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Non homogeneous differential equation - power series solution

  1. Dec 2, 2009 #1
    I am having trouble getting to a solution for this differential equation

    [tex]2(x^2+2x)y' - y(x+1) = x^2+2x[/tex] -------- 1

    for a series solution, we have to assume y = [tex]\sum a_{n}x^n[/tex] ---------- 2

    if we divide equation 1 by [tex] x^2 + 2x [/tex], we get [tex] (x+1)/(x^2+2x) [/tex] for the y term, which is where my problem is, i can't get that term to common out in the series and always leaves me with an x term.

    may be i need to keep going with expression, what do you math geniuses think?
    Last edited: Dec 2, 2009
  2. jcsd
  3. Dec 2, 2009 #2
    x=0 is a regular singular point ???
    Never see anywhere people solve first order DE with Frobenius series!!!
  4. Dec 2, 2009 #3
    The indicial equation is s-1/4=0 so your trial series solution for the homogenous equation should be multiplied by x1/4 and remember to use a0=1.

    Once you have the homogeneous solution you just need to get the particular solution which can be obtained by substituting a general series form [tex]\sum_{n=0}^{\infty} a_n x^n [/tex]into equation 1 and balancing powers of x on each side of the equation.

    For the particular solution don't divide through by the first factor -- just multiply through by that factor and you'll see that all terms are powers of x. You should be able to collect terms and get your recurrence relations for the particular solution since you've got polynomial coefficients.
  5. Dec 2, 2009 #4
    Thanks for the suggestions/comments.

    @coldatoms - to get the indicial equation, how do we get rid of the right hand side term in equation 1.

    i am still left out with a x or an [tex] x^2 [/tex] when i use the frobenius method.
  6. Dec 2, 2009 #5
    You should just set the right hand side to zero and solve the remaining homogeneous equation using the Frobenius method. This gives you just the homogeneous part of the solution.

    You then determine a power series solution for the particular solution with the right hand side in place. Remember the particular solution plus the homogeneous solution give you the general solution. For this first order equation you will have one constant (C) that will be determined by a boundary condition. The general solution will be y(x)=C*yhomogeneous(x) +yparticular(x).
  7. Dec 3, 2009 #6
    Ok. I understand and could get to common out the [tex]x[/tex] terms, but am stuck at this step.

    [tex](4n-5)/2\sum a_{n-1}x^{(4n+1)/4}+4n\sum a_{n}x^{(4n+1)/4} = x^{2}+2x[/tex]

    I was looking up the answers to the values for coefficients and they come out to be(for non-homogeneous part)

    [tex]a_{0} = 0,
    a_{1} = 2/3,
    a_{2} = 1/21,
    a_{n} = -a_{n-1} (2n-3)/(4n-1)

    i have tried solving for the coefficients but they don't seem to come out right. do you think its calculation mistake or if my equation is wrong?
  8. Dec 3, 2009 #7
    You shouldn't be taking the n's out of the summation sigma.

    I would handle your problem this way. Make the coefficient of y' as 1

    [tex] y\prime{} - \frac{1}{4x} (x+1)(1+\frac{x}{2})^{-1}y=\frac{1}{2}[/tex]

    The coefficient of y can be simplifies as

    [tex] - \frac{1}{4x} (1+\frac{x}{2} + \frac{x}{2})(1+\frac{x}{2})^{-1}

    =- \frac{1}{4x} (1+\frac{x}{2}(1+\frac{x}{2} )^{-1}) [/tex]

    [tex]= - \frac{1}{4x} (1 + \sum_{n=0} (-1)^n (\frac{x}{2})^{n+1}) [/tex]

    May be you could continue from here.
  9. Dec 4, 2009 #8
    I just checked that for you in Maple. For the particular solution, the first three terms of the series solution about x=0 are exactly what you have derived (0, 2/3, 1/21,-1/77) so I'd assume your general term is correct as well. Also for the homogeneous part I get x1/4 ( 1+x/8 -3 x2 /128+ ...).
  10. Dec 4, 2009 #9
    Just adding to what matmatikawan has said ... if that equation with the summations was just supposed to be the result of substituting the series form into the differential equation you should not have pulled the n's out of the summation. You need to adjust the summation indices in each individual sum until you can write the whole left hand side as a single summation over n. At that point you can compare powers of x on each side of the equation and derive the coefficients of the particular solution.

    If the index manipulation is bothering you just take the first few terms of the assumed series solution as a polynomial and plug that in for y. Then try and find the coefficients. You should see what's happening in the sums that way.
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