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Reducing 2nd order to 1st order type question

  1. Oct 13, 2004 #1
    Help! I'm just starting this class and I have no idea what's going on. What I don't understand is, what answer are you supposed to give? My question says "Find the general solution and also the singular solution, if it exists". What the hell does that mean?

    Can someone tell me if this is right?

    Question: [tex]x(y')^2 - (2x + 3y)y' + 6y = 0[/tex]

    so let [tex]p = y'[/tex]

    [tex]xp^2 - (2x + 3y)p + 6y = 0[/tex]
    [tex]xp^2 - 2xp - 3yp + 6y = 0[/tex]
    [tex]-3yp + 6y = -xp^2 + 2xp[/tex]
    [tex]y = \frac{(-xp)(p-2)}{(-3)(p-2)}[/tex]
    [tex]y = \frac{xp}{3} [/tex]

    so then,

    [tex]3y' = 3p = xp' + p[/tex]
    [tex]2p = xp'[/tex]
    [tex]2p = x \frac{dp}{dx}[/tex]
    [tex]\frac{dp}{p} = 2\frac{dx}{x}[/tex]
    [tex]\ln{P} = 2\ln{x} + \ln{C}[/tex]
    [tex]p = cx^2[/tex]
    [tex]y' = cx^2[/tex]
    [tex]y = cx^3 + b[/tex], where c and b are constants

    and that's as far as I got...now what?
    Last edited: Oct 13, 2004
  2. jcsd
  3. Oct 13, 2004 #2


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    Did you check your solution? Unless I'm making a mistake, plugging in that expression for y doesn't yield 0 on the LHS of the original equation.

    As for "singular", I presume your professor is referring to the fact that one of the steps you take will not necessarily be valid... the singular solution corresponds to the case where that step is not legal.
  4. Oct 13, 2004 #3
    hmm...I didn't plug in anything, but my lecture notes says plug in back to original equation so I think you're on to something. If you're talking about the step where I went from

    y = (xp)/3, I moved the 3 to the LHS so it will be
    3y = xp

    then I differentiated both sides

    3y' = xp' + p (product rule)
    and since y' = p,
    3p = xp' + p
  5. Oct 13, 2004 #4


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    I meant that y = c x^3 + b doesn't seem to work in the original differential equation.
  6. Oct 14, 2004 #5


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    Try solving
    [tex]xp^2 - (2x + 3y)p + 6y = 0[/tex]
    for p using the quadratic equation. That will give two different differential equations for y.

    One of them will be your 3y= xp which is a simple separable equation:
    [tex] x\frac{dy}{dx}= 3y[/tex]

    and the other is even simpler!

    I understand
    "3y = xp
    then I differentiated both sides
    3y' = xp' + p (product rule)"

    but I don't see WHY you would do that. The original 3y= xp is much simpler than 3p= xp'+ p.

    Also differentiating again may introduce "extraneous solutions" which appears to be what happened to you.

    And, by the way, your title was "Reducing 2nd order to 1st order type question " so it is very strange that you should take a 1st order d.e. and change it to 2nd order!
    Last edited by a moderator: Oct 14, 2004
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