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Particular Solutions of Differential Equations

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\frac{d^{2}y}{dx^{2}}[/itex] = [itex]y\frac{dy}{dx}[/itex]

    2. Relevant equations

    Let [itex] v = \frac{dy}{dx}[/itex] and [itex]v\frac{dv}{dy} = \frac{d^{2}y}{dx^{2}}[/itex]


    3. The attempt at a solution

    The question can be rewritten as:

    [itex]v\frac{dv}{dy} = yv[/itex]
    [itex]\frac{dv}{dy} = y[/itex]. (v =/=0 )

    This is very easy to solve since it's basically a normal integral. I get v and substitute in [itex]\frac{dy}{dx}[/itex] to get an implicit expression for y:

    [itex]C+\frac{x}{2}= D Tan^{-1}(Dy)[/itex]

    However, the problem is when I divided v on both sides, and I noted that v can't be 0 because division by 0 is not allowed.

    Thus, v = 0 is a particular solution to the DE, so y equals a constant is not a solution to this DE?

    I really hope someone can give me a better understanding of particular solutions, and what I should do with them in a problem such as this.

    Thank you very much.
     
  2. jcsd
  3. Nov 17, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, that says just the opposite! v= 0 is a solution to the equation after you had "reduced" it so y equal to a constant is a solution to the DE.

    If y is a constant, both first and second derivatives are 0 so your equation just becomes 0= y(0).
     
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