Closed form of a power series solution to a differential equation Given the differential equation $$f''(x)-2xf'(x)-2f(x)=2$$ I've managed to find the power series solution  $$\sum_{n=0}^\infty \frac{x^{2(n+1)+1}}{(n+1)!}.$$ Now how can I find the closed form of this function? It seems a bit similar to $$\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ which is the exponential function. But with a difference of shifted powers. Substitution didn't seem to get me anywhere. Does it require integrating or differentiating the expression? Also, is there a general strategy for tackling this kind of problems? What should be the first thing to look at? 
 A: You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:
$$\frac{x^{2(n+1)+1}}{(n+1)!} \overset{m=n+1}{=} \frac{x^{2m+1}}{m!} = x \frac{(x^2)^m}{m!}$$
So we get
$$\sum_{n=0}^\infty \frac{x^{2(n+1)+1}}{(n+1)!} = x \sum_{m=1}^\infty \frac{(x^2)^m}{m!} = x (\exp(x^2) - 1)$$
Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.
EDIT:  Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html
A: Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
$$\sum_{n=0}^\infty \frac{x^{2(n+1)+1}}{(n+1)!}=x\cdot\sum_{n=0}^\infty \frac{(x^2)^{n+1}}{(n+1)!}=x\cdot\sum_{n=1}^\infty \frac{(x^2)^n}{n!}=x\cdot\left(\sum_{n=0}^\infty \frac{(x^2)^n}{n!}-1\right)=\cdots$$
