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Differential equations: Elimination of arbitrary constants

  1. Jun 23, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the differential equation of ln y = ax^2 + bx + c by eliminating the arbitrary constants a, b and c.

    2. Relevant equations

    Wrosnkian determinant.

    3. The attempt at a solution

    I've solved a similar problem (y=ax^2+bx+c --> y'''=0), but couldn't do the same with this one.
    All what I could is taking the exponent of both sides --> y=e^(ax^2 + bx + c).
     
  2. jcsd
  3. Jun 23, 2014 #2
    I do not see a DE...
     
  4. Jun 23, 2014 #3

    Simon Bridge

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    I think this is the calc form of Jeopardy ... you are given the solution to a DE, and you have to find the DE.

    Taking the exponential of both sides looks promising - you can use you knowledge of how powers combine to simplify it further or investigate what happens as you differentiate it.

    note. y=e^x comes from y'=y
     
  5. Jun 23, 2014 #4

    pasmith

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    Does it not follow from [itex]\ln y = ax^2 + bx + c[/itex] and [tex]
    \frac{d^3}{dx^3}(ax^2 + bx + c) = 0[/tex] that [tex]
    \frac{d^3}{dx^3}(\ln y) = 0?[/tex] Some would regard that as an adequate ODE for [itex]y[/itex]; others might insist that you expand the left hand side and re-arrange the result into the form [tex]
    y''' = F(y, y', y'').
    [/tex]
     
    Last edited: Jun 23, 2014
  6. Jun 24, 2014 #5
    I'm using a method in which I differentiate depending on the number of constants. So, in this case I shall differentiate three times.

    [itex]y=e^{ax^{2}+bx+c}[/itex]
    [itex]y'=e^{ax^{2}+bx+c} (2ax+b)[/itex]
    [itex]y''=e^{ax^{2}+bx+c} (2ax^{2}+4axb+b^{2}+2a)[/itex]
    [itex]y'''=e^{ax^{2}+bx+c} (4a^{2}x^{3}+8^{2}x^{2}b+2axb^{2}+4a^{2}x+2ax^{2}b+4axb^{2}+b^{3}+2ab)[/itex]

    And then I shall put the coefficients of a, b and c in Wrosnkian and then find determinant.

    [itex]\begin{matrix}
    y & * & * & * \\
    y' & * & * & * \\
    y'' & * & * & *\\
    y''' & * & * & *
    \end{matrix}[/itex]

    Where the stars (*) are the coefficients. And that my problem right now, in some terms of y'' and y''' there are a and b together. How can I deal with it?
     
  7. Jun 24, 2014 #6

    ehild

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    First check if the equations are correct.

    Note that you do not have a linear system of equations for a,b,c.
    The equations can be divided by y so the exponential factors cancel, and there are three equations to solve.
    [tex]y'/y=2ax+b[/tex]
    [tex]y''/y=4a^2x^{2}+4axb+b^{2}+2a[/tex]

    y'''/y=....

    Isolate b from the first one, substitute for b into the second one, and isolate a. Substitute a and b in terms of y'/y and y"/y into the third one.

    But it is much simpler to follow pasmith's hint.


    ehild
     
    Last edited: Jun 24, 2014
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