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James Woolley permalink

January 19, 2010

I am intrigued by your method for generating a series solution to most any form of DE. While your case studies are enlightening, I would enjoy seeing a presentation of the overview of the methodology. I think I understand the calculate and balance methodology. I did get lost in Example 2.4 when you substituted y=cx into the DE and set the resultant DE equal to x^3! I thought the DE should equal zero! Perhaps I didn’t understand what you were doing with the DE with only the y+ result.

Jim W

admin permalink*

January 19, 2010

I much appreciate your interest in my method. Thanks a lot.
Here I try to answer your question:
We are looking for functions that satisfy the differential equation as mentioned in article 2.4, always we are looking for n values that eliminate as much terms as possible. so we try n=2 and n=3. for n=2 we get y+=1. If we try n=3 we get a series that is not convergent. If you substitute n=3 into the equation we get a series like this:
y+=x-1.x-0+1^2.x+0…
If you proceed substituting n=3 in the equation you will end up with a non convergent series that have x of power 1. This solution is misleading because it does not converge, but it tells us that it is a function of x. so our inspected function equals f(x) or cx, where c is a constant. so y+=cx. now we need to get this constant c. to get it, we substitute this solution into the differential equation which gives us c=1/2.
Now comes why I equated the positive function to x^3. If you look at our function, it is y=y+ + y-, and we started y+ by +x^n (positive seed), and y- with -x^n (negative seed) so that the result will be equal to 0. i.e. y+ + y- = +x^n – x^n = 0, and since we are using n=3 then y+ should be equated to x^3. Please note that y+ is just half the solution, so it should be equated only to its positive seed which is +x^3 and not to 0.
I hope this answers your question.
Again thank you very much for your interest.

Yousif.