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The D operator

Solving Differential Equations using the D operator
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Theory Of Differential Operator (differential Module)


A differential operator is an operator defined as a function of the differentiation operator.

It is helpful, as a matter of notation first, to consider differentiation as an abstract operation, accepting a function and returning another (in the style of a higher-order function in computer science).

The most commonly used differential operator is the action of taking the derivative itself. Common notations for this operator include:

\inline \displaystyle D\equiv\frac{d}{dx} and if generalize \inline \displaystyle D^n\equiv\frac{d^n}{dx^n}

\inline D is an operator and must therefore always be followed by some expression on which it operates.

Simple Equivalents

  • \inline Du means \inline \displaystyle Du\equiv \frac{du}{dx} but \inline uD\equiv u\frac{d}{dx}
  • \inline \displaystyle D^2y\equiv D\times Dy\equiv \frac{d}{dx}\left(\frac{dy}{dx} \right) = \frac{d^2y}{dx^2}
  • Similarly \inline \displaystyle D^2\equiv \frac{d^2}{dx^2} and \inline D^3\equiv \frac{d^3}{dx^3}

The D Operator And The Fundamental Laws Of Algebra

The following differential equation:
2\,\frac{d^2y}{dx^2} + 5\,\frac{dy}{dx} + 2\,y = 0

may be expressed as: \inline \left(2\,D^2+5\,D+y \right) y=0 or \inline 2\,D^2+5\,D+2=0

This can be factorised to give:
(2D+1)(D+2) = 0
  • D(x^2+2x)=2x+2
  • D(e^{\alpha x})=\alpha e^{\alpha x}
  • D(ln(x))=\frac{1}{x}
  • D(\sqrt{x})=\frac{1}{2\sqrt{x}}
  • D(sin(x))=cos(x)

But is it justifiable to treat D in this way?

Algebraic procedures depend upon three laws.
  • The Distributive Law: \inline \displaystyle m(a + b) = ma + mb
  • The Commutative Law: \inline \displaystyle a b = b a
  • The Index Law: \inline \displaystyle a^{m}\times a^{n} = a^{(m\,+\,n)}

If D satisfies these Laws, then it can be used as an Algebraic operator(or a linear operator). However:
  • \inline D(u + v)=Du+Dv
  • \inline D^m(D^n\,u)=D^{(m+n)}\;u
  • \inline D(uv) = u Dv only when u is a constant.

Thus we can see that D does satisfy the Laws of Algebra very nearly except that it is not interchangeable with variables.

In the following analysis we will write
F(D)\;\equiv \;p_0D^n + p_1D^{n\,-\,1} + ....p_{n\,-\,1}D + p_n

\inline p_i are constants and \inline n is a positive integer. As has been seen, we can factorise this or perform any operation depending upon the fundamental laws of Algebra.

We can now apply this principle to a number of applications.

The Use Of The D Operator To Find The Complementary Function For Linear Equations

It is required to solve the following equations:

Example - Simple example
Solve the following equation:-

Using the D operator this can be written as:-

Integrating using as the factor

Three Useful Formulae Based On The Operator D

Equation A

Let \inline F(D) represent a polynomial function

\mathbf{F(D)\;e^{ax} = e^{ax}\;F\;(a)}
D\;e^{ax} = a\;e^{ax}
D^2\;e^{ax} = a^2\;e^{ax}
From which it can be seen that:
F(D)\;e^{ax}\;= \;\left( p_0D^n + p_1D^{n\,-\,1} + ....p_{n\,-\,1}D + p_n \right)e^{ax}
= \;\left( p_0a^n + p_1a^{n\,-\,1} + ....p_{n\,-\,1}a + p_n \right)e^{ax}

Example - Equation A example
This can be re-written as:

We can put D = 4

Equation B

\mathbf{F(D)\left<e^{ax}V \right> = e^{ax}F(D + a)V}
Where \inline V is any function of x

Applying Leibniz's theorem for the \inline n{th} differential coefficient of a product.

D^n\left<e^{ax}V \right> = (D^ne^{ax})V + n(D^{n-1}e^{ax})(DV) + \frac{1}{2}n(n-1)(D^{n-2}e^{ax})(D^2V) + .....e^{ax}(D^nV)
= a^ne^{ax}V + na^{n-1}e^{ax}DV + \frac{1}{2}n(n-1)a^{n-2}e^{ax}D^2V + .....e^{ax}D^nV
= e^{ax}(a^n + na^{n-1}D + \frac{1}{2}n(n-1)a^{n-2}D^2 + .....+\:D^n)V
= e^{ax}\;(D + a)^n\,V

Similarly \inline {\displaystyle D^{n-1}\left<e^{ax}V\right>= e^{ax}\;(D + )^{n-1}\,V and so on
F(D)\left<e^{ax}V \right>\;=\:\left(p_0D^n + p_1D^{n-1} + .........+\;p_{n-1}D + D \right)\left<e^{ax}V \right>
= e^{ax}\left<p_0(D+a)^n + p_1(D+a)^{n-1} + .........+\;p_{n-1}(D+a) + p_n \right>V
F(D)\left<e^{ax}V \right> = e^{ax}\;F\;(D+a)\;V

Example - Equation B example
Find the Particular Integral of:

We have used D as if it were an algebraic constant but it is in fact an operator where

Equation C - Trigonometrical Functions

\mathbf{F(D^2)\;cos \,ax= F(-\,a^2)\;cos\,ax}
D^2\;cos \,ax= -\,a^2\;cos\,ax}
D^4\;cos \,ax= (-\,a^2)^2\;cos\,ax}

And so on
F(D^2)\;cos\,ax = \left(p_0D^n + p_1D^{n-1} + .......+p_{n-1}D + p_n \right)cos\;ax
= \left<p_0(-\,a^2)^n + p_1(-\,a^2)^{n-1} + ........+p_{n-1}(\,-\,a^2) + p_n \right>cos\;ax
F(D^2)\;cos\;ax = F(-\,a^2)\;cos\;ax


\mathbf{\therefore\;\;\;\;\;\;F(D^2)\;sin\;ax = F(-\,a^2)\;sin\;ax}

Example - Trigonometric example
Find the Particular Integral of:-
This can be re-written as:-

Using equation 1 we can put

If we multiply the top and bottom of this equation by


But since

Linear First Order D Equations With Constant Coefficients

These equations have \inline 0 on the right hand side

(D - \alpha )y = 0

This equation is
\frac{dy}{dx} - \alpha \,y = 0

Using an Integrating Factor of \inline \displaystyle e^{-\alpha x} the equation becomes:-
\frac{d}{dx}\left(y\,e^{-\alpha x} \right) = 0
\therefore\;\;\;\;y\,e^{-\alpha x}  = C
\mathbf{Thus\;\;\;\;y = C\,e^{\alpha x}}
Which is the General Solution.

Linear Second Order D Equations With Constant Coefficients

p_0\,\frac{d^2y}{dx^2} + p_1\,\frac{dy}{dx} + p_2\,y = 0\;\;\;\;\;where\;\;\;\;p_0\neq 0
or\;\;\;\;\left(p_0\,D^2 + p_1\,D + p_2 \right)\,y = 0
i.e.\;\;\;\;\;p_0(D - \alpha )(D - \beta )\,y = 0

Where \inline  \apha\;and\;\beta are the roots of the quadratic equation. i.e. the auxiliary equation.

p_0\,m^2 + p_1\,m + p_2 = 0
(D - \alpha)[(D - \beta)]\,y = 0
\therefore\;\;\;\;(D - \beta)\,y = C\,e^{\alpha x}

Where \inline C is an arbitrary Constant
\therefore\;\;\;\;\frac{dy}{dx} - \beta\,y = C\,e^{\alpha x}

This equation can be re-written as:-
\frac{d}{dx}(y\,e^{-\beta x}) = C\,e^{\alpha\,x}\times e^{-\beta\,x} = C\,e^{(\alpha - \beta)}

y\,e^{-\beta\,x} = \frac{C\,e^{(\alpha\,-\,\beta)}}{\alpha\,-\,\beta} + K
\therefore\;\;\;\;y = \frac{C}{\alpha - \beta}\;e^{\alpha\,x} + K\,e^{\beta\,x}
  • Thus when \inline \displaystyle \alpha\neq \beta we can write the General Solution as:-
\mathbf{y = A\,e^{\alpha\,x} + B\,e^{\beta\,x}}

Where A and B are arbitrary Constants.
Example - Linear second order example


The roots of this equation are:-

Therefore the General Solution is

  • The Special Case where

From Equation (41)

  • The roots of the Auxiliary Equation are complex.

If the roots of the are complex then the General Solution will be of the form , and the solution will be given by:-

The roots of this equation are :-

Physical Examples

Example - Small oscilations
Show that if satisfies the differential equation with k < n and if when

The complete period of small oscillations of a simple pendulum is 2 secs. and the angular retardation due to air resistance is 0.04 X the angular velocity of the pendulum. The bob is held at rest so the the string makes a small angle with the downwards vertical and then let go. Show that after 10 complete oscillations the string will make an angle of about 40' with the vertical.(LU)


Using the "D" operator we can write

When t = 0 = 0 and = 0
At t = 0

We have been given that k = 0.02 and the time for ten oscillations is 20 secs.