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Solving Differential Equations using the D operator

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Theory of Differential Operator (Differential Module)

Definition

Let S be the space of differentiable functions $\f$ .

We call a differential operator $\displaystyle D\equiv \frac{d}{dx}$ .

Let (D,+,*) be the ring of differential operators over S and the operations:

+ : representing the sum of two differentiable functions (we know for sure that the result will also be a differentiable function)
* : representing the multiplication of differentiable functions .

This result is very important because it shows that for each differential equation (which can be represented with the D operator) can be uniquely attached to a polynomial equation !

A simple example on how to use the D-operator

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

(1)

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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

Problem

Solve the following equation:-

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

Workings

Using the D operator this can be written as:-

$(D^2 - 2D + 1)\,y = 0$ $Or\;\;\;\;(D - 1)^2\,y = 0$ $Let\;\;\;\;(D - 1)\,y = u$ $Then\;\;\;\;(D - 1)\,u = 0$ $\therefore\;\;\;\;u = A\,e^{x}$ $\therefore\;\;\;\;(D - 1)\,y = A\,e^{x}$ $\;\;\;\;\frac{dy}{dx}\;-\,y = A\,e^{x}$

Solution

Integrating using $e^{-\,x}$ as the factor $y\,e^{-x} = Ax + B$ $\mathbf{\therefore\;\;\;\;y = (Ax + B)\,e^{x}}$

Three useful formulae based on the Operator D

Equation A

Let F(D) represent a polynomial function

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

(12)

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

(13)

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

(14)

From which it can be seen that:-

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

(15)

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

(16)

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

(17)

Example 1

Problem

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

Workings

This can be re-written as:

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

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

Solution

We can put D = 4

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

Equation B

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

(18)

Where V is any function of x

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

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

(19)

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

(20)

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

(21)

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

(22)

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

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

(23)

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

(24)

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

(25)

Example 1

Problem

Find the Particular Integral of: $(D^2 - 5D + 6)\,y = x^2$

Workings

$y = \frac{x^2}{(D^2 - 5D + 6)}=\left(\frac{1}{(2 - D)} - \frac{1}{(3 - D)} \right)\,x^2$ $= \frac{1}{2}(1+\frac{1}{2}D+\frac{1}{4}D^2+\frac{1}{8}D^3+.....)x^2 - \frac{1}{3}(1+\frac{1}{3}D+\frac{1}{9}D^2+\frac{1}{27}D^3+...)x^2$

We have used D as if it were an algebraic constant but it is in fact an operator where $D\,(x^2) = 2x\;and\;D^2\,(x^2) = 2.$

Solution

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

Equation C - Trigonometrical functions

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

(26)

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

(27)

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

(28)

And so on

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

(29)

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

(30)

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

(31)

similarly

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

(32)

Example 1

Problem

Find the Particular Integral of:- $\frac{d^2y}{dx^2} - 5\,\frac{dy}{dx} + 6y = sin\,2x$

Workings

This can be re-written as:-

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

(33)

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

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

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

But $D^2;=-4$

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

Solution

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

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

Linear First Order D equations with Constant Coefficients.

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

$(D - \alpha ) = 0$

(34)

This equation is

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

(35)

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

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

(36)

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

(37)

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

(38)

Which is the General Solution.

Linear Second Order D equations with constant Coefficients

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

(39)

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

(40)

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

(41)

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

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

(42)

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

(43)

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

(44)

Where C is an arbitrary Constant

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

(45)

This equation can be re-written as:-

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

(46)

Integrating

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

(47)

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

(48)

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

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

(49)

Where A and B are arbitrary Constants.

Example 1

Problem

$2\,\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 2y = 0$ $Or\;\;\;\;(2D^2 + 5D + 2)\,y = 0$ $\therefore\;\;\;\;(2D + 1)(D + 2)\,y = 0$ $y = A\,e^{-\frac{1}{2}x} + B\,e^{-2x}$

Workings

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

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

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

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

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

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

The roots of the Auxiliary Equation are complex.

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

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

Solution

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

Physical Examples

Example 1

Problem

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

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 $\alpha\;=\;1^0$ 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

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

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

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

Solution

At t = 0 $\theta \;=\;1^0\;=\;\frac{\pi }{180}\;rds.$

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

The D operator

Theory of Differential Operator (Differential Module)

Definition

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.

Example 1

Example 2

Example 3

Example 4

Three useful formulae based on the Operator D

Equation A

Example 1

Example 2

Example 3

Example 4

Equation B

Example 1

Example 2

Example 3

Example 4

Equation C - Trigonometrical functions

Example 1

Example 2

Linear First Order D equations with Constant Coefficients.

Linear Second Order D equations with constant Coefficients

Example 1

Physical Examples

Example 1