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Hard power series and initial value problem question

  1. Dec 12, 2011 #1
    1. The problem statement, all variables and given/known data

    We know that y = Aex is the solution to the initial value problem dy/dx = y; y(0) = A.
    This can be shown by solving the equation directly. The goal of this problem is to reach the same conclusion using power series.
    Method: Let y be a solution to the initial value problem, and suppose y has the
    power series representation
    y(x) =the sum from n= 0 to n= infinity of (an)(x^n)
    which equals a0 + a1x + a2x^2 + ...

    First finnd a0. Next, take the term-by-term deriviative of the series to fi nd a power
    series representation for dy/dx . Using the fact that dy/dx = y, obtain a formula which
    shows how an is related to an-1 for n >1. From this, find an explicit formula for an.
    Finally, use the known power series representation for e^x to conclude that y(x) = Ae^x


    3. The attempt at a solution

    I know an = f(n)(0)/n!
    And I know the function is a power series centered at 0.
    but I don't really know where to go from there?
    i just don't know where to start on this. please help. if I see how to get started, I will be able to understand. I did try, but I don't know what to do!
     
  2. jcsd
  3. Dec 12, 2011 #2
    If the solution can be represented as:

    [tex]y(x)=a_0+a_1 x+a_2 x^2+\cdots[/tex]

    and you know y(0)=A, then you know what a_0 is then and when you take the derivative of y(x), it's

    [tex] y'(x)=a_1+2a_2 x+3 a_3x^3+\cdots[/tex]

    and since you're given y'=y, then what about equating the respective power series for y'=y then equating coefficients? Doing that, can't you find an explicit expression for a_n? I'll do two:

    [tex]a_1=A[/tex]
    [tex]2a_2=a_1[/tex]

    but since a_1=A then [itex]a_2=\frac{a_1}{A}[/itex]. Now do a few more and notice the trend. Then come up with the general expression for a_n.
     
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