Solve $2y''+y'=0$ using power series I got the answer $C_0e^{-x/2}$ by doing 
$$\sum_{n=1}^{\infty }\:\left(x^{n-1}\right) \left(2n\left(n+1\right)C_{n+1}\:\right)+\left(nC_n\right) = 0$$
thus giving 
$$C_{n+1}= -\dfrac{C_n}{2(n+1)}.$$
This leads me to the answer $y = C_0e^{-x/2}$.  

However, the answer in my textbook is $C_1e^{-x/2} + C_0$.  I've found this to be true, but still couldn't figure out a method of solving the equation so that there are two constants.

Edit: what my textbook did was  $$c_{n+2}= -\dfrac{1}{2(n+2)}c_{n+1}.$$  , and then  substitute $$c_{0}=C_0-2c_1$$  and $$c_1=\dfrac{-1}{2}C_1$$    into $y=c_0 +c_1 - \dfrac{1}{2*2!}c_{1} x^2. + \dfrac{1}{2^2*3!}c_{1} x^3........$ 

Where did the substitutions come from?
$$ \frac{\mathrm{d^2}y }{\mathrm{d} x^2}=-\frac{1}{2} \frac{\mathrm{d} y}{\mathrm{d} x} \\
\frac{\mathrm{d} }{\mathrm{d}x}\left ( \frac{\mathrm{d}y }{\mathrm{d} x} \right ) = -\frac{1}{2} \frac{\mathrm{d} y}{\mathrm{d} x} \\
\ln{\frac{\mathrm{d}y}{\mathrm{d}x}} = \frac{-x}{2}+c \\
\frac{\mathrm{d}y}{\mathrm{d}x} = c_{1}e^{-\frac{x}{2}}
\\ y=c_{1}e^{-\frac{x}{2}}+c_{2}
$$
$$ \frac{\mathrm{d^2}y }{\mathrm{d} x^2}=-\frac{1}{2} \frac{\mathrm{d} y}{\mathrm{d} x} \\
\frac{\mathrm{d} }{\mathrm{d}x}\left ( \frac{\mathrm{d}y }{\mathrm{d} x} \right ) = -\frac{1}{2} \frac{\mathrm{d} y}{\mathrm{d} x} \\
\ln{\frac{\mathrm{d}y}{\mathrm{d}x}} = \frac{-x}{2}+c \\
\frac{\mathrm{d}y}{\mathrm{d}x} = c_{1}e^{-\frac{x}{2}}
\\ y=c_{1}e^{-\frac{x}{2}}+c_{2}
$$