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Math Help - cos(x)y'' + xy' - 2y = 0

  1. #1
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    cos(x)y'' + xy' - 2y = 0

    Find the first three non-zero terms in each of two linearly independent power series solutions centred at x = 0 of the equation

    cos(x)y''+xy' - 2y = 0

    What do you expect the radius of convergence of the solutions to be?


    I tried to right cos(x) in its taylor series form but I got stuck somehow... x=0 is a regular singular point that I've checked...please help thanks!
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  2. #2
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    I think zero is an ordinary point, not singular. I can derive a series solution with two arbitrary constants but can't figure out how to extract two linearly independent solutions from it. Maybe someone here can help. Writing it as:

    y''+\frac{x}{\cos(x)}y'-\frac{2}{\cos(x)}y=0

    where:

    P(x)=\frac{x}{\cos(x)}=\sum_{n=0}^{\infty} b_n x^n

    Q(x)=-\frac{2}{\cos(x)}=\sum_{n=0}^{\infty} d_n x^n

    and if we assume a solution of the form y=\sum_{n=0}^{\infty} a_n x^{n+c}

    and substitute all the sums into the DE and shift the indexes I get:

    \begin{aligned}<br />
&\sum_{n=0}^{\infty} (n+c)(n+c-1)a_n x^{n+c-2} \\<br />
&\hspace{20pt}+\sum_{n=0}^{\infty}\sum_{k=0}^{n-1} (k+c)a_k b_{n-k-1}z^{n+c-2} \\<br />
&\hspace{20pt}+\sum_{n=0}^{\infty}\sum_{k=0}^{n-2} a_k d_{n-k-2}z^{n+c-2}=0<br />
\end{aligned}<br />

    collecting terms then:

    c(c-1)a_0=0

    c[(1+c)a_1+a_0 b_0]=0

    and if I choose c=0 then for n\geq 2 we have:

    a_n=\frac{n}{n-1}\left(-\sum_{k=1}^{n-1} k a_k b_{n-k-1}-\sum_{k=0}^{n-2} a_k d_{n-k-2}\right)

    and I can let a_0 and a_1 be arbitrary and it's not hard to code this in Mathematica and show at least graphically, a good fit between this series solution and a numerical solution for some IVP such as:

    \cos(x)y''+x y'-2y=0,\quad y(0)=1,\quad y'(0)=0

    The plot below superimposes a numerical solution to this IVP against the series solution using the first 20 terms. The agreement is not bad for my money.

    Code:
    pseries = Series[x/Cos[x], {x, 0, 25}]; 
    Subscript[b, n_] := SeriesCoefficient[
       pseries, n]
    qseries = Series[-2/Cos[x], {x, 0, 25}]; 
    Subscript[b, n_] := SeriesCoefficient[
       pseries, n]
    Subscript[d, n_] := SeriesCoefficient[
       qseries, n]
    Subscript[a, 0] = 1; 
    Subscript[a, 1] = 0; 
    mycoef = Table[Subscript[a, n] = 
         (1/(n*(n - 1)))*
          (-Sum[k*Subscript[a, k]*Subscript[
               b, n - k - 1], {k, 1, 
              n - 1}] - Sum[Subscript[a, k]*
             Subscript[d, n - k - 2], 
            {k, 0, n - 2}]), {n, 2, 20}]; 
    myfun[x_] = Sum[Subscript[a, n]*x^n, 
       {n, 0, 20}]
    series = Plot[myfun[x], {x, 0, 2}, 
       PlotStyle -> Red]
    sol = NDSolve[
       {Cos[x]*Derivative[2][y][x] + 
          x*Derivative[1][y][x] - 2*y[x] == 
         0, y[0] == 1, Derivative[1][y][0] == 
         0}, y, {x, 0, 2}]
    numeric = Plot[Evaluate[y[x] /. sol], 
       {x, 0, 2}, PlotStyle -> Blue]
    Show[{series, numeric}]
    Attached Thumbnails Attached Thumbnails cos(x)y'' + xy' - 2y = 0-cosine-de.jpg  
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  3. #3
    Super Member
    Joined
    Aug 2008
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    903
    Look,this is what I'm gonna' do with this then: I'm going to let a_0=1 and a_1=0. Then:

    f_1(x)=1+x^2+\frac{x^4}{12}+\frac{x^6}{120}+\frac{  19 x^8}{20160}+\cdots

    Likewise, I'll let a_0=0 and a_1=1. Then:

    f_2(x)=x + x^3/6 + x^5/60 + x^7/560 + x^9/4480+\cdots

    and these are obviously linearly-independent so then I can (probably) write the solution as:

    y(x)=c_1 f_1(x)+c_2 f_2(x)

    . . . there's your five. Now, note he asked what do you "think" the radius of convergence is and not what "IS" the radius of convergence. Then that part is easy. I think the radius of convergence is \pi/2 since \cos(\pi/2)=0 and P(x) and Q(x) are singular there even though my plot goes past this point. And if you wanted to be more rigorous, then I'd recommend "Intermediate Differential Equations" by Rainville where I got most of this and where it is shown that this type of problem expressed as the series above has a radius of convergence at least as large as \frac{r}{M} where this expression is derived in the text.

    If that's not it then I bet it's close . . . B is ok with me.
    Last edited by shawsend; January 28th 2010 at 05:47 AM.
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