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

1. ## 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!

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

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}]

3. 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.