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Differential equation: 4sqrt(xy)dy/dx=1, y(1)=1

  1. Feb 11, 2013 #1
    The differential equation I'm working on is:

    4(√(xy))dy/dx=1, y(1)=1

    (√(xy)dy/dy)2 = (1/4)2

    ((xy)dy/dx)*dx = (1/8)*dx

    (xy)dy = (1/8)dx

    (y)dy = (1/(8x))dx

    ...So I think this is right so far.

    Now I'm going to take the integral of both sides.

    ∫(y)dy = ∫(1/(8x))dx <------∫(1/x)(1/8)dx

    1/2y2 = ln|x|*(1/8) + C

    √[y2] = √[2(ln|x|*(1/8) + C)]

    y= √(1/4*ln|x| + C)

    ...Substitution in the y(1)=1

    1 = √(1/4*ln|1| + C)

    4 = 0 + C

    C = 4

    y(x)= √(1/4*ln|x| + 4)

    This is counted as the wrong answer, but I don't know what I'm doing wrong here... #=_=
    Thank you so much!
     
  2. jcsd
  3. Feb 11, 2013 #2

    Dick

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    Your doing a lot of stuff wrong. Don't square it. Just separate the variables.
     
  4. Feb 11, 2013 #3
    Would you please tell me where exactly I squared wrongly?

    Do you mean in the very beginning- my 2nd step?

    "(√(xy)dy/dy)2 = (1/4)2"

    If this is where you mean, isn't it necessary to square first in order to get rid of the square root and then isolate the x from the y?
     
  5. Feb 12, 2013 #4

    SteamKing

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    dy and dx are not immune to being squared.
     
  6. Feb 12, 2013 #5
    Ha! Oh! ...I actually wasn't sure if they'd be squared or not! But I guess they ARE squared! Thanks! ;)
     
  7. Feb 12, 2013 #6
    So my 2nd step really should be: 4√(y)dy = 1/√(x)dx?


    4*√(x)*√(y)dy/dx = 1

    4√(y) dy = 1/√(x)dx
     
  8. Feb 12, 2013 #7
    ugh, I don't know.

    I'm still not getting the right answer... =_=

    If this is correct: ∫4√(y)dy = ∫1/(√(x))dx, it seems it should be worked out like this:

    (2/3)(4y3/2) = 2x1/2 + C

    = (8/3)y3/2 = 2√(x) + C

    = y3/2 = 3/8(2√(x) + C)

    = y= √((3/8(2√(x) + C))3)


    Solving for C:

    (1)^(3/2) = (2sqrt(1) + C)*3/8

    1 = .75 + 3/8C

    .25 = 3/8C

    C= 2/3


    Seems that f(x)= sqrt[(3/8(sqrt(x)) + 2/3)^3]...

    But this is incorrect...
     
  9. Feb 12, 2013 #8

    haruspex

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    As Dick said, separate the variables: get all the y terms on one side and all the x on the other. This includes dy and dx. Then you can just integrate.
     
  10. Feb 12, 2013 #9

    ehild

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    You made an error again. If y3/2=A then y=A2/3

    And it is easiest to get C from the equation in bold.

    ehild
     
  11. Feb 12, 2013 #10
    Thanks very much, ehild! :)
    You showed me where exactly I went wrong!

    y3/2 = 3/8(2√(x) + C)

    y= (3/8(2√(1) + C))2/3

    1 = 3/8(2 + C)

    C= 2/3

    f(x) = (3/8(2√(x) + 2/3))2/3

    So this is counted as the right answer! Thanks! :)
     
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