## Solution of xy'‘ + y’ + xy = 0 using series [closed]

Hello,

I've tried to solve the differential equation xy'' + y' + xy = 0 using series. First, I assumed the solution:

$y=a_0+a_1x+a_2x^2+a_3x^3+...$

And from that I tried to solve it by finding out the equality between the coefficients once I substituted my solution in the differential equation. After that I get that the solution is:

$y=a_0(1-\frac{x^2}{2^2}+\frac{x^4}{2^24^2}-\frac{x^6}{2^24^46^2}+...)$

I've just started to learn about this so I'm probably missing something, but the thing is that the answer should be the Bessel function of order 0. I plotted both my solution and y = J0(x) and they appear to be almost the same.

In case my solution is wrong, why is it? Or is there a way to work my solution to make it the Bessel function of order 0?

Thanks.

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Not appropriate for MO. Use College Playground on AoPS for such questions. – fedja Jan 29 2010 at 20:30
In any case, what are you asking, really? That solution you've got is the $J_0$ function up to a scalar. – Mariano Suárez-Alvarez Jan 29 2010 at 20:33
This sort of thing is covered in many, many books on differential equations. See e.g. Mathews and Walker, Mathematical methods of physics section 7.2 which does a very thorough job of deriving everything you described in your answer. – jc Jan 29 2010 at 20:35
Closed, see jc's and fedja's answers. (The mathoverflow site is only intended to serve 'research mathematicians', loosely defined, and this question isn't of interest to our target audience. See the FAQ for more details.) – Scott Morrison Jan 29 2010 at 20:46