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Linear with Constant Coefficient

A guide to linear equations of second and higher degrees
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Definition

The Equations in this section are of the form:

where f(x) is a function of x but all of the P'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

If n - 1 and f(x) = 0 equation (2) becomes

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 ln A 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

If n = 2 and f(x) = 0

Equation (2) can now be written as: \displaystyle y = A\;e^{\alpha x}\;\;\;and\;\;\;y = B\;e^{\beta x}

The solution to equation (3) suggests that \displaystyle y = A\;e^{mx} where m is some constant may satisfy equation (4). With this value for y equation (4) reduces to:

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

Thus if m is a root of:

\displaystyle y = A\;e^{mx} is a solution of equation (4) whatever the value of A

Let the roots of equation (5) be \alpha and \beta.

If the roots are unequal we will have two solutions to equation (4) 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 (4) 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 (5) 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 m = -2 or - 1/2

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 (5) has roots of the form (p + iq) and (p - iq) where

i = \sqrt{-1}

it is best to modify the solution

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 (6) becomes

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

Writing E for (A + B) and F for i(A - B)
y = e^{px}\left(E\;cos\;qx + F\;sin\;qx \right)

E and F are arbitrary constants as were A and B. It might look as if F must be imaginary but this is not necessarily so . Thus if A = 1 + 2i and B = 1 - 2i then E = 2 and F = -4.

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)

Or

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 (2) whatever the value of n provided that f(x) = 0
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 f(x) of equation (2) has been zero. It will now be shown that the relation between the solution of the equation when f(x) is not zero and the solution of a simpler equation derived from it by replacing f(x) 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 y = (x + v) 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.

If y = u is a particular integral of :

So that:

Putting y = u + v in equation (7) and subtracting equation (8) 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 v = F(x) contains n arbitrary constants then the general solution to equation (7) is :

y = u + F(x)

and F(x) is called the Complementary Function.