I am having trouble coming up with series solutions to differential equations I have a problem that needs to be solved by Power Series Method. The equation is $$y'+y=2$$
I know this is trivial to use seperable equations so I know what the answer is: $$y(x)=c_1e^{-x}+2$$ but I can't derive it using the Taylor Series expansion. using power series the DE becomes $$x^n\cdot\sum{[c_{n+1}\cdot(n+1)+c_n]}=2$$ My recurrence relationship is $$c_{n+1}=\frac{2-c_n}{n+1}$$ I used out to $n=5$, I have the general solution as $$c_{m+1}=\frac{(-1)^m(c_0+2[what\ I\ can't\ find])}{(m+1)!}$$
at $n=5$ $$[what\ I\ can't\ find]=2(4!-3!+2!)$$ If you increase n then the number of factorial terms increases. Any help is appreciated.

You are playing extremely fast and loose with the summation notation. After plugging in the power series, you should arrive at
$$
0=-2+\sum_{n=0}^{\infty}x^n(a_n+(n+1)a_{n+1})\,.
$$
(You cannot pull the $x^n$ out of the sum!) Then, what we're using is a theorem from complex analysis that if a power series is equal to zero, i.e.,
$$
0 = \sum_{n=0}^{\infty}c_nx^n\,,
$$
then necessarily the coefficients $c_n$ are all zero.
Here, the $-2$ is outside the sum, but we can fix this situation in one of two ways. One is to just extract the constant terms from the sum, yielding
$$
0=-2+a_0+a_1+\sum_{n=1}^{\infty}x^n(a_n+(n+1)a_{n+1})\,,
$$
and recognize that along with the coefficients in the infinite sum being zero, i.e.,
$$
0=a_n+(n+1)a_{n+1}\,,
$$
we must have that the constant term must be zero as well, i.e.,
$$
0=-2+a_0+a_1\,.
$$
This recurrence relation allows you to write everything in terms of $a_0$.
