1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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 :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook