# Why does factorizing linear differential operators work?

When a 2nd-order differential equation is represented as something like: $$\big(D^2+(a+b)D+ab\big)(y) = \big((D+a)(D+b)\big)y$$.

I don't understand how it can then by solved as: $$Du+au$$ where $$u = (D+b)y$$.

I don't understand why this works because when there is a function like: $$f(x) = \big(a(x)+b(x)\big)c(x) = \big((a+b)c\big)(x)$$, it can't just become $$a(u)+b(u)$$ where $$u = c(x)$$. Therefore, why shouldn't this logic apply to the $$D$$ operator? Does it have something do do with the fundamental difference between operators and functions?

• Welcome to MSE. Please read how to ask a good question and try to edit your post accordingly. Otherwise, you may not find others interested in answering your question, and your question may get closed. Jul 9, 2020 at 9:10
• What is meant by $((a+b)c)(x)$ ?
– user65203
Jul 9, 2020 at 9:16
• I think you are mixing composition of functions in $(D + a)(D + b)$ with pointwise multiplication in an expression like $(a + b)c$. Jul 9, 2020 at 9:20

This works because addition and multiplication by a constant and differentiation commute:

$$D(ay+bz)=aDy+bDz.$$

That makes the $$D$$ operator behave like a polynomial variable.

$$(D+a)y=Dy+ay,$$ $$(D+a)(D+b)y=(D+a)(Dy+by) \\=D(Dy+by)+a(Dy+by) \\=D^2y+Dby+aDy+aby \\=D^2y+bDy+aDy+aby \\=D^2y+(b+a)Dy+aby \\=(D^2+(b+a)D+ab)y.$$

This is not more true when the coefficients are not constant,

$$D(a(x)y)\ne a(x)Dy.$$

An operator is a function as well. Just like how a real-valued function maps a real number to another, an operator maps a real-valued function to another. The differential operator has the following properties

$$D(af(x)) = aD(f(x))$$ $$D(f(x) + g(x)) = D(f(x)) + D(g(x))$$

This means the Differential operator is linear. It is analogous to a linear function. Hence, just how you have polynomial functions as a product of affine variables, you can break down an operator into factors and solve each individually.