The Differential Operator D
Chapter One Hundred Sixteen
Syllabus topic Module 2, "2.3 Linear Differential Equations with Constant Coefficients"
Pages 259 to 260 of 303
In one line
Write D for d/dx, and a differential equation becomes a polynomial in D acting on y.
MU's label
"The Differential Operator". Her later labels are written in this notation, so it has to come first.
The definition
D is the operation of differentiating with respect to x.
D y = dy/dx
Applying it twice is written D squared.
D^2 y = d2y/dx2
And in general D to the n means differentiating n times.
Why bother
Because an equation written in D can be treated as algebra.
Take y'' minus 5y' plus 6y = 0. In operator notation it is:
(D^2 - 5D + 6) y = 0
And the bracket factorises, exactly as an ordinary quadratic does.
(D - 2)(D - 3) = D^2 - 5D + 6
That factorisation is not a formal trick: applying (D minus 2) then (D minus 3) to a function really does give the same result as applying D squared minus 5D plus 6. The next chapter checks that, and it is what the whole method depends on.
The notation f(D)
A polynomial in D is written f(D), so the general linear equation with constant coefficients is:
f(D) y = X
MU writes her labels exactly that way: "Linear Differential Equation f(D) y = 0" and "Linear differential equation f(D) y = X".
Worked translations
| Equation | In operator form |
|---|---|
| y'' + 3y' + 2y = 0 | (D^2 + 3D + 2)y = 0 |
| y''' - y = e^x | (D^3 - 1)y = e^x |
| y'' + 4y = sin 2x | (D^2 + 4)y = sin 2x |
| y'' - 2y' + y = x^2 | (D - 1)^2 y = x^2 |
The last row shows a factorised f(D), which is how it will usually be written once the auxiliary equation has been solved.
What D does to the standard functions
These are the facts the particular-integral rules are built from, and they are worth having in front of you.
diff(e^(a x), x) = a e^(a x)
diff(sin(a x), x) = a cos(a x)
diff(cos(a x), x) = -a sin(a x)
diff(x^n, x) = n x^(n - 1)
So D acting on an exponential multiplies it by a, which is the single most useful fact in the section: it means an exponential is left in its own shape, only scaled, and that is why f(D) acting on e to the ax gives f(a) times e to the ax.
diff(diff(e^(3x), x), x) = 9 e^(3x)
Nine is three squared, and in general D to the n acting on e to the ax gives a to the n times it.
For a sine or a cosine, D twice multiplies by minus a squared.
diff(diff(sin(a x), x), x) = -a^2 sin(a x)
The Differential Operator D
So D squared acting on a sine or a cosine of ax is the same as multiplying by minus a squared, which is the second most useful fact and the basis of the trigonometric rule for particular integrals.
The inverse operator, in advance
One over f(D) is defined as the operation that undoes f(D): if f(D)y equals X then y equals one over f(D) acting on X.
That is exactly how integration relates to differentiation, and one over D is integration.
(1/D) X = integrate(X, x)
The chapter on the inverse operator makes this precise. It is mentioned here because MU's label "The inverse operator 1/f(D)" is what the last eight chapters of this section are about.
A warning in advance
D behaves like an algebraic symbol in almost every respect, and the next chapter lists exactly which. But there is one thing it does not do, and every mistake made with operator methods comes from forgetting it.
D does not commute with a function of x. Applying D and then multiplying by x is not the same as multiplying by x and then applying D, because of the product rule.
diff(x f(x), x) = f(x) + x diff(f(x), x)
The extra f(x) is exactly the failure to commute. The next chapter treats it properly.
Check yourself
Write each in operator form.
| Equation | Operator form |
|---|---|
| y'' - 4y = 0 | (D^2 - 4)y = 0 |
| y'' + 2y' + y = e^x | (D + 1)^2 y = e^x |
| y''' + y' = 0 | D(D^2 + 1)y = 0 |
| y'' - 6y' + 9y = x | (D - 3)^2 y = x |
And check these facts about D.
diff(e^(5x), x) = 5 e^(5x)
diff(diff(cos(3x), x), x) = -9 cos(3x)
diff(x^4, x) = 4 x^3
diff(diff(e^(-2x), x), x) = 4 e^(-2x)
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.