lines?

    • one year ago
  104. TuringTest Group Title
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    oh crap I see a mistake

    • one year ago
  105. wasiqss Group Title
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    wait you told me w=xe^x!

    • one year ago
  106. MathSofiya Group Title
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    dude this whole discussion was based on finding what W equaled

    • one year ago
  107. wasiqss Group Title
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    ok just ask lana what is the right answer

    • one year ago
  108. TuringTest Group Title
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    @MathSofiya yes, because the entire problem we were having was keeping our solutions linearly independent, which we check by finding W

    • one year ago
  109. wasiqss Group Title
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    waittttttttttttt what the....... i was solving by another method not by variation of parameter lol

    • one year ago
  110. TuringTest Group Title
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    \[y_c=c_1+c_2e^x\]\[W=e^x\]\[y_p=-\int e ^x dx+e^x\int dx=-e^x+xe^x\]\[y=y_c+y_p=c_1+c_2e^x-e^x+xe^x=c_1+e^x(x-1+c_2)\]

    • one year ago
  111. TuringTest Group Title
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    there we go... stupid algebra tripped me up :/

    • one year ago
  112. MathSofiya Group Title
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    yeah that's what I get \[-e^x+e^xx\]

    • one year ago
  113. TuringTest Group Title
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    so yc+yp gives what I have, and what wolf has so we're good :)

    • one year ago
  114. MathSofiya Group Title
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    good. Hi 5

    • one year ago
  115. TuringTest Group Title
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    back atchya!

    • one year ago
  116. MathSofiya Group Title
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    As always thank you!

    • one year ago
  117. TuringTest Group Title
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    As always you're welcome :D

    • one year ago
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