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# The D operator

Solving Differential Equations using the D operator

## Theory Of Differential Operator (differential Module)

### Definition

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:

$\inline&space;\displaystyle&space;D\equiv\frac{d}{dx}$ and if generalize $\inline&space;\displaystyle&space;D^n\equiv\frac{d^n}{dx^n}$

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

### Simple Equivalents

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

## The D Operator And The Fundamental Laws Of Algebra

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

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

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

But is it justifiable to treat D in this way?

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

If D satisfies these Laws, then it can be used as an Algebraic operator(or a linear operator). However:
• $\inline&space;D(u&space;+&space;v)=Du+Dv$
• $\inline&space;D^m(D^n\,u)=D^{(m+n)}\;u$
• $\inline&space;D(uv)&space;=&space;u&space;Dv$ 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
$F(D)\;\equiv&space;\;p_0D^n&space;+&space;p_1D^{n\,-\,1}&space;+&space;....p_{n\,-\,1}D&space;+&space;p_n$

$\inline&space;p_i$ are constants and $\inline&space;n$ 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:
##### Example - Simple example
Problem
Solve the following equation:-

$\frac{d^2y}{dx^2}&space;-&space;2\,\frac{dy}{dx}&space;+&space;y&space;=&space;0$
Workings
Using the D operator this can be written as:-

$(D^2&space;-&space;2D&space;+&space;1)\,y&space;=&space;0$
$Or\;\;\;\;(D&space;-&space;1)^2\,y&space;=&space;0$
$Let\;\;\;\;(D&space;-&space;1)\,y&space;=&space;u$
$Then\;\;\;\;(D&space;-&space;1)\,u&space;=&space;0$
$\therefore\;\;\;\;u&space;=&space;A\,e^{x}$
$\therefore\;\;\;\;(D&space;-&space;1)\,y&space;=&space;A\,e^{x}$
$\;\;\;\;\frac{dy}{dx}\;-\,y&space;=&space;A\,e^{x}$
Solution
Integrating using $\inline&space;&space;e^{-\,x}$ as the factor
$y\,e^{-x}&space;=&space;Ax&space;+&space;B$
$\mathbf{\therefore\;\;\;\;y&space;=&space;(Ax&space;+&space;B)\,e^{x}}$

## Three Useful Formulae Based On The Operator D

### Equation A

Let $\inline&space;F(D)$ represent a polynomial function

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

Example:
##### Example - Equation A example
Problem
$\frac{d^2y}{dx^2}&space;-&space;5\,\frac{dy}{dx}&space;+&space;6y&space;=&space;e^{4x}$
Workings
This can be re-written as:

$(D^2&space;-&space;5D&space;+&space;6)\,y&space;=&space;e^{4x}$

$\therefore\;\;\;\;y&space;=&space;e^{4x}\times\frac{1}{D^2&space;-&space;5D&space;+&space;6}$
Solution
We can put D = 4

$\therefore\;\;\;\;y&space;=&space;e^{4x}\times\frac{1}{4^2&space;-&space;5\times4&space;+&space;6}&space;=&space;\frac{1}{2}\,e^{4x}$

### Equation B

$\mathbf{F(D)\left&space;=&space;e^{ax}F(D&space;+&space;a)V}$
Where $\inline&space;V$ is any function of x

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

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

Similarly $\inline&space;{\displaystyle&space;D^{n-1}\left=&space;e^{ax}\;(D&space;+&space;)^{n-1}\,V$ and so on
$F(D)\left\;=\:\left(p_0D^n&space;+&space;p_1D^{n-1}&space;+&space;.........+\;p_{n-1}D&space;+&space;D&space;\right)\left$
$=&space;e^{ax}\leftV$
therefore
$F(D)\left&space;=&space;e^{ax}\;F\;(D+a)\;V$

Example:
##### Example - Equation B example
Problem
Find the Particular Integral of:
$(D^2&space;-&space;5D&space;+&space;6)\,y&space;=&space;x^2$
Workings
$y&space;=&space;\frac{x^2}{(D^2&space;-&space;5D&space;+&space;6)}=\left(\frac{1}{(2&space;-&space;D)}&space;-&space;\frac{1}{(3&space;-&space;D)}&space;\right)\,x^2$
$=&space;\frac{1}{2}(1+\frac{1}{2}D+\frac{1}{4}D^2+\frac{1}{8}D^3+.....)x^2&space;-&space;\frac{1}{3}(1+\frac{1}{3}D+\frac{1}{9}D^2+\frac{1}{27}D^3+...)x^2$

We have used D as if it were an algebraic constant but it is in fact an operator where $\inline&space;D\,(x^2)&space;=&space;2x\;and\;D^2\,(x^2)&space;=&space;2.$
Solution
$y=&space;\frac{1}{6}x^2&space;+&space;\frac{5}{18}x&space;+&space;\frac{19}{108}}$

### Equation C - Trigonometrical Functions

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

And so on
$F(D^2)\;cos\,ax&space;=&space;\left(p_0D^n&space;+&space;p_1D^{n-1}&space;+&space;.......+p_{n-1}D&space;+&space;p_n&space;\right)cos\;ax$
$=&space;\leftcos\;ax$
Therefore
$F(D^2)\;cos\;ax&space;=&space;F(-\,a^2)\;cos\;ax$

similarly

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

Example:
##### Example - Trigonometric example
Problem
Find the Particular Integral of:-
$\frac{d^2y}{dx^2}&space;-&space;5\,\frac{dy}{dx}&space;+&space;6y&space;=&space;sin\,2x$
Workings
This can be re-written as:-
$y&space;=&space;\frac{1}{D^2&space;-&space;5D&space;+&space;6}\;sin\,2x$

Using equation 1 we can put $\inline&space;&space;D^2&space;=&space;-\,4$
$\therefore\;\;\;\;y&space;=&space;\frac{1}{-\,4\;-\,5D&space;+&space;6}sin\,2x$
$=\frac{1}{2&space;-&space;5D}\;sin\,2x$

If we multiply the top and bottom of this equation by $\inline&space;2&space;+&space;5D$

$=\frac{2&space;+&space;5D}{4&space;-&space;25D^2}\;sin\,2x$

But $\inline&space;D^2;=-4$

$\therefore\;\;\;\;y\;=\frac{2&space;+&space;5D}{104}\;sin\,2x&space;=&space;\frac{1}{104}\left(2\;sin\,2x&space;+&space;5D\;sin\,2x&space;\right)$
Solution
But since $\inline&space;&space;D\;sin\,2x&space;=&space;2\;cos\,2x$

$\mathbf{y&space;=&space;\frac{1}{104}(2\;sin\,2x&space;+&space;10\;cos\,2x)}$

## Linear First Order D Equations With Constant Coefficients

These equations have $\inline&space;0$ on the right hand side

$(D&space;-&space;\alpha&space;)y&space;=&space;0$

This equation is
$\frac{dy}{dx}&space;-&space;\alpha&space;\,y&space;=&space;0$

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

## Linear Second Order D Equations With Constant Coefficients

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

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

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

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

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

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

Where A and B are arbitrary Constants.
Example:
##### Example - Linear second order example
Problem
$2\,\frac{d^2y}{dx^2}&space;+&space;5\frac{dy}{dx}&space;+&space;2y&space;=&space;0$
$Or\;\;\;\;(2D^2&space;+&space;5D&space;+&space;2)\,y&space;=&space;0$
$\therefore\;\;\;\;(2D&space;+&space;1)(D&space;+&space;2)\,y&space;=&space;0$
$y&space;=&space;A\,e^{-\frac{1}{2}x}&space;+&space;B\,e^{-2x}$

Workings
$(D^2&space;+&space;3D&space;+&space;1)y&space;=&space;0$

The roots of this equation are:-
$\frac{-\,3\;\pm&space;\sqrt{9&space;-&space;4}}{2}&space;=&space;\frac{-\,3\;\pm&space;\sqrt{5}}{2}$

Therefore the General Solution is
$\;y&space;+&space;A\;e^{\frac{-3&space;+&space;\sqrt{5}}{2}x}&space;+&space;B\;e^{\frac{-3\,-&space;\sqrt{5}}{2}x$

• The Special Case where $\inline&space;\displaystyle&space;\alpha&space;=&space;\beta$

From Equation (41)
$\frac{d}{dx}(y\,e^{-\alpha\,x})&space;=&space;C$
$\therefore\;\;\;\;y\,e^{-&space;\alpha\,x}&space;=&space;Cx&space;+&space;K$
or
$\mathbf{y&space;=&space;(Cx&space;+&space;K)\,e^{\alpha\,x}}$

$9\;\frac{d^2y}{dx^2}&space;-&space;6\,\frac{dy}{dx}&space;+&space;1&space;=&space;0$
$Or\;\;\;(9D^2&space;-&space;6D&space;+&space;1)\;y&space;=&space;0$
$\therefore\;\;\;\;(3D&space;-&space;1)(3D&space;-&space;1)&space;=&space;0$
$\therefore\;\;\;\;y&space;=&space;(A\,x&space;+&space;B)\;e^{\frac{1}{3}x}$

• The roots of the Auxiliary Equation are complex.

If the roots of the are complex then the General Solution will be of the form $\inline&space;\displaystyle&space;p\;\pm&space;j\,q$, and the solution will be given by:-
$\mathbf{y&space;=&space;A\;e^{(p&space;+&space;jq)x}&space;+&space;B\;e^{(p&space;-&space;jq)x}}$

$\frac{d^2y}{dx^2}&space;+&space;4\,\frac{dy}{dx}&space;+&space;8\,y&space;=&space;0$
$(D^2&space;+&space;4D&space;+&space;8)\,y&space;=&space;0$

Solution
The roots of this equation are :-
$-\frac{-\,4\;\pm&space;\sqrt{16&space;-&space;20}}{2}=2\;\pm&space;\sqrt{-\,1}$
$\therefore\;\;\;\;y&space;=&space;A\;e^{(-2\,+\;j)x}+B\,e^{(-2&space;-&space;j)x}$

## Physical Examples

Example:
##### Example - Small oscilations
Problem
Show that if $\inline&space;theta$ satisfies the differential equation $\inline&space;\displaystyle&space;\frac{&space;d^2\theta}{dt^2}\;+\;2k\;\frac{d\theta&space;}{dt}\;+\;n^2\,\theta&space;\;=\;0$ with k < n and if when $\inline&space;\displaystyle&space;t\;+\;0\;:\;\theta&space;\;+\;\alpha&space;\;and\;\frac{d\theta&space;}{dt}\;=\;0$
$Then\;\;\;\;\theta&space;\;=\;e^{-kt}\left(\alpha&space;\,cos\,pt\;+\;\frac{k\,\alpha&space;}{p}&space;\;sin\,pt\right)$
$where\;\;\;\;p^2\;=\;n^2\;-\;k^2$

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 $\inline&space;\alpha\;=\;1^0$ 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)

Workings
$\frac{d^2\theta&space;}{dt^2}\;+\;2k\;\frac{d\theta&space;}{dt}\;+\;n^2\,\theta&space;\;=\;0$
$\therefore\;\;\;\;\frac{-\;2k\;\pm&space;\;\sqrt{4k^2\;-\;4n^2}}{2}$
$\therefore\;\;\;\;D\;=\;-k\;\pm&space;\sqrt{k^2\;-\;n^2}$

Using the "D" operator we can write
$D^2\;+\;2kD\;+\;n^2\;=\;0$
$=\;-k\;\pm&space;jp\;\;\;\;\;where\;\;\;\;p^2\;=\;k^2\;-\;n^2$
$\theta&space;\;=\;e^{-kt}\left(A\;cos\,pt\;+\;B\;sin\,pt&space;\right)$
$\dot{\theta&space;}\;=\;-\,k\,e^{-kt}\,A\,cos\,pt\;-\;e^{-kt}\;A\;p\,sin\,pt\;-\;ke^{-kt}\;B\,sin\,pt\;+\;e^{-kt}\;B\,pcos\,pt$

When t = 0 $\inline&space;\alpha$ = 0 and $\inline&space;\dot{\theta}$ = 0
$\therefore\;\;\;\;A\;=\;\alpha$
and
$0=-\;k\;\alpha&space;\;+\;Bp\;\;\;\;\therefore\;\;\;\;B\;=\;\frac{k\alpha&space;}{p}$
$\therefore\;\;\;\;\theta&space;\;=\;e^{-kt}\left(\alpha&space;\;cos\,pt\;+\;\frac{k\alpha&space;}{p}\;sin\,pt&space;\right)$
$Periodic\;Time\;=\;2\;secs.\;=\;\frac{2\,\pi&space;}{p}\;\;\;\;\;\therefore\;\;\;\;p\;=\;\pi$
Solution
At t = 0
$\theta&space;\;=\;1^0\;=\;\frac{\pi&space;}{180}\;rds.$

We have been given that k = 0.02 and the time for ten oscillations is 20 secs.
$\therefore\;\;\;\;\theta&space;_{10&space;cycles}\;=\;e^{-0.02\times20}\left(cos\,\pi&space;\times20\;+\;\frac{0.02}{20}\;sin\,\pi&space;\times20}&space;\right)\frac{\pi&space;}{180}\f]&space;\f[=\;e^{-0.4}\times1\times\frac{\pi}{&space;180}\;=\;\frac{1}{1.49182}\;\times\;\frac{\pi&space;}{180}times\;60\approx&space;40\,seconds$