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Maths › Calculus › Differential ›

Linear with Constant Coefficient

A guide to linear equations of second and higher degrees

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Contents

Definition
Equation Of The First Order
Equations Of The Second Order
Modifications When The Auxiliary Equation Has Imaginary Or Complex Roots
The Extension To Orders Higher Than The Second
The Complementary Function And The Particular Integral
Page Comments

Definition

The Equations in this section are of the form:

$P_0\,\frac{d^ny}{dx^n} + P_1\,\frac{d^{n-1}y}{dx^{n-1}}\;+........+\;P_{n-1}\frac{dy}{dx} + P_ny = f(x)$

(1)

where

is a function of x but all of the

's are Constants.

These equations are of the utmost importance in the study of vibrations of all kinds(Mechanics; Acoustics and Electrical). The methods given are chiefly due to Euler and D'Alembert.

Equation Of The First Order

and

equation (1) becomes

$P_o\;\frac{dy}{dx} + P_1\;y = 0$

(2)

Therefore

$P_0\;\frac{dy}{y}\;+ P_1\;dx = 0$

Integrating

$P_o\;ln\,y + P_1\;x = Constant$

Therefore

$ln\,y\;= - \frac{P_1\;x}{P_0} + Constant$

Let the Constant equal

Thus

$ln\,y\;= - \frac{P_1\;x}{P_0} + ln\,A$

Therefore $\displaystyle y = A\;e^{- \frac {P_1\,x}{P_0}$ is the general solution for the first order differential equation .

Equations Of The Second Order

and

Equation (1) can now be written as: $\displaystyle y = A\;e^{\alpha x}\;\;\;and\;\;\;y = B\;e^{\beta x}$

$P_0\;\frac{d^2y}{dx^2} + P_1\;\frac{dy}{dx} + P_2\;y = 0$

(3)

The solution to equation (2) suggests that $\displaystyle y = A\;e^{mx}$ where m is some constant may satisfy equation (3). With this value for

equation (3) reduces to:

$A\,e^{mx}\.(P_0m^2 + P_1m + P_2) = 0$

Thus if

is a root of:

(4)

$\displaystyle y = A\;e^{mx}$ is a solution of equation (3) whatever the value of A

Let the roots of equation (4) be $\alpha$ and $\beta$ .

If the roots are unequal we will have two solutions to equation (3) namely $y = A\;e^{\alpha x}$ and $y = B\;e^{\beta x}$
Then the general solution will be $y = A\;e^{\alpha x} + B\;e^{\beta x}$

If the roots are equal we will also have two solutions to equation (3) namely $y = A\;e^{\alpha x}$ and $y = x\;e^{\alpha x}$
Then the general solution will be $y = A\;e^{\alpha x} + B\;xe^{\alpha x}$

Equation (4) is called the "Auxiliary Equation"

As an example, to solve

$2\,\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 2y = 0\;\;\;\;\;try\;\;y = Ae^{mx}\f$ as a trial solution.$

Therefore

$A\,e^mx{(2m^2 + 5m + 2)} = 0$

This equation is satisfied by

The General Solution is therefore given by:

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

Modifications When The Auxiliary Equation Has Imaginary Or Complex Roots

When the auxiliary equation (4) has roots of the form

and

where

$i = \sqrt{-1}$

it is best to modify the solution

$y = A\,e^{(p+iq)x} + Be^{(p-iq)x}$

(5)

so that it does not contain imaginary quantities.. To do this use the following trigonometrical identities:

$e^{iqx} = cos\,qx\;+i\,sin\;qx$

$e^{-iqx} = cos\,qx\;-i\,sin\;qx$

Thus equation (5) becomes

$y = e^{px}\left(A\;(cos\;qx + i\;sin\;qx) + B(cos\;qx - i\;sin\;qx) \right)$

Writing

for

and

for

$y = e^{px}\left(E\;cos\;qx + F\;sin\;qx \right)$

and

are arbitrary constants as were

and

. It might look as if F must be imaginary but this is not necessarily so . Thus if

and

then

and

Example:

Example - Second degree equation

Problem

$\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 13\,y = 0$

Workings

From this the auxiliary equation is:

$m^2\;-6m + 13\;=0$

and the roots are $\displaystyle m = 3\;\pm 2i$

The solution can be written as;-

$y = A\;e^{(3+2i)x}\;+B\;e^{(3-2i)x}$

Solution

or in a more useful form:

$y = e^{3x}(E\;cos\,2x + F\;sin\,2x)$

$y = C\;e^{3x}\;cos(2x\;-\alpha )$

Where $\displaystyle C\;cos\,\alpha = E\;\;\;\;and\;\;\;\;C\;sin\,\alpha = F$

So that $\displaystyle C = \sqrt{(E^2 + F^2)}\;\;\;\;and\;\;\;\;tan\,\alpha = \frac{F}{E}$

The Extension To Orders Higher Than The Second

The methods discussed in this section apply to equation (1) whatever the value of n provided that

Example:

Example - Third degree equation

Problem

$\frac{d^3y}{dx^3} - 6\,\frac{d^2y}{dx^2} + 11\frac{dy}{dx} - 6\,y = 0$

Workings

The Auxiliary Equation is:

$m^3 - 6m^2 + 11m - 6 = 0$

Solution

Thus m = 1, 2, or 3 Therefore

$;y = A\,e^x + B\,e^{2x} + C\,e^{3x}$

The Complementary Function And The Particular Integral

So far we have only dealt with examples where the

of equation (1) has been zero. It will now be shown that the relation between the solution of the equation when

is not zero and the solution of a simpler equation derived from it by replacing

by zero.

Consider the equation:

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

By inspection it can be seen that y = x is one solution. Such a solution containing no arbitrary constants is called a Particular Integral

Now substitute

in the equation which becomes:

$2\,\frac{d^2v}{dx^2} + 5\,\left(1 + \frac{dv}{dx} \right) + 2(x + v) = 5 + 2x$

$2\,\frac{d^2v}{dx^2} + 5\,\frac{dv}{dx} + 2\;v = 0$

From this it can be shown that :

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

Note
The general solution of a linear differential equation with constant coefficients is the sum of a Particular Integral and the Complementary Function, the latter being the solution of the equation obtained by substituting zero for the function of x occurring.

The terms containing the arbitrary constants are called the Complementary Function

This can be expressed in a general form.

is a particular integral of :

$P_0\,\frac{d^ny}{dx^n} + P_1\,\frac{d^{n\,-\,1}y}{dx^{n\,-\,1}} + .......P_{n\,-\,1}\frac{dy}{dx} + P_n\,y = f(x)}$

(6)

So that:

$P_0\,\frac{d^nu}{dx^n} + P_1\,\frac{d^{n\,-\,1}u}{dx^{n\,-\,1}} + .......P_{n\,-\,1}\frac{du}{dx} + P_n\,u = f(x)}$

(7)

Putting

in equation (6) and subtracting equation (7) gives:

$P_0\,\frac{d^nu}{dx^n} + P_1\frac{d^{n\,-\,1}v}{dx^{n\,-\,1}} + .......+\;P_{n\,-\,1}\;\frac{dv}{dx} + p_n\,v = 0$

If the solution to this equation is

contains n arbitrary constants then the general solution to equation (6) is :

and

is called the Complementary Function.

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