Linear with Constant Coefficient

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.

A guide to linear equations of second and higher degrees

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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 $f(x)$ 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 (#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 $ln A$ Thus $ln\,y\;= - \frac{P_1\;x}{P_0} + ln\,A$

Blank

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 (#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 $y$ equation (#3) reduces to:

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

Thus if $m$ is a root of:

$(P_0m^2 + P_1m + P_2) = 0$

(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 $m = -2$ 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 (#4) has roots of the form $(p + iq)$ and $(p - iq)$ 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 (#8) 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 1

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 $f(x) = 0$

Example 1

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 (#1) 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:

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