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A faster way to solve this ODE?

  1. May 15, 2008 #1

    That ODE took me a little while to solve... I had to switch from dy/dx to dx/dy, do a Bernoulli substitution, two applications of integration by parts, and then some algebraic manipulations in order to solve it.

    The final result?
    y = (x^3-4)/12

    This really simple equation which satisfies the ODE leads me to believe that there is an easier method for solving this... A few of these might appear on my final tomorrow, and I don't want to spend all of that time if there is an easier method.

  2. jcsd
  3. May 17, 2008 #2


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    Whilst this reply will come a little too late for your final, it may still be useful. Notice that if we assume that [itex]x\neq0[/itex] we may write,

    [tex]\frac{dy}{dx} - \frac{12}{x^3}y=\frac{1}{x}[/tex]

    Which is a much easier to solve.
  4. May 17, 2008 #3


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    What happened to the dy/dx term in your [tex]- \frac{12}{x^3} y[/tex]?
  5. May 17, 2008 #4
    Ya... The derivative was supposed to be distributed.
  6. May 17, 2008 #5
    Looking at the de, if you were to power expand y (heck you could even Laurent expand it) in order to preserve the right hand side y would have to be cubic, that is the powers can't be less than 0 nor can they be greater than 3.

    So say there exists a,b,c,d such that [tex]y = ax^3 + bx^2 + cx + d[/tex] then we have

    [tex]\frac{dy}{dx}(x^3-12y) = (3 ax^2 + 2bx + c)[(1 - 12a)x^3 -12bx^2-12cx-12d][/tex]

    but that has to equal x^2 so the x^5 term must go away
    and clearly a is not zero so [tex]a=1/12[/tex].

    The x^4 term must also go away
    -12b(3a) = 0
    and we know a is not zero so that means that [tex]b=0[/tex].

    Let's restate the equation again

    The x^3 term goes away so
    or [tex]c=0[/tex].

    Alright finally the x^2 term should have a coefficient of 1 so
    or [tex]d=-1/3[/tex].

    And we have the solution [tex]y=\frac{1}{12}x^3-\frac{1}{3}[/tex]

    This is not a simpler method to solve it, but I think it shows more clearly why the solution is so simple.
  7. May 17, 2008 #6


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    Hi scorpion990! :smile:

    The way I'd do it is to assume that y is a polynomial in x.

    If its highest term is x^n, then you start with nx^(n-1)[x^3 - 12x^n], so the highest terms are 3+n-1 and/or n+n-1, and so … :smile:
  8. May 17, 2008 #7


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    Sorry guys, I misread the parenthesise, my bad.
  9. May 22, 2008 #8
    Make it exact?

    Why wouldn't you make it exact first.

    [tex] (x^3 - 12y)dy + (-x^2)dx = 0 [/tex]

    By inspection one can find [itex]e^\left-3y\right[/itex] makes this exact.

    [tex] e^\left-3y\right\right((x^3 - 12y)dy + e^\left-3y\right\right((-x^2)dx = 0 [/tex]
    [tex] \smallint e^\left-3y\right\right((x^3 - 12y)dy [/tex]
    [tex] \smallint e^\left-3y\right\right((-x^2)dx [/tex]

    After the one integration by parts that is required, you end up with:
    [tex] C = \frac{e^\left-3y\right}{3}(4 - x^3 + 12y) [/tex]

    I guess you have an initial condition that tells you that C is zero.

    This may or may not be considered quicker, but it was the only way I remembered after taking DE 10+ years ago. :)

    Keep these problems coming. I'm going to go back to school and get a master's, assuming the wife lets me. :)
  10. May 22, 2008 #9


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    Well I certainly couldn't see that would have worked! I mean, how did you know that some integrating factor existed which could make it exact?
  11. May 22, 2008 #10

    D H

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    Nice work. Everyone else has found a particular solution. You found the general solution.
  12. May 22, 2008 #11
    When I took DE in college, I would always look to make it exact first. Since the @/@y of the second term is zero and the @/@x of the first term is in terms of x only, you know that the integrating factor is probably a function of y only. You end up with

    [tex] F(y) = -F'(y)/3 [/tex]
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