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.

Write your $2$ as $$2=2\cdot x^0+0\cdot x^1+0\cdot x^2+...$$
Since your equation is valid for any $x$, all the coefficients on the left hand side must equal to the coefficients on the right hand side. So for $n=0$ you have $$c_1(0+1)+c_0=2$$
For every other $n$ you have $$c_{n+1}(n+1)+c_n=0$$
The recurrence relationship is then $$c_n=\frac{(-1)c_{n-1}}{n}=\frac{(-1)^2c_{n-2}}{n(n-1)}=\frac{(-1)^3c_{n-3}}{n(n-1)(n-2)}=...$$
