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Creating series solutions for a non-constant coefficient ODE

  1. May 23, 2016 #1
    1. The problem statement, all variables and given/known data
    This is for differential equations with nonconstant coefficients and I wasn't so great at series and sequences in calculus so when I came across this example problem I wasn't sure how they got to their final form. If someone could explain it to me that would be really helpful.

    2. Relevant equations
    The ODE is:
    [tex] y"-2xy+2y=0[/tex]

    3. The attempt at a solution
    We assume there is some solution of the form:
    [tex]y(x)=\sum_{n=0}^{\infty}a_nx^n [/tex]
    The derivatives are:
    [tex]y'(x)=\sum_{n=1}^{\infty}na_{n}x^{n-1}\:\: \text{and}\:\: y''(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}[/tex]
    A change of index gives
    [tex]y'(x)=\sum_{n=0}^{\infty}na_nx^{n-1}\:\:\text{and}\:\:y''(x)=\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n[/tex]
    Putting these into the ODE gives:
    [tex]\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n-2x\sum_{n=0}^{\infty}na_nx^{n-1}+2\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}\Big((n+2)(n+1)a_{n+2}-2na_n+2a_n\Big)x^n=0[/tex]
    The Identity Theorem gives:
    [tex](n+2)(n+1)a_{n+2}-2na_n+2a_n=0[/tex]
    Solving this for ##a_{n+2}## gives:
    [tex]a_{n+2}=\frac{2(n-1)a_n}{(n+2)(n+1)}[/tex]
    Odd coefficients are zeros except for ##a_1## because:
    [tex]a_3=\frac{2(1-1)a_1}{(n+2)(n+1)}=0\:\: \text{and all other coefficients will depend on a multiple of}\:a_3[/tex]
    The even coefficients are given by:
    [tex]a_{2k+2}=\frac{2(2k-1)a_{2k}}{(2k+2)(2k+1)}[/tex]
    This is a worked example from the book and the answer is:
    [tex]\frac{2(2k!)}{k!}[/tex]

    I'm not sure how they got this final form, but I think it has something to do with the even number recursion formula, which can be written as:
    [tex]a_{2k+2}=\frac{2(2k-1)a_{2k}}{(2k+2)(2k+1)}=\frac{2(2k-1)}{(2k+2)(2k+1)}\frac{2(2k-3)a_{2k-2}}{2k(2k-1)}[/tex]
    This is where I am having the problem. The book says the ##a_{2k+2}## can be written as:
    [tex]\frac{2^{k+1}(2k-1)(2k-3)...3\cdot2\cdot1}{(2k+2)!}a_0[/tex]
    I understand where the ##2^{k+1}## and ##a_0## are coming from, but the other terms I am unsure of.
     
  2. jcsd
  3. May 24, 2016 #2
    Hi rmiller:

    I have not yet digested all of your calculations, but I noticed a typo in your ODE. You are missing an apostrophe. The second term should be
    -2xy' .​

    Since the equation is linear, any constant multiple of a solution is also a solution. This means that a0 is an arbitrary constant.
    Since this is a second order equation, there should be two independent solutions.

    It has been decades since I worked with ODEs so I may be mis-remembering the following.

    I think you can also choose a1 to be another arbitrary constant. Then find how a2 relates to a1 and a0. You then need a recursion relationship between
    a2k+2 and a2k.​

    The solution given
    a2k = 2(2k!)/k!​
    seems to be based on assuming that a0 = 4 and ignoring a1.

    You should be able to find the second solution by assuming a0 = 0 and a1 = 1.

    ADDED
    I notice that the solution given above is inconsistent with the previous even coefficients expression for a2k+2.
    Assuming a0 = 1, the expression for a2k+2 calculates
    a2 = -a0.​
    However, the book answer 2(2k!)/k! is always positive.

    Hope that this helps.

    Regards,
    Buzz
     
    Last edited: May 24, 2016
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