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MathSofiya

  • 2 years ago

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?

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  1. MathSofiya
    • 2 years ago
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    No I did that wrong, one moment

  2. MathSofiya
    • 2 years ago
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    no it equals \[e^{2x}\]

  3. MathSofiya
    • 2 years ago
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    \[y_p(x)=-e^xx^2+x^2e^x\]

  4. TuringTest
    • 2 years ago
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    wrong complimentary

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

  6. MathSofiya
    • 2 years ago
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    \[r^2-r=0\]

  7. MathSofiya
    • 2 years ago
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    r=0 and r=1

  8. TuringTest
    • 2 years ago
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    so the complimentary is...?

  9. MathSofiya
    • 2 years ago
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    \[y_c=c_1+c_2e^x\]

  10. TuringTest
    • 2 years ago
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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

  11. MathSofiya
    • 2 years ago
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    sure \[y_c=c_1+c_2xe^x\]

  12. TuringTest
    • 2 years ago
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    \[y_c=c_1x+c_2xe^x\]

  13. MathSofiya
    • 2 years ago
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    why on both c_1 and c_2, how did come to that conclusion?

  14. TuringTest
    • 2 years ago
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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...

  15. MathSofiya
    • 2 years ago
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    I'm not sure... DunnBros is about to kick me out, they close at 11pm

  16. MathSofiya
    • 2 years ago
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    I'll be back on in like 15 mins when I get home

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

  18. MathSofiya
    • 2 years ago
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    Make that 7 mins, ha!

  19. MathSofiya
    • 2 years ago
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    \[W(y_1,y_2)=x^2e^x\]

  20. MathSofiya
    • 2 years ago
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    \[y_1=x\] \[y_2=xe^x\]

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

  22. wasiqss
    • 2 years ago
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    well ik it

  23. wasiqss
    • 2 years ago
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    my twin :D

  24. MathSofiya
    • 2 years ago
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    hi twin

  25. TuringTest
    • 2 years ago
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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!!!!!

  26. wasiqss
    • 2 years ago
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    well i can do it

  27. wasiqss
    • 2 years ago
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    tell me equation for which i have to solve\y complimentary

  28. MathSofiya
    • 2 years ago
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    y''-y'=e^x

  29. TuringTest
    • 2 years ago
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    we need variation of parameters @wasiqss

  30. wasiqss
    • 2 years ago
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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

  31. TuringTest
    • 2 years ago
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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

  32. TuringTest
    • 2 years ago
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    ok @wasiqss let's see what you got :)

  33. wasiqss
    • 2 years ago
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    tell me equation man completely cause this seems messy

  34. TuringTest
    • 2 years ago
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    y''-y'=e^x

  35. wasiqss
    • 2 years ago
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    easy one lol

  36. TuringTest
    • 2 years ago
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    then do it, by all mean!

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

  38. wasiqss
    • 2 years ago
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    y complimentary is y= c1 +c2e^-x

  39. TuringTest
    • 2 years ago
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    ^yes, answer not important, we need the process...

  40. TuringTest
    • 2 years ago
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    complimentary is y=c1+c2e^x

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

  42. wasiqss
    • 2 years ago
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    yes cause roots are zero and -1

  43. TuringTest
    • 2 years ago
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    no they are not

  44. TuringTest
    • 2 years ago
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    r^2-r=0 r(r-1)=0 r={0,1}

  45. MathSofiya
    • 2 years ago
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    do you think my W is right? x^2 e^x

  46. TuringTest
    • 2 years ago
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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

  47. TuringTest
    • 2 years ago
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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

  48. TuringTest
    • 2 years ago
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    w=x^2e^x *

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

  50. MathSofiya
    • 2 years ago
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    yeah that's what I had

  51. TuringTest
    • 2 years ago
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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 :)

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

  53. TuringTest
    • 2 years ago
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    so let's try that

  54. TuringTest
    • 2 years ago
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    otherwise we get W=0 which is a failure

  55. TuringTest
    • 2 years ago
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    @lalaly I know you can help us, please where are you?

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

  57. MathSofiya
    • 2 years ago
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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?...

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

  59. TuringTest
    • 2 years ago
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    *\[\int\frac{e^x}xdx\]

  60. TuringTest
    • 2 years ago
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    here is the problem... have you taken linear algebra?

  61. MathSofiya
    • 2 years ago
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    no

  62. MathSofiya
    • 2 years ago
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    @lalaly is here

  63. TuringTest
    • 2 years ago
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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

  64. wasiqss
    • 2 years ago
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    now listen

  65. wasiqss
    • 2 years ago
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    for y particular

  66. TuringTest
    • 2 years ago
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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

  67. wasiqss
    • 2 years ago
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    is it xe^x

  68. wasiqss
    • 2 years ago
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    so as the denominator becomes zero

  69. TuringTest
    • 2 years ago
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    and would you care to explain how/why ?

  70. wasiqss
    • 2 years ago
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    we multiply by x

  71. wasiqss
    • 2 years ago
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    cofficeint of e^1x

  72. wasiqss
    • 2 years ago
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    =1

  73. wasiqss
    • 2 years ago
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    [xe^]/[(-1)^2-1)]

  74. wasiqss
    • 2 years ago
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    HENCE we get zero in denominator

  75. TuringTest
    • 2 years ago
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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

  76. wasiqss
    • 2 years ago
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    now we need to apply binomeal expansion

  77. TuringTest
    • 2 years ago
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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

  78. wasiqss
    • 2 years ago
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    arghh i wish i was good at making you understand it

  79. wasiqss
    • 2 years ago
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    we need it!

  80. TuringTest
    • 2 years ago
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    you cannot have 0 in the denom, that shows you that you have already made a mistake !!!!

  81. wasiqss
    • 2 years ago
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    well to counter this zero!! we need to multiply x

  82. TuringTest
    • 2 years ago
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    ...as I was saying...

  83. wasiqss
    • 2 years ago
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    plz no offense but plz go revise DE s

  84. TuringTest
    • 2 years ago
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    ._.

  85. wasiqss
    • 2 years ago
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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

  86. TuringTest
    • 2 years ago
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    yes, but I think it is way more work than is necessary now I just wanna hear @lalaly

  87. wasiqss
    • 2 years ago
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    kay

  88. TuringTest
    • 2 years ago
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    oh dear, we seem to have lost her I am going to try this op paper I guess

  89. wasiqss
    • 2 years ago
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    hahaha lana dear

  90. wasiqss
    • 2 years ago
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    \[[1+(D(D+1)-1)^-1[x^2e^x\]

  91. wasiqss
    • 2 years ago
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    we have to expand binomeally [1+(D(D+1)−1)−1

  92. wasiqss
    • 2 years ago
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    well i have a solution on page i wonder how to give you that

  93. MathSofiya
    • 2 years ago
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    take a picture with your cellphone and attach it

  94. MathSofiya
    • 2 years ago
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    or use \[\LaTeX\]

  95. wasiqss
    • 2 years ago
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    good idea twin can i do it tomorow because i would need to upload it too?

  96. TuringTest
    • 2 years ago
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    okay got it :)

  97. wasiqss
    • 2 years ago
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    latex is something not my type

  98. wasiqss
    • 2 years ago
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    turing my approach?

  99. TuringTest
    • 2 years ago
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    do everything normally, then multiply the particular by x at the end that is all

  100. MathSofiya
    • 2 years ago
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    get used to it, it's a very helpful Latex is a very useful tool

  101. wasiqss
    • 2 years ago
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    turing binomeal is essential in it!

  102. TuringTest
    • 2 years ago
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    oh yeah?, just watch...

  103. TuringTest
    • 2 years ago
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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?

  104. TuringTest
    • 2 years ago
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    oh crap I see a mistake

  105. wasiqss
    • 2 years ago
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    wait you told me w=xe^x!

  106. MathSofiya
    • 2 years ago
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    dude this whole discussion was based on finding what W equaled

  107. wasiqss
    • 2 years ago
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    ok just ask lana what is the right answer

  108. TuringTest
    • 2 years ago
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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

  109. wasiqss
    • 2 years ago
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    waittttttttttttt what the....... i was solving by another method not by variation of parameter lol

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

  111. TuringTest
    • 2 years ago
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    there we go... stupid algebra tripped me up :/

  112. MathSofiya
    • 2 years ago
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    yeah that's what I get \[-e^x+e^xx\]

  113. TuringTest
    • 2 years ago
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    so yc+yp gives what I have, and what wolf has so we're good :)

  114. MathSofiya
    • 2 years ago
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    good. Hi 5

  115. TuringTest
    • 2 years ago
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    back atchya!

  116. MathSofiya
    • 2 years ago
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    As always thank you!

  117. TuringTest
    • 2 years ago
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    As always you're welcome :D

  118. Mendicant_Bias
    • one month ago
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    I don't understand part of the very end of your post, near the solutions. I'll come back and ask about it later, but I do not understand how you obeyed the laws of calculus in factoring out an e^x when integrating with respect to x.

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