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Need help on Power Series Question

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data

    y''+t^2*y'-y=1-t^2


    2. Relevant equations

    y(0)=-2
    y'(0)=1

    Find the first 6 coefficients

    C1=-2
    C2=1
    C3=? (-1?)
    C4=?
    C5=?
    C6=?

    3. The attempt at a solution

    Okay so I tried to do this but I'm not used to having anything on the RHS of the equation.

    I got down to [tex]\Sigma[/tex](n+1)(n+2)an+2tn+[tex]\Sigma[/tex]an-1(n-1)tn-[tex]\Sigma[/tex]antn-a0+2*a2

    Where the sigma's have n=2 on the bottoms of them. Don't know what to do next tho!
     
  2. jcsd
  3. Dec 4, 2009 #2

    Mark44

    Staff: Mentor

    Set what you have equal to 1 - t^2.
    Expand the summations on the left side up to about 5th degree terms.
    Bring the two terms on the right side over to the left side.
    Group all constants together, group all terms in t together, all terms in t^2 together, and so on up to the t^5 terms.
    The expression on the left is identically equal to zero, so the constant term has to be zero, the coefficient of t has to be zero, and so on. These should allow you to solve for c0, c1, c2, ..., c5. (BTW, switch your coefficients from a_i to c_i so you don't confuse yourself.)
     
  4. Dec 4, 2009 #3
    Write the equation as

    [tex] \sum [...] t^n = 1 - t^2 [/tex]

    and then work with each n separately. For example, for n=0, you'd get something like 2a2 - a0 = 1, which gives you 2a2 with the help of the initial conditions. Then move to n=1, 2 ...

    Alternatively, you can cheat a little; just looking at the differential equation you can immediately see what y''(0) has to be with those initial conditions. Then you can differentiate it once to find y'''(0) and so forth. Differentiating the series expression, you can immediately see the connection between the coefficients and derivatives at t=0.
     
  5. Dec 4, 2009 #4
    So for the n=0 you subbed the 0 into the t term as well?
     
  6. Dec 4, 2009 #5
    Bump!!!!
     
  7. Dec 4, 2009 #6
    Nvm I got it guys thanks!
     
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