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Homework Help: Partial Differential du/dx Equations

  1. Nov 10, 2006 #1
    Trying to solve the two questions attached, for the first one:

    du/dx - x.du/dy = 0

    Assume u = X.Y
    Y.dX/dx - x.X.dY/dy = 0
    Dividing by x.X.Y and taking one term over to the other side:
    dX/dx.1/(x.X) = dY/dy.1/Y

    These can be equated to a constant m

    dX/X = m.x.dx
    ln(X/C) = 1/2.m.x^2
    X = C.e^((1/2.m.x^2))

    dY/Y = m.dy
    ln(Y/C) = m.y
    Y = C.e^((y.m))

    u = C^2.e^((1/2.x^2 + y).m)

    However the answer given is C.e^((x^2 + 2y).m)

    Have they absorbed the two constants earlier into one? Since the equation is a first order one. Also they seem to have multiplied the exponent by 2 throughout, surely this isn't allowed so I assume I've made a mistake?

    Second question:
    x.du.dx - 2y.du/dy = 0
    Assume u = X.Y
    x.Y.dX/dx - 2y.X.dY/dy = 0

    Dividing by 2.X.Y and taking one term to the other side:

    x/2X.dX/dx = y/Y.dY/dy
    These can be equated to a constant m

    y/Y.dY/dy = m
    dY/Y = m.dy/y
    ln(Y/C) = m.ln(y)
    ln(Y/C) = ln(y^m)
    Y = C.y^m

    x/2X.dX/dx = m
    dX/2X = m.dx/x
    1/2.ln(X/C) = m.ln(x)
    ln(X/C) = 2.m.ln(x)
    ln(X/C) = ln(x^2m)
    X = C.x^2m

    XY = C^2.(x^2.y)^m

    However the book has it as:


    Again have they just combined the two constants into one since the equation is only first order?

    Thanks in advance for any help :)

    Attached Files:

    Last edited: Nov 10, 2006
  2. jcsd
  3. Nov 11, 2006 #2


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    Science Advisor

    The only "mistake" you made was in assuming that the constant C in was the same in each case. you are really saying that
    [tex]X= C_1e^{\frac{1}{2}mx^2}[/tex]
    [tex]Y= C_2e^{my}[/tex]
    and so
    [tex]u= (C_1C_2)e^{\frac{1}{2}mx^2+ my}[/tex]
    Since C1 and C2 are unknown constants, their product is also an unknown constant: let C= C1C2.
    As far as the factor of 2 is concerned, what is m? Suppose you had said "These can be equated to a constant 2m" at the beginning? (Someone just doesn't like fractions!)

    Exactly. You have no reason to assume that C is the same in both parts. You really have
    [tex]XY= C_1C_2(x^2y)^m[/tex]
    Since C1 and C2 are just unknown constants, so is there product. Let C= C1C2.
    Last edited by a moderator: Nov 11, 2006
  4. Nov 11, 2006 #3
    Ah okay, all makes sense. Thank you very much :)
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