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Homogeneous Problem

  1. Oct 8, 2006 #1
    Hi all,

    I have been given this question:

    Find the initial value problem of the homogeneous equation:

    [tex](x^2 - y^2) y' = xy \ , \ y(1) = 1[/tex]

    Now I know, from my lessons, I have to get it in the form of:

    [tex]\frac{dy}{dx} = f(\frac{y}{x})[/tex]

    I have managed to get close but nothing is working out.

    If I make the substitution x = vy then I can get [tex]y' = \frac{v}{1 - v^2}[/tex] but at this state I have made the substitution twice, not once. I then get to a place where I can seperate the variables:

    [tex]\int \frac{1 - v^2}{v^3} dv = \int \frac{1}{x} dx[/tex]

    [tex]\int \frac{1}{v^3} dv - \int \frac{1}{v} dv = \int \frac{1}{x} dx[/tex]

    [tex] \frac{-1}{2v^2} - ln \vert v \vert = ln \vert x \vert + c[/tex]

    Then from here my calculations go all funny. Does anyone know if I am barking up the correct tree?


    The Bob (2004 ©)
  2. jcsd
  3. Oct 11, 2006 #2


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    You mean find the solution to this initial value problem!

    Actually, if you made that substitution, then the right hand side would be [itex]\frac{v}{v^2-1}[/itex]. But you are making the wrong substitution- you don't want to substitute for the independent variable x, you want to substitute for the dependent variable y. The reason for that is that your
    [tex]y'= \frac{v}{1- v^2}[/itex]
    has only replaced x/y by v on the right side: the left side, y', is still differentiation with respect to x, not v.

    Let y= vx so that v= y/x. Divide both sides of the equation by [itex]x^2- y^2[/itex] to get [itex]y'= \frac{xy}{x^2- y^2}[/itex] and then divide both numerator and denominator by x2 to get the right hand side to be
    [tex]\frac{\frac{y}{x}}{1- \left(\frac{y}{x})^2}= \frac{v}{1-v^2}[/tex]
    But since y= xv, y'= xv'+ v so your equation is really [itex]xv'+ v= \frac{v}{1- v^2}[/itex] and so [itex]xv'= \frac{v}{1-v^2}-v= \frac{v^3}{1-v^2}[/itex]. That is a separable equation and can be solved for v.

  4. Oct 12, 2006 #3
    Cheers HallsofIvy. I didn't think anyone was going to reply. :rofl:

    I cannot see how, from [tex]\frac{-1}{2v^2} - ln|v| = ln|x| + c[/tex], I can be solved for v. I say that because there is a term in the logarithm and one out of it. Unfortunately, I have not covered this so I am still stuck. I do eventually get to [tex]y^{2 - y^2} = x^2[/tex] but I cannot do anything with that either.

    Thanks for all your help. I did allow me to see how I should have made the substitution. :biggrin:


    The Bob (2004 ©)
  5. Oct 12, 2006 #4
    This is wrong. I got to [tex]\frac{e^{y^2}}{y^{y^2}} = e^{x^2}[/tex].

    The Bob (2004 ©)
  6. Oct 13, 2006 #5


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    I don't see how you could get

    I get [itex]-\frac{1}{2}\frax{x^2}{y^2}- ln\left(\frac{y}{x}\right)= ln -\frac{1}{2}[/itex]. Are you required to solve for y? In general, in problems like this, that is impossible.
  7. Oct 13, 2006 #6
    I do not know what I got off hand, unfortunately I am in a rush. I do agree that [tex]y^{y^2}[/tex] is not correct.


    The Bob (2004 ©)
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