Series Sol'n to a DE with trig coefficient?

My computer is not letting me have access to that document but I got a suggestion for you: muscle through it. But start at the end and work backwards. Since it's singular at x=0, then the solution would be valid for x>0 so say work towards solving:

[tex]4xy''+2y'+\cos(x)y=0,\quad y(1)=0,\quad y'(1)=1[/tex]

and since the roots of the indical equation are 0 and 1/2 and the difference non-integer, the solution is:

[tex]y(x)=C_1 \sum_{n=0}^{\infty} a_n x^n+C_2 \sum_{n=0}^{\infty} b_n x^{n+1/2}[/tex]

and we obtain that solution by substituting each power series into the DE, and letting the first coefficient equal to 1 and compute all the rest recursively even though there's a double sum in there and you just have to learn how to work with a "Cauchy product of double sums".

I'll do c=0 for you, you do the messy one with all the fractional powers of x:

[tex]\sum_{n=0}^{\infty} a_n(4n^2-2n)x^{n-1}+\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(-1)^k}{(2n)!} a_{n-k} x^{n+k}=0[/tex]

Ok, that's it. You're really not expected to reduce that further manually are you? That's just not reasonable to do in my opinion in the 21's century. It's just so simple with a CAS like Mathematica to get all the terms recursively: set a_0=1, extract all the coefficients of powers of x, solve for a_n:

Code (Text):

nmax = 50;
Remove[a, b]

myalist = Sum[Subscript[a, n]*(4*n^2 - 2*n)*x^(n - 1), {n, 0, nmax}] + Sum[((-1)^k/(2*k)!)*Subscript[a, n - k]*x^(n + k),
     {n, 0, nmax}, {k, 0, n}];

Subscript[a, 0] = 1;

myatable = Table[Subscript[a, n + 1] = First[Subscript[a, n + 1] /. N[Solve[Coefficient[myalist, x, n] == 0,
         Subscript[a, n + 1]]]], {n, 0, nmax - 1}];

myatable = Prepend[myatable, 1];

f1[x_] := c1*Sum[Subscript[a, n]*x^n, {n, 0, nmax - 1}]
 

Once you do that for c=1/2 and find:

[tex]f2(x)=C_2 \sum_{n=0}^{\infty}b_n x^{n+1/2}[/tex]

Then solve for c1 and c2 using the initial conditions:

Code (Text):

they[x_] := f1[x] + f2[x]
thec = NSolve[{they[1] == 0, they'[1] == 1}, {c1, c2}] // First
pa = Plot[they[x] /. thec, {x, 1, 2}]
 

 

Ok. I'm sorry. I shouldn't have told you to cheat. You need the practice learning how to do it manually. I mean really, I found a_1, a_2, a_3 by hand initially but got dizzy tryin' to find a_4. Tell you what, how about I just use Mathematica to find the recursive expressions for the first six (this is in Mathematica Latex so I don't have to type it in):
[tex]
\left\{\left\{a_1\to -\frac{a_0}{2}\right\},\left\{a_2\to -\frac{a_1}{12}\right\},\left\{a_3\to \frac{1}{60} \left(a_0-2 a_2\right)\right\},\left\{a_4\to \frac{1}{112} \left(a_1-2 a_3\right)\right\},\left\{a_5\to \frac{-a_0+12 a_2-24 a_4}{2160}\right\},\left\{a_6\to \frac{-a_1+12 a_3-24 a_5}{3168}\right\}\right\}[/tex]

See what I mean. I don't think that's happening for me.

Also, this is what I get for the fractional part.[tex]\text{myexp2}=\left(\sum _{n=0}^{\text{nmax}} b_n(4(n+c)(n+c-1)+2(n+c))x^{n+c-1}+\sum _{n=0}^{\text{nmax}} \sum _{k=0}^n \frac{(-1)^kx^{2k}}{(2k)!} b_{n-k}x^{n+c-k}\right)\text{/.}c\to 1/2;[/tex]

and here's what the first 4 terms look like:

[tex]\sqrt{x} b_0-\frac{1}{2} x^{5/2} b_0+\frac{1}{24} x^{9/2} b_0-\frac{1}{720} x^{13/2} b_0+6 \sqrt{x} b_1+x^{3/2} b_1-\frac{1}{2} x^{7/2} b_1+\frac{1}{24} x^{11/2} b_1+20 x^{3/2} b_2+x^{5/2} b_2-\frac{1}{2} x^{9/2} b_2+42 x^{5/2} b_3+x^{7/2} b_3[/tex]

. . . worst

Also keep in mind I compared this analytic solution with the numerical solution of the IVP above and got excellent agreement in the interval [1,5] using the first 50 terms and so that gives me high confidence the series expressions above are correct.

 

That's ok for regular easy problems. When you have a double sum, or something else complicated, need to just learn to use what you have. So don't try to bring everything to the same power. I don't think you can do that with this one but maybe I'm wrong. Also, I'm approaching it the way I was taught in Rainville and Bedient. Perhaps there is another way to approach it that would make it easier to determine the recursion relations.

 

I used Mathematica:

[tex]\text{myatable}=\text{Table}\left[a_{n+1}=a_{n+1}\text{/.}\text{Solve}\left[\text{Coefficient}[\text{myaexp},x,n]==0,a_{n+1}\right]\text{//}\text{First},\{n,0,\text{nmax}-1\}\right][/tex]

That is, find the coefficient of x^n, set the resulting expression equal to zero, solve for a_(n+1).

Really, I believe in doing things by hand to understand them. That's important. But I've already done a bunch of these that way and understand the concept well enough to do that. Also I'm practical: if I can't figure it out, understand or no understantd I'm still going to use Mathematica to help me find the answer.

 

Ok, I looked at it. Suppose I write the DE then as:

[tex]y''+\frac{1/2}{x}y'+\frac{1}{x^2}\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{4(2n)!}y=0[/tex]

or now:

[tex]y''+\frac{P(x)}{x} y'+\frac{Q(x)}{x^2}y=0[/tex]

can we then now use that formula for z_n to compute the coefficients? Keep in mind that:

[tex]Q(x)=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{4(2n)!}=\sum_{n=0}^{\infty}a_n x^n[/tex] with a_n=0 for even n though.