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.

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    $\begingroup$ Do you mean to factor is into first order operators whose composite acts on $y$? Or, do you mean to use reduction of order to replace the one second order ODE with a system of first order ODEs? $\endgroup$ – James S. Cook Jul 2 '17 at 23:35
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    $\begingroup$ Also, if $b$ and $c$ are functions of $t$ I think your algebra fails. $\endgroup$ – James S. Cook Jul 2 '17 at 23:36
  • $\begingroup$ Yeah, I was trying to factor into first-order operators. $\endgroup$ – Kimonaize Jul 3 '17 at 6:03
  • $\begingroup$ Well, I suppose you could find it by taking the solutions and working backwards. But, given nonconstant coefficients, the operators do not commute and to solve it we use substitution... $\endgroup$ – James S. Cook Jul 3 '17 at 14:09

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


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