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Homework Help: Differential equations/checking an answer - help, please!

  1. Apr 8, 2005 #1
    Problem from my textbook - Solve:

    [tex]\frac{dy}{dx} = \frac{4y - 3x}{2x - y}[/tex]

    Answer provided:

    [tex]\left| y - x \right| = \mathrm{C} \left| y + 3x \right| ^5 \mbox{.} \mbox{ Also, } y = -3x \mbox{.}[/tex]

    My answer is pretty close to that one, but it seems like I missed some step. Anyhow, here's what's I've got:

    Left-hand side:

    [tex] \frac{dy}{dx} = \frac{d}{dx} \left( xv \right)[/tex]

    [tex]\frac{dy}{dx} = xv^{\prime} + v[/tex]

    [tex]\frac{dy}{dx} = x\frac{dv}{dx} + v[/tex]

    Right-hand side:

    [tex]\frac{dy}{dx} = \frac{4y - 3(y/v)}{2(y/v) - y}[/tex]

    [tex] \frac{dy}{dx} = -\left( \frac{4v-3}{v-2} \right)[/tex]

    Hence, this equation is homogeneous. As a result, we find

    [tex] x\frac{dv}{dx} + v = -\left( \frac{4v-3}{v-2} \right) [/tex]

    [tex]x\frac{dv}{dx} = \frac{-v^2 -2v + 3}{v-2} \mbox{,}[/tex]

    which is separable. Thus, we obtain

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

    [tex]\frac{1}{4} \ln \left| v - 1 \right| - \frac{5}{4} \ln \left| v + 3 \right| = \ln \left| x \right| + \mathrm{C}[/tex]

    [tex]\ln \left| v - 1 \right| - 5 \ln \left| v + 3 \right| = 4 \ln \left| x \right| + \mathrm{C}[/tex]

    [tex]\ln \frac{\left| v - 1 \right|}{\left| v + 3 \right| ^5} = \ln \left( x^4 \right) + \mathrm{C}[/tex]

    [tex]\frac{\left| v - 1 \right|}{\left| v + 3 \right| ^5} = \mathrm{C} x^4[/tex]

    [tex]\frac{\left| (y/x) - 1 \right|}{\left| (y/x) + 3 \right| ^5} = \mathrm{C} x^4[/tex]

    [tex]\frac{\left| y - x \right|}{\left| y + 3x \right| ^5} = \mathrm{C} x^4[/tex]

    [tex]\left| y - x \right| = \mathrm{C} x^4 \left| y + 3x \right| ^5 \mbox{,}[/tex]

    which is different from

    [tex]\left| y - x \right| = \mathrm{C} \left| y + 3x \right| ^5 \mbox{.}[/tex]

    Again, it seems like I missed some step. Can anyone help me out with that?

    By the way, I can't see why [tex]y=-3x[/tex] also is an answer. Could you please clarify it?

    Any help is highly appreciated.
    Last edited: Apr 8, 2005
  2. jcsd
  3. Apr 8, 2005 #2

    Don't you have a sign error in your "Right-hand side" equation (second line)? I'm not sure if that will get rid of the x^4 factor or not.
  4. Apr 8, 2005 #3
    here is your problem
    [tex]\frac{\left| (y/x) - 1 \right|}{\left| (y/x) + 3 \right| ^5} = \mathrm{C} x^4[/tex]

    will turn into [tex] \frac{\frac{y-x}{x}}{\frac{(y+3)^5}{x^5}} = Cx^4 [/tex] which cancels out the x^4
  5. Apr 8, 2005 #4
    Oh... sure!!! I certainly skipped that step. About the y=-3x, don't bother explaining. I've just figured it out.

    Thank you so much.

  6. Apr 9, 2005 #5
    One more question!!!

    I've just noticed that [tex]y=x[/tex] also seems to work. Is it really possible?

  7. Apr 9, 2005 #6
    what do you mean y=x seems to work? Do you mean for the substitution part? Or do you mean for the equilibrium points?
  8. Apr 9, 2005 #7
    As an equilibrium solution, maybe. I'm not sure. Probably, something like y=-3x is.
  9. Apr 9, 2005 #8
    isnt hte equilibrium solution obtained when dy/dx = 0?? in which case y = 3/4x?
  10. Apr 9, 2005 #9


    User Avatar
    Science Advisor
    Homework Helper

    Yes, y=x is a solution, but it doesn't need to be stated in addition, since it is simply the solution you get when C = 0. Well, actually, C is of the form exp(A^4) where A is the constant of integration, and this will be a positive number (i.e. C cannot be 0). So perhaps y=x does need to be written separately.
  11. Apr 9, 2005 #10
    That makes sense. There still is a question I'd like to ask, though. I need to check something. My procedure with respect to [tex]y=-3x[/tex] is to plug into the given DE:

    [tex]\frac{dy}{dx} = \frac{4y - 3x}{2x - y}[/tex]

    which gives [tex]-3[/tex], which is the expected result. Thus, it has been verified as an additional solution. A mistake that I made previously was to do the same thing with [tex]y=x[/tex], because it also implies that [tex]\mathrm{C}=0[/tex], which is false.

    My question is: does [tex]y=-3x[/tex] work because it satisfies the procedure above and is not dependent on the value of the constant [tex]\mathrm{C}[/tex]?


    I literally was dozing off when I suggested otherwise. Sorry about that. :rofl: Thanks for reminding me of the definition.
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