The Inverse Operator 1 Over f(D)
Chapter One Hundred Twenty-Five
Syllabus topic Module 2, "2.3 Linear Differential Equations with Constant Coefficients"
Pages 278 to 279 of 303
In one line
One over f(D) is the operation that undoes f(D), and applying it to X gives a particular integral.
MU's label
"The inverse operator 1/f(D) and the symbolic expression for the particular integral". Her sentence is cut off after the comma in the printed circular, and this book's contract records that; the standard set of rules that follows is the subject of the next eight chapters.
The definition
y = (1/f(D)) X
means: y is a function with f(D)y equal to X.
So one over f(D) acting on X is a particular integral, by definition. The work of the next chapters is finding what it actually equals for each shape of X.
It is not unique, and that is fine
If y is one such function, so is y plus anything the CF contains, because f(D) kills those. So one over f(D) acting on X is determined only up to a CF term.
That is exactly what the complete-solution chapter said: any PI will do, and the CF absorbs the difference. So the inverse operator is allowed to be sloppy about constants of integration, and by convention they are omitted: a constant of integration from one over D would be a CF term anyway.
One over D is integration
The simplest case. If f(D) is D, then y must satisfy Dy = X, so y is the integral of X.
(1/D) X = integrate(X, x)
And no constant, by the convention above.
integrate(x^2, x) = x^3/3
integrate(e^(3x), x) = e^(3x)/3
integrate(sin(2x), x) = -cos(2x)/2
So one over D squared is integrating twice, and one over D to the n is integrating n times.
integrate(integrate(x, x), x) = x^3/6
The rules it obeys
Linearity. One over f(D) acting on a sum is the sum of the results, and constants pass through. So X can be split into its terms and each handled separately, which is what makes the five shape rules useful.
Factorisation. If f(D) factorises, one over f(D) can be applied one factor at a time, in any order.
(1/((D - 2)(D - 3))) X = (1/(D - 2))((1/(D - 3)) X)
That is the basis of the general method of its own chapter, and it is also what partial fractions in D exploits.
Partial fractions in D. Since one over f(D) is a rational function of D, it can be split by partial fractions exactly as a rational function of s was in Module 1.
1/((D - 2)(D - 3)) = -1/(D - 2) + 1/(D - 3)
That identity is about polynomials in D, so it is legitimate, and it turns one hard inverse operator into two easy ones. It is the same algebra as the inverse Laplace transform's partial fractions, on the same kind of object.
The Inverse Operator 1 Over f(D)
What is coming: the five shapes of X
| X | Rule | Chapter |
|---|---|---|
| e^(ax) | replace D by a | the next one |
| sin ax or cos ax | replace D^2 by -a^2 | third from here |
| x^n | expand 1/f(D) as a series in D | fifth |
| e^(ax) V | shift: e^(ax) times 1/f(D + a) applied to V | sixth |
| x V | a term with the derivative of the operator | seventh |
And two failure cases, each with its own chapter, for when the first two rules divide by zero. And a general method for when none of them fits.
The warning that applies to all of them
One over f(D) is an operator, not a fraction. You may not cancel it against something containing x, and you may not move a function of x through it. The D-laws chapter's single prohibition applies in full.
What you may do is what the three rules above permit: split sums, take constants out, factorise, and use partial fractions in D.
Worked, with what is available so far
Solve y'' minus 5y' plus 6y = x, using only the inverse operator and integration.
The PI is one over (D squared minus 5D plus 6) acting on x. Partial fractions in D give minus one over (D minus 2) plus one over (D minus 3) acting on x.
Each of those needs the rule for one over (D minus a), which is an integration and is the general method's content. Rather than anticipate it, here is the answer, and the polynomial rule of its own chapter gets it in two lines.
d2y/dx2 - 5 dy/dx + 6y = x
y = C1 e^(2x) + C2 e^(3x) + x/6 + 5/36
Check yourself
integrate(x^3, x) = x^4/4
integrate(cos(3x), x) = sin(3x)/3
integrate(integrate(e^(2x), x), x) = e^(2x)/4
1/((D - 1)(D + 1)) = 1/(2(D - 1)) - 1/(2(D + 1))
The last line is partial fractions in D, and it is checked as an algebraic identity in the symbol D, which is exactly what it is.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.