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The D operator

Solving Differential Equations using the D operator

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Contents

Overview
Simple Equivalents
The D Operator And The Fundamental Laws Of Algebra
The Use Of The D Operator To Find The Complementary Function For Linear Equations.
Three Useful Formulae Based On The Operator D
Linear First Order D Equations With Constant Coefficients.
Linear Second Order D Equations With Constant Coefficients.
Page Comments

Overview

When a particular integral is not obvious by inspection it is often convenient to use methods involving the D operator. We define $\displaystyle D\equiv \frac{d}{dx}$ so that :-

$D(x^2\;+\;3x)\;=\;2x\;+\;3$

(1)

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

Simple Equivalents

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

The D Operator And The Fundamental Laws Of Algebra

The following differential equation:-

$2\,\frac{d^2y}{dx^2}\;+\;5\,\frac{dy}{dx}\;+\;2\,y\;=\;0$

(2)

may be expressed as:-

$\left(2\,D^2+5\,D+2y \right) y=0$

(3)

$2\,D^2\,y+5\,D\,y+2y=0$

(4)

This can clearly be factorised to give:-

$(2D+1)(D+2)\;=\;0$

(5)

But is it justifiable to treat D in this way?

Algebraic procedures depend upon three laws.

The Distributive Law: $\displaystyle m(a + b) = ma + mb$
The Commutative Law: $\displaystyle a b = b a$
The Index Law: $\displaystyle a^{m}\times a^{n}\;=\;a^{(m\,+\,n)}$

If D satisfies these Laws, then it can be used as an Algebraic operator. However:-

$D(u\;+\;v)=Du+Dv$
$D^m(D^n\,u)=D^{(m+n)}\;u$
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. However it does mean that it is permissible to factorise equation (4) to give (5).

In the following analysis we will write

$F(D)\;\equiv \;p_0D^n\;+\;p_1D^{n\,-\,1}\;+\;....p_{n\,-\,1}D\;+\;p_n$

(6)

The "p's" are constants and "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:-

$\displaystyle \frac{d^2y}{dx^2}\;-\;3\;\frac{dy}{ax}\;+\;2\,y\;=\;0$

This can be re-written in terms of the D operator as:-

$D^2-3D\;+\;2y=\;0\;=(D-1)(D-2)y$

(7)

Let $\displaystyle (D-2)y=u$ Then $\displaystyle [(D-1)u=0$

$\therefore\;\;\;\;u=A\,e^{x}$

(8)

$\therefore\;\;\;\;(D-2)\;y=A\,e^{x}$

(9)

or $\displaystyle \frac{dy}{dx}\;-\;2\,y\;=\;A\,e^{x}$

Integrating using $e^{-x}$ as the factor

$y\,e^{-2x}=-\,A\,e^{-X}+B$

(10)

$\mathbf{y=\B\,e^{2X}-A\,e^{X}}$

(11)

Example 1:

Solve the following equation:-

$\frac{d^2y}{dx^2}\;-\;2\,\frac{dy}{dx}\;+\;y\;=\;0$

(12)

Using the D operator this can be written as:-

$(D^2\;-\;2D\;+\;1)\,y\;=\;0$

(13)

$Or\;\;\;\;(D\;-\;1)^2\,y\;=\;0$

(14)

$Let\;\;\;\;(D\;-\;1)\,y\;=\;u$

(15)

$Then\;\;\;\;(D\;-\;1)\,u\;=\;0$

(16)

$\therefore\;\;\;\;u\;=\;A\,e^{x}$

(17)

$\therefore\;\;\;\;(D\;-\;1)\,y\;=\;A\,e^{x}$

(18)

$\;\;\;\;\frac{dy}{dx}\;-\,y\;=\;A\,e^{x}$

(19)

Integrating using $e^{-\,x}$ as the factor

$y\,e^{-x}\;=\;Ax\;+\;B$

(20)

$\mathbf{\therefore\;\;\;\;y\;=\;(Ax\;+\;B)\,e^{x}}$

(21)

Three Useful Formulae Based On The Operator D

Equation A

Let F(D) represent a polynomial function

$\mathbf{F(D)\;e^{ax}\;=\;e^{ax}\;F\;(a)}$

(22)

$Since\;\;\;\;\;\;\; D\;e^{ax}\;=\;a\;e^{ax}$

(23)

$and\;\;\;\;\;\;\;D^2\;e^{ax}\;=\;a^2\;e^{ax}$

(24)

From which it can be seen that:-

$F(D)\;e^{ax}\;= \;\left( p_0D^n\;+\;p_1D^{n\,-\,1}\;+\;....p_{n\,-\,1}D\;+\;p_n \right)e^{ax}$

(25)

$= \;\left( p_0a^n\;+\;p_1a^{n\,-\,1}\;+\;....p_{n\,-\,1}a\;+\;p_n \right)e^{ax}$

(26)

$e^{ax}\,F\;(a)$

(27)

Example 2:

$\frac{d^2y}{dx^2}\;-\;5\,\frac{dy}{dx}\;+\;6y\;=\;e^{4x}$

(28)

This can be re-written as:-

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

(29)

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

(30)

Using equation (7) we can put D = 4

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

(31)

Equation B

$\mathbf{F(D)\left<e^{ax}V \right>\;=\;e^{ax}F(D\;+\;a)V}$

(32)

Where V is any function of x

Applying Leibniz's theorem for the

differential coefficient of a product.

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

(33)

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

(34)

$=\;e^{ax}(a^n\;+\;na^{n-1}D\;+\;\frac{1}{2}n(n-1)a^{n-2}D^2\;+\;.....+\:D^n)V$

(35)

$=\;e^{ax}\;(D\;+\;a)^n\,V$

(36)

Similarly ${\displaystyle D^{n-1}\left<e^{ax}V\right>=\;e^{ax}\;(D\;+\;)^{n-1}\,V$ and so on

$\therefore\;\;\;\;\;\;F(D)\left<e^{ax}V \right>\;=\:\left(p_0D^n\;+\;p_1D^{n-1}\;+\;.........+\;p_{n-1}D\;+\;D \right)\left<e^{ax}V \right>$

(37)

$=\;e^{ax}\left<p_0(D+a)^n\;+\;p_1(D+a)^{n-1}\;+\;.........+\;p_{n-1}(D+a)\;+\;p_n \right>V$

(38)

$\therefore\;\;\;\;\;\;F(D)\left<e^{ax}V \right>\;=\;e^{ax}\;F\;(D+a)\;V$

(39)

Example 3:

Find the Particular Integral of:-

$(D^2\;-\;5D\;+\;6)\,y\;=\;x^2$

(40)

$\therefore\;\;\;\;y\;=\;\frac{1}{(D^2\;-\;5D\;+\;6)}\;x^2\;=\;\left(\frac{1}{(2\;-\;D)}\;-\;\frac{1}{(3\;-\;D)} \right)\,x^2$

(41)

$=\;\frac{1}{2}(1+\frac{1}{2}D+\frac{1}{4}D^2+\frac{1}{8}D^3+.....)x^2\;-\;\frac{1}{3}(1+\frac{1}{3}D+\frac{1}{9}D^2+\frac{1}{27}D^3+...)x^2$

(42)

$=\;\frac{1}{6}x^2\;+\;\frac{5}{18}x\;+\;\frac{19}{108}}$

(43)

Note: If you are having problems with the move from equation (28) to (36) the following should help.

Click here to expand this hidden section

Equation C - Trigonometrical functions

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

(44)

$D^2\;cos \,ax\[=\;-\,a^2\;cos\,ax}$

(45)

$D^4\;cos \,ax\[=\;(-\,a^2)^2\;cos\,ax}$

(46)

And so on

$F(D^2)\;cos\,ax\;=\;\left(p_0D^n\;+\;p_1D^{n-1}\;+\;.......+p_{n-1}D\;+\;p_n \right)cos\;ax$

(47)

$=\;\left<p_0(-\,a^2)^n\;+\;p_1(-\,a^2)^{n-1}\;+\;........+p_{n-1}(\,-\,a^2)\;+\;p_n \right>cos\;ax$

(48)

$\therefore\;\;\;\;\;\;F(D^2)\;cos\;ax\;=\;F(-\,a^2)\;cos\;ax$

(49)

similarly

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

(50)

Example 4:

Find the Particular Integral of:-

$\frac{d^2y}{dx^2}\;-\;5\,\frac{dy}{dx}\;+\;6y\;=\;sin\,2x$

(51)

This can be re-written as:-

$y\;=\;\frac{1}{D^2\;-\;5D\;+\;6}\;sin\,2x$

(52)

Using equation 37 we can put $D^2\;=\;-\,4$

$\therefore\;\;\;\;y\;=\;\frac{1}{-\,4\;-\,5D\;+\;6}sin\,2x$

(53)

$=\frac{1}{2\;-\;5D}\;sin\,2x$

(54)

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

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

(55)

But

$\therefore\;\;\;\;y\;=\frac{2\;+\;5D}{104}\;sin\,2x\;=\;\frac{1}{104}\left(2\;sin\,2x\;+\;5D\;sin\,2x \right)$

(56)

But since $D\;sin\,2x\;=\;2\;cos\,2x$

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

(57)

Linear First Order D Equations With Constant Coefficients.

(These equations have "0" on the right hand side)

$(D\;-\;\alpha )\;=\;0$

(58)

This equation is

$\frac{dy}{dx}\;-\;\alpha \,y\;=\;0$

(59)

Using an Integrating Factor of $\displaystyle e^{-\alpha x}$ the equation becomes:-

$\frac{d}{dx}\left(y\,e^{-\alpha x} \right)\;=\;0$

(60)

$\therefore\;\;\;\;y\,e^{-\alpha x} \;=\;C$

(61)

$\mathbf{Thus\;\;\;\;y\;=\;C\,e^{\alpha x}}$

(62)

Which is the General Solution.

Linear Second Order D Equations With Constant Coefficients.

$p_0\,\frac{d^2y}{dx^2}\;+\;p_1\,\frac{dy}{dx}\;+\;p_2\,y\;=\;0\;\;\;\;\;where\;\;\;\;p_0\neq 0$

(63)

$or\;\;\;\;\left(p_0\,D^2\;+\;p_1\,D\;+\;p_2 \right)\,y\;=\;0$

(64)

$i.e.\;\;\;\;\;p_0(D\;-\;\alpha )(D\;-\;\beta )\,y\;=\;0$

(65)

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

$p_0\,m^2\;+\;p_1\,m\;+\;p_2\;=\;0$

(66)

$(D\;-\;\alpha)[(D\;-\;\beta)]\,y\;=\;0$

(67)

$\therefore\;\;\;\;(D\;-\;\beta)\,y\;=\;C\,e^{\alpha x}$

(68)

Where C is an arbitrary Constant as in equation (34)

$\therefore\;\;\;\;\frac{dy}{dx}\;-\;\beta\,y\;=\;C\,e^{\alpha x}$

(69)

This equation can be re-written as:-

$\frac{d}{dx}(y\,e^{-\beta x})\;=\;C\,e^{\alpha\,x}\times e^{-\beta\,x}\;=\;C\,e^{(\alpha\;-\;\beta)}$

(70)

Integrating

$y\,e^{-\beta\,x}\;=\;\frac{C\,e^{(\alpha\,-\,\beta)}}{\alpha\,-\,\beta}\;+\;K$

(71)

$\therefore\;\;\;\;y\;=\;\frac{C}{\alpha\;-\;\beta}\;e^{\alpha\,x}\;+\;K\,e^{\beta\,x}$

(72)

Thus when $\displaystyle \alpha\neq \beta$ we can write the General Solution as:-

$\mathbf{y\;=\;A\,e^{\alpha\,x}\;+\;B\,e^{\beta\,x}}$

(73)

Where A and B are arbitrary Constants.

Example 5:

$2\,\frac{d^2y}{dx^2}\;+\;5\frac{dy}{dx}\;+\;2y\;=\;0$

(74)

$Or\;\;\;\;(2D^2\;+\;5D\;+\;2)\,y\;=\;0$

(75)

$\therefore\;\;\;\;(2D\;+\;1)(D\;+\;2)\,y\;=\;0$

(76)

$y\;=\;A\,e^{-\frac{1}{2}x}\;+\;B\,e^{-2x}$

(77)

Example 6:

$(D^2\;+\;3D\;+\;1)y\;=\;0$

(78)

The roots of this equation are:-

$\frac{-\,3\;\pm \sqrt{9\;-\;4}}{2}\;=\;\frac{-\,3\;\pm \sqrt{5}}{2}$

(79)

Therefore the General Solution is

$\;y\;+\;A\;e^{\frac{-3\;+\;\sqrt{5}}{2}x}\;+\;B\;e^{\frac{-3\,-\;\sqrt{5}}{2}x$

(80)

The Special Case where $\displaystyle \alpha\;=\;\beta$

From Equation (41)

$\frac{d}{dx}(y\,e^{-\alpha\,x})\;=\;C$

(81)

$\therefore\;\;\;\;y\,e^{- \alpha\,x}\;=\;Cx\;+\;K$

(82)

$\mathbf{y\;=\;(Cx\;+\;K)\,e^{\alpha\,x}}$

(83)

Example 7:

$9\;\frac{d^2y}{dx^2}\;-\;6\,\frac{dy}{dx}\;+\;1\;=\;0$

(84)

$Or\;\;\;(9D^2\;-\;6D\;+\;1)\;y\;=\;0$

(85)

$\therefore\;\;\;\;(3D\;-\;1)(3D\;-\;1)\;=\;0$

(86)

$\therefore\;\;\;\;y\;=\;(A\,x\;+\;B)\;e^{\frac{1}{3}x}$

(87)

The roots of the Auxiliary Equation are complex.

If the roots of the are complex then the General Solution will be of the form $\displaystyle p\;\pm j\,q$ , and the solution will be given by:-

$\mathbf{y\;=\;A\;e^{(p\;+\;jq)x}\;+\;B\;e^{(p\;-\;jq)x}}$

(88)

Example 8:

$\frac{d^2y}{dx^2}\;+\;4\,\frac{dy}{dx}\;+\;8\,y\;=\;0$

(89)

$(D^2\;+\;4D\;+\;8)\,y\;=\;0$

(90)

The roots of this equation are :-

$-\frac{-\,4\;\pm \sqrt{16\;-\;20}}{2}=2\;\pm \sqrt{-\,1}$

(91)

$\therefore\;\;\;\;y\;=\;A\;e^{(-2\,+\;j)x}+B\,e^{(-2\;-\;j)x}$

(92)

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