Power series solution to $y’ + 2xy = 0$ 
So I'm stuck. I'm pretty sure the constant $c_1$ equals zero, which makes the equation easy to solve by using the identity principle. But how do I show that $c_1=0$.
 A: Yes, $c_1=0$. Since the coefficients on the left-side have to be zero (the right-side is identically zero), we have that
$$c_1=0\quad \text{and}\quad c_{n+1}=-\frac{2c_{n-1}}{n+1},\quad\text{for $n\geq 1$.}$$ 
Note that $c_1=0$ and the recurrence imply that $c_{2k+1}=0$ for all $k\in\mathbb{N}$ which means that the solutions are an even functions.
Are you able to find a closed formula for the "even" coefficients $c_{2k}$?
P.S. As noted by LordVader007, if we multiply both sides of the ODE by $e^{x^2}$, we have that 
$$e^{x^2}y'(x) + 2xe^{x^2}y(x) =D(e^{x^2}y(x))= 0$$ 
and after integration we get 
$$y(x)=c_0e^{-x^2}=c_0\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k}}{k!}.$$
A: Let's repeat the calculation with @LordVader007's observation that only even powers are needed, viz. $y=\sum_{n\ge 0}a_n x^{2n}$ with $a_n:=c_{2n}$. This gives the boundary condition $a_0=y_0$. Since$$0=\sum_{n\ge 0}(2na_nx^{2n-1}+2a_n x^{2n+1})=2\sum_{n\ge 0}((n+1)a_{n+1}+a_n)x^{2n+1}$$we have$$a_{n+1}=-\frac{a_n}{n+1}\implies a_n=y_0\frac{(-1)^n}{n!}\implies y=y_0\exp -x^2.$$
