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# Linear with Constant Coefficient

A guide to linear equations of second and higher degrees

## Definition

The Equations in this section are of the form:
$P_0\,\frac{d^ny}{dx^n}&space;+&space;P_1\,\frac{d^{n-1}y}{dx^{n-1}}\;+........+\;P_{n-1}\frac{dy}{dx}&space;+&space;P_ny&space;=&space;f(x)$

where $\inline&space;f(x)$ is a function of x but all of the $\inline&space;P$'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 $\inline&space;n&space;-&space;1$ and $\inline&space;f(x)&space;=&space;0$ equation (1) becomes

$P_o\;\frac{dy}{dx}&space;+&space;P_1\;y&space;=&space;0$
Therefore
$P_0\;\frac{dy}{y}\;+&space;P_1\;dx&space;=&space;0$

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

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

Let the Constant equal $\inline&space;ln&space;A$ Thus
$ln\,y\;=&space;-&space;\frac{P_1\;x}{P_0}&space;+&space;ln\,A$

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

## Equations Of The Second Order

If $\inline&space;n&space;=&space;2$ and $\inline&space;f(x)&space;=&space;0$

Equation (1) can now be written as: $\inline&space;\displaystyle&space;y&space;=&space;A\;e^{\alpha&space;x}\;\;\;and\;\;\;y&space;=&space;B\;e^{\beta&space;x}$
$P_0\;\frac{d^2y}{dx^2}&space;+&space;P_1\;\frac{dy}{dx}&space;+&space;P_2\;y&space;=&space;0$

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

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

Thus if $\inline&space;m$ is a root of:

$(P_0m^2&space;+&space;P_1m&space;+&space;P_2)&space;=&space;0$

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

Let the roots of equation (4) be $\inline&space;&space;\alpha$ and $\inline&space;\beta$.

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

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

Equation (4) is called the "Auxiliary Equation"
As an example, to solve
$2\,\frac{d^2y}{dx^2}&space;+&space;5\frac{dy}{dx}&space;+&space;2y&space;=&space;0\;\;\;\;\;try\;\;y&space;=&space;Ae^{mx}\f&space;&space;as&space;a&space;trial&space;solution.$
Therefore
$A\,e^mx{(2m^2&space;+&space;5m&space;+&space;2)}&space;=&space;0$

This equation is satisfied by $\inline&space;m&space;=&space;-2$ or $\inline&space;-&space;1/2$

The General Solution is therefore given by:

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

## Modifications When The Auxiliary Equation Has Imaginary Or Complex Roots

When the auxiliary equation (4) has roots of the form $\inline&space;(p&space;+&space;iq)$ and $\inline&space;(p&space;-&space;iq)$ where

$i&space;=&space;\sqrt{-1}$

it is best to modify the solution

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

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

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

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

Thus equation (5) becomes

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

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

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

Example:
##### Example - Second degree equation
Problem

$\frac{d^2y}{dx^2}&space;-&space;6\frac{dy}{dx}&space;+&space;13\,y&space;=&space;0$
Workings
From this the auxiliary equation is:

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

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

The solution can be written as;-

$y&space;=&space;A\;e^{(3+2i)x}\;+B\;e^{(3-2i)x}$
Solution
or in a more useful form:

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

Or

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

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

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

## 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 $\inline&space;f(x)&space;=&space;0$
Example:
##### Example - Third degree equation
Problem

$\frac{d^3y}{dx^3}&space;-&space;6\,\frac{d^2y}{dx^2}&space;+&space;11\frac{dy}{dx}&space;-&space;6\,y&space;=&space;0$
Workings
The Auxiliary Equation is:

$m^3&space;-&space;6m^2&space;+&space;11m&space;-&space;6&space;=&space;0$
Solution
Thus m = 1, 2, or 3 Therefore
$;y&space;=&space;A\,e^x&space;+&space;B\,e^{2x}&space;+&space;C\,e^{3x}$

## The Complementary Function And The Particular Integral

So far we have only dealt with examples where the $\inline&space;f(x)$ of equation (1) has been zero. It will now be shown that the relation between the solution of the equation when $\inline&space;f(x)$ is not zero and the solution of a simpler equation derived from it by replacing $\inline&space;f(x)$ by zero.

Consider the equation:
$2\,\frac{d^2y}{dx^2}&space;+&space;5\,\frac{dy}{dx}&space;+&space;2y&space;=&space;5&space;+&space;2x$

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 $\inline&space;y&space;=&space;(x&space;+&space;v)$ in the equation which becomes:

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

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

From this it can be shown that :

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

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 $\inline&space;y&space;=&space;u$ is a particular integral of :

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

So that:

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

Putting $\inline&space;y&space;=&space;u&space;+&space;v$ in equation (6) and subtracting equation (7) gives:

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

If the solution to this equation is $\inline&space;v&space;=&space;F(x)$ contains n arbitrary constants then the general solution to equation (6) is :

$y&space;=&space;u&space;+&space;F(x)$

and $\inline&space;F(x)$ is called the Complementary Function.