Q:

Reducing homogeneous second order differential equation to first order (Operator factorisation)

I need to reduce the homogeneous second-order differential equation  
$\ y'' + by' + cy = 0$
to a first-order one using operator factorisation, where$\ b, c$ and$\ y$ are functions of t.  
I began by rewriting it in operator form and completing the square, getting
$\ [(D + \dfrac{b}{2})^2 + (c - \dfrac{b^2}{4})]y = 0$. 
I'm basically stumped from here. I could try applying$\ D^2$ to both sides to get zero on the right-hand side and then substitute, but then I run into more dead-ends that give me no hints to either proceed or point to a different approach (at least in my mind). 
Can someone help out with suggestions? Thank you in advance.

A:

Write 
$$(D+\alpha)(D+\beta)y=0$$
This gives
$$D^{2}y+(\alpha+\beta)Dy+\alpha\beta{y}=0$$
Comparing with your equation, you get 
$$\alpha+\beta=b$$
and 
$$\alpha\beta=c$$
Then you let $g(x)=(D+\beta)y(x)$ and you are left with a first order system
$$g'+\alpha{g}=0, \ y'+\beta{y}=g$$
The first equation is solved by
$$g(x)=c_{1}e^{-\alpha{x}}$$
The second equation is solved by the integrating factor technique 
$$y(x)=c_{2}e^{-\beta{x}}+c_{1}\frac{e^{-\alpha{x}}}{\beta-\alpha}$$

