Using a series to find a differential equation?

Question

So we had a homework assignment where we are asked to solve a differential equation through a series,I am not sure where to begin for this as we have not actually gone over this. If someone could help guide me through the process I would greatly appreciate it. $$y=\sum_{n=0}^\infty\frac{(-1)^n2^n(n!)x^{2n+1}}{(2n+1)}$$ Also I tried searching for the input for factorial but no luck if someone could edit it or let me know how to add it I would appreaciate that.

Answer 1

I'm assuming you know how to derive a series solution to a differential equation and are just struggling to work backwards to find the correct DE for this series.

There is a very powerful method that can be used here: trial and error (with some intuition) :D

$$y= \sum_{n=0}^{\infty} \frac {(-1)^n 2^n n!}{2n+1} x^{2n+1}$$ This looks like an integral of x^2n so differentiate it: $$y'= \sum_{n=0}^{\infty} (-1)^n 2^n n! x^{2n}$$ This is immediately nicer to work with: $$x^2y'= \sum_{n=0}^{\infty} (-1)^n 2^n n! x^{2n+2}$$ $$\frac {d}{dx}[x^2y']= \sum_{n=0}^{\infty} (-1)^n 2^n n!(2n+2) x^{2n+1}$$ $$2xy'+x^2y''= \sum_{n=0}^{\infty} (-1)^n 2^{n+1} (n+1)! x^{2n+1}$$ This is beginning to look a lot like the series for y': $$-x(2xy'+x^2y'')= \sum_{n=0}^{\infty} (-1)^{n+1} 2^{n+1} (n+1)! x^{2n+2}$$ $$-x(2xy'+x^2y'')= \sum_{n=0}^{\infty} (-1)^{n} 2^{n} n! x^{2n} -1$$ $$ \therefore -x(2xy'+x^2y'')=y'-1 \\ $$

$$ \\ x^2y''-(2x^2+1)y'+1=0 \\$$ This is your DE unless I made any calculation errors.