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Solving differential equations

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data
    If (x^2+1)dy/dx = xy for all y>0, and y(2) = 5, then y(3) = ?


    2. Relevant equations



    3. The attempt at a solution
    dy/y = xdx/(x^2+1) -- use substitution with u=x^2
    lny = 2(u+1)/du
    lny = 2ln(x^2+1)+c
    y = e^(2ln(x^2+1)+c)
    5 = e^(2ln(5)+c)
    5 = e^(ln25+c)
    5 = 25+c
    c = 20

    y(3) = e^(2ln10+20)
    = e^ln100+20
    = 100+20
    = 120



    .... the answer should be 5sqrt(2), so I am very off, but I can't find why. help?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 9, 2011 #2

    CompuChip

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    If [itex]u = x^2[/itex], then [itex]du = 2 x dx[/itex].
    So instead of
    [tex]\ln y = 2(u+1)/du [/tex]
    you get
    [tex]d(\ln y) = du/[2(u + 1)] [/tex]
    and, integrating,
    [tex]\ln y = \ln[ 2(u + 1) ] + c[/tex].
     
    Last edited: Aug 9, 2011
  4. Aug 9, 2011 #3

    Ray Vickson

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    We have exp[(1/2)*ln(x^2+ 1) + c] = k*sqrt(x^2+1), where k =exp(c). Note: x*dx/(1+x^2) = (1/2)du/(u+1).

    RGV
     
  5. Aug 9, 2011 #4
    if y = e^(1/2)ln(x^2+1)+c
    5 = e^lnsqrt(5)+c
    then i end up with c = 5-sqrt(5), and then
    y(3) = e^1/2ln10+5-sqrt(5)
    = sqrt(10)+5-sqrt(5)
    .... confused?
     
  6. Aug 9, 2011 #5

    I like Serena

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    @CompuChip: you forgot a factor (1/2) in your last line...
     
  7. Aug 9, 2011 #6

    I like Serena

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    You forgot a couple of brackets...

    You should use:
    [tex]5 = e^{[\ln \sqrt 5+c]}[/tex]
    You'll see that you'll get a different value for c. :smile:
     
  8. Aug 9, 2011 #7
    do you mean then i should get
    5 = sqrt5 +e^c?
    how do i simplify that?
     
  9. Aug 9, 2011 #8

    I like Serena

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    No, that is not right.
    Let's take it one step back.
    Let me put the braces slightly diffently for clarity.
    [tex]5 = e^{(\ln \sqrt 5)+c}[/tex]
    Here you need to simplify by applying the rules for powers.
    That is [itex]e^{a+b}=e^a \cdot e^b[/itex].
     
  10. Aug 9, 2011 #9
    so then don't i get
    5 = e^ln(sqrt5) + e^c?
    and e^ln(sqrt5) = sqrt5?
     
  11. Aug 9, 2011 #10

    I like Serena

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    Yes! :smile:


    No. :frown:
     
  12. Aug 9, 2011 #11
    oh, durrr
    5 = sqrt5*e^c?
     
  13. Aug 9, 2011 #12

    I like Serena

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    Yes! :wink:
     
  14. Aug 9, 2011 #13
    YAY!
    so from there, i get
    5/sqrt5 = e^c?
    so c = ln(5/sqrt5)?
     
  15. Aug 9, 2011 #14

    I like Serena

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    Yes, so... :rolleyes:
     
  16. Aug 9, 2011 #15
    hahaha
    f(3) = e^lnsqrt10+ln5/sqrt5
    = sqrt10+5/sqrt5

    .... annnd i'm off track again lol
     
  17. Aug 9, 2011 #16

    I like Serena

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    Err... can you do the "oh, durrr" thing again? :shy:
     
  18. Aug 9, 2011 #17
    well that was dumb.
    e^lnsqrt10+ln5/sqrt5 = e^lnsqrt10*e^ln5/sqrt5
    = sqrt10*5/sqrt5

    I GOT IT YAYYYY :):)
    thanksyou :)
     
  19. Aug 9, 2011 #18

    I like Serena

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    Congratulations! :biggrin:


    One thing though...

    You started with:
    [tex]y = e^{(1/2)\ln(x^2+1)+c} [/tex]
    But that is not the equation you had in your first post... o:)

    How did you get it?
     
  20. Aug 9, 2011 #19
    :D

    i changed the first equation to
    dy/y = xdx/(x^2+1), then used the substitution u=x^2 on the right side :)
     
  21. Aug 9, 2011 #20

    I like Serena

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    Yeah, I got that...
    So did you follow CompuChip's correction and then my correction on CompuChip's correction? :confused:
     
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