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Power Series/series solutions near a point

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data
    given the equation (2+x2)y''-xy'+4y determine the first four terms of the coresponding independent solutions (a(0),a(1))
    given the solution is [itex]\Sigma[/itex]a(n)xn

    2. Relevant equations
    when I calculated the series i got a(n+2)= (n(n-2)+4)/2(n+2)(n+1)


    3. The attempt at a solution
    so when I used values of 0-7 for n i got a(0),a(1)/4,....ect I'm not sure if my series equations is incorrect any help would be great
     
  2. jcsd
  3. Jun 27, 2011 #2

    lanedance

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    show your steps

    There should be 2 independent solutions as it is a 2nd order DE. These can be presented in terms of the first coefficients a(0) and a(1)

    Generally you should find a general recurrence relation for general n, however you may need specific cases for the first few terms
     
  4. Jun 27, 2011 #3

    lanedance

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    show your steps

    There should be 2 independent solutions as it is a 2nd order DE. These can be presented in terms of the first coefficients a(0) and a(1)

    Generally you should find a general recurrence relation for general n, however you may need specific cases for the first few terms
     
  5. Jun 27, 2011 #4
    I find two series with a(0) and a(1) but its wrong and no matter how many times I try to equate my equation for a(n+2) i get the same equation I posted so I'm not sure where I am screwing up
     
  6. Jun 28, 2011 #5

    lanedance

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    I can't be sure either without seeing what you've tried... ;) if you show your attempt I can have a look though
     
    Last edited: Jun 28, 2011
  7. Jun 28, 2011 #6

    lanedance

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    also thats not quite an equation, do you mean
    (2+x2)y''-xy'+4y=0?
     
  8. Jun 28, 2011 #7
    from the beginning:
    since the answer is of the form y=[itex]\Sigma[/itex]anxn so the first and second order derivatives are y'=[itex]\Sigma[/itex]nanxn-1 and y''=[itex]\Sigma[/itex]n(n-1)anxn-2 so plugging those into the original equation we get
    (2+x2)[itex]\Sigma[/itex]n(n-1)anxn-2-x[itex]\Sigma[/itex]nanxn-1+4[itex]\Sigma[/itex]anxn and from that equation you can see that the zero terms for the first order and second order for x2 will be zero so you wont need to change the order, and for the 2y'' term setting the sigma to 0 gives the form
    2[itex]\Sigma[/itex](n+2)(n+1)an+2xn
    so you can then factor out the [itex]\Sigma[/itex] and the xn terms so whats left i used to formulate the equation for an+2 and then subbed in values for n to get terms for a2....a10 in terms of a0 and a1
    thats how I did it and it follows my instructors example completely but I am unable to get the correct answer hope this is helpful
     
  9. Jun 28, 2011 #8

    lanedance

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    ok so following along (note you can write a whole equation in tex)

    first in these problems its worth including the sum as terms will be a little different at low n as the derivative of a constant is zero
    [tex] y = \sum_{n=0}a_nx^n[/tex]
    [tex] y' = \sum_{n=1}a_nnx^{n-1}[/tex]
    [tex] y'' = \sum_{n=2}a_nn(n-1)x^{n-2}[/tex]
     
    Last edited: Jun 29, 2011
  10. Jun 28, 2011 #9

    lanedance

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    the sums and powers can get a little confusing so here's a reasonably foolproof recipe for putting it all together

    now calculating each term in the DE
    [tex] 4y = 4\sum_{n=0}a_nx^n[/tex]
    [tex] xy' = x\sum_{n=1}a_n n x^{n-1} = \sum_{n=1}a_n n x^{n}[/tex]
    [tex] 2y'' = 2\sum_{n=2}a_n n (n-1)x^{n-2}[/tex]
    [tex] x^2y'' = x^2\sum_{n=2}a_nn (n-1)x^{n-2}= \sum_{n=2}a_nn (n-1)x^{n}[/tex]

    the next trick I find useful is to re-write each sum so the sum has the power of x expressed in the same form and change the indexes to align.

    so we can group three terms together, those with x to the power of n, and these need no change but we change m to n for clarity
    [tex] m = n [/tex]
    [tex] 4y = 4\sum_{m=0}a_m x^m[/tex]
    [tex] xy'= \sum_{m=1}a_m m x^{m}[/tex]
    [tex] x^2y'' = \sum_{m=2}a_m m (m-1) x^{m}[/tex]

    now for the n-2 power term we do the following shift
    [tex] m = n-2\implies n=m+2 [/tex]
    [tex] 2y'' = 2\sum_{n=2}a_nn(n-1)x^{n-2}= 2\sum_{m=0}a_{m+2}(m+2)(m+1)x^{m} [/tex]

    from here it should be a simple case to substitute into the original and compare terms,

    note the sums start at different m values so you may want to write out the terms for m<2 explicitly and have sums form m=2 up
     
    Last edited: Jun 29, 2011
  11. Jun 28, 2011 #10
    That is the same as what I did except there are no constants when you took the derivative's of the series how come you dont add any new terms when you took your derivatives
     
  12. Jun 28, 2011 #11

    lanedance

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    sorry i don't understand the question?
    [tex] \frac{d}{dx} (a_0x^0) = \frac{d}{dx} (a_0.1) = 0 [/tex]
     
  13. Jun 28, 2011 #12
    when you took the answer to the ode
    y=[itex]\sum[/itex]anxn and then derived it to get
    y'=[itex]\sum[/itex]anxn-1 shouldn't you bring down the n as a constant? and if not then wouldn't your solution for am+2=am?
     
  14. Jun 29, 2011 #13

    lanedance

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    woops that one slipped through cutting & pasting tex, sorry, have corrected above
     
    Last edited: Jun 29, 2011
  15. Jun 29, 2011 #14
    so then the equation for am+2 would be
    am+2=(-m(m-1)+m-4)am/(m+2)(m+1) which would be the same one I got before
     
  16. Jun 29, 2011 #15
    Nvm I am an idiot lol thanks alot you've really helped
     
  17. Jun 29, 2011 #16

    lanedance

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    yeah so i think you got it but it should become
    [tex] (2+x^2)y''-xy'+4y=0[/tex]
    [tex]
    =2\sum_{m=0}a_{m+2}(m+2)(m+1)x^{m}+
    \sum_{m=2}a_m m (m-1) x^{m}+
    \sum_{m=1}a_m m x^{m}+
    4\sum_{m=0}a_m x^m
    [/tex]
    [tex]
    =2\sum_{m=0}a_{m+2}(m+2)(m+1)x^{m}+
    \sum_{m=2}a_m m (m-1) x^{m}+
    \sum_{m=1}a_m m x^{m}+
    4\sum_{m=0}a_m x^m
    [/tex]
    [tex]
    =2a_{2}+6a_{3}x+
    a_1 x+
    a_0 +a_1 x+
    +\sum_{m=2}(a_{m+2}(m+2)(m+1)+a_m m (m-1)+a_m m +a_m )x^{m}
    =0[/tex]
     
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