I am a math tutor, and had a student come to me with the ordinary differential equation
$$t^2y''+3ty'+y=0$$
The problem asked them to rewrite this in terms of the differential operator $D$, the identity operator $I$, and functions of $t$, and then factor the resulting second order operator into two linear operators. One of the factors was given as $(D+2t^{-1}I)$.
I can certainly write this as a second order operator, as $$(t^2D^2 +3tD+I)[y]$$ with square brackets designating what the resulting operator is operating on.
I also know how to solve the differential equation once I have the other operator, as two first order differential equations, but I never learned this method or anything similar, so the question is just how to "factor" that operator? A more general explanation alongside this particular one would also be appreciated.
The exceptionally tricky thing about operators is that they are not commutative: the operator $(tD)$ is totally different from the operator $(Dt).$ If you are given one linear factor as $(D+2t^{-1}I),$ it matters whether you consider this the left factor or the right factor. The end result is, I think, relatively clear: we want a linear factor such that the product of it and the given operator yields the total operator. Now operators are associative, so you'll see a sloppiness with my parentheses in the calculations below that is justified by associativity. So let's try an example, and see where it leads. Perhaps we can figure it out by trial and error.
Suppose we try $(D+2t^{-1}I)(t^2D+I)[y]=0.$ We certainly need the $t^2$ out front, but let's see where the computation leads us. Note that it's usually safer to use a test function (I'll just use $y$) to make sure the non-commutativity is correct. One other note: to see what we really have, we have to get each derivative operator $D$ on the far right of its expression, because that's the way the original DE reads. \begin{align*} (D+2t^{-1}I)(t^2D+I)[y] &=(D+2t^{-1}I)(t^2Dy+y)\\ &=D(t^2Dy)+Dy+2t^{-1}t^2Dy+2t^{-1}y\\ &=2tDy+t^2D^2y+Dy+2tDy+2t^{-1}y\\ &=t^2Dy+(2tD+D+2tD)y+2t^{-1}y. \end{align*} This is certainly not going to give us what we want. But notice one thing: if we think of the given operator as the left operator, and we use $t^2D$ on the right, we'll get $2tD$, which is part of the $3tD$ we want. Also note that multiplying the plain $y$ was $2t^{-1},$ which we don't want. We can correct that by trying a right operator of something like $t^2D+nt,$ where we need to determine $n.$ Let's work this out: \begin{align*} (D+2t^{-1}I)(t^2D+nt)y&= (D+2t^{-1}I)(t^2Dy+nty)\\ &=D(t^2Dy)+nD(ty)+2t^{-1}t^2Dy+2ny\\ &=2tDy+t^2D^2y+n(y+tDy)+2tDy+2ny\\ &=t^2D^2y+2tDy+ntDy+2tDy+3ny\\ &=t^2D^2y+(4+n)tDy+3ny. \end{align*} Here we'd need $4+n=3$ and $3n=1.$ This is not solvable, but you can see that we're much closer. Let's try the given factor as a right operator, and see where that lands: \begin{align*} (t^2D+nt)(D+2t^{-1}I)y&= (t^2D+nt)(Dy+2t^{-1}y)\\ &=t^2D^2y+2t^2D(t^{-1}y)+ntDy+2ny\\ &=t^2D^2y+2t^2(-t^{-2}y+t^{-1}Dy)+ntDy+2ny\\ &=t^2D^2y-2y+2tDy+ntDy+2ny\\ &=t^2D^2y+(2+n)tDy+2(n-1)y. \end{align*} So we would need $2+n=3$ and $2(n-1)=1.$ This is also not solvable!
I'm going to ditch the given operator, and start from scratch. Let's try something a bit more symmetrical: $(atD+bI)(ctD+dI)[y]=0.$ We have \begin{align*} (atD+bI)(ctDy+dy) &=atD(ctDy)+atD(dy)+bctDy+bdy\\ &=act(Dy+tD^2y)+adtDy+bctDy+bdy\\ &=act^2D^2y+(ac+ad+bc)tDy+bdy. \end{align*} From here, we need \begin{align*} ac&=1\\ ac+ad+bc&=3\\ bd&=1. \end{align*} The simpler, the better. Try $a=b=c=d=1,$ and it works! So, evidently, $$t^2D+3tD+I=(tD+I)(tD+I)=(tD+I)^2. $$