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.
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 $$y(x)=c_{2}e^{-\beta{x}}+c_{1}\frac{e^{-\alpha{x}}}{\beta-\alpha}$$