Higher Order Linear Differential Equations with Constant Coefficients

 

 

Introduction

1st Stage: The Solution of the Homogeneous Equation

Case A: The characteristic equation has n real roots:

Case B: The characteristic equation has n multiple real roots of multiplicity q:

Case C: The characteristic equation has complex roots:

Case D: Combination of the above cases

2nd Stage: The Partial Solution

Case A: Polynomial

Case B: Exponential-Polynomial Function

Case C: Trigonometric Function

Case D: Combination of the previous cases

 

 

Introduction

Consider the following linear differential equation of order n, with constant coefficients:

Our purpose is to find the unknown function y that satisfies the equation. To do this we will follow a two stage procedure. In the first stage we will calculate the solution of the Homogeneous equation and at the second stage we will calculate the partial solution of the differential equation, which is the solution of the right part of the differential equation. Finally we will add these two solutions to get the general solution of the differential equation.

 

1st Stage: The Solution of the Homogeneous Equation

The Homogeneous higher order linear differential equation with constant coefficients has the following form:

The question is which function y could be the solution of this equation. One thing is for sure, this function y must not change its form when it is differentiated, so that when its different order of differentiations are added, the result could be zero. It can be shown that the solution of this Homogeneous differential equation has the following general form:

In order to calculate the coefficients of the solution, we will create from the Homogeneous equation the following polynomial:

This equation is called the characteristic equation of the homogeneous one. Depending on the form of the roots of the characteristic equation, we have the following four cases:

Case A: The characteristic equation has n real roots:

Case B: The characteristic equation has n multiple real roots of multiplicity q:

Case C: The characteristic equation has complex roots:

Case D: Combination of the above cases

 

Case A: The characteristic equation has n real roots:

In this case the solution of the homogeneous linear differential equation is:

 

Example:

The characteristic equation is:

The roots of the characteristic equation are:

The solution of the homogeneous differential equation is:

 

Case B: The characteristic equation has n multiple real roots of multiplicity q:

In this case that the characteristic equation has p real roots and each root is multiple by a number qp, then the solution of the characteristic equation is:

 

Example:

The characteristic equation is:

The roots of the characteristic equation are:

The first root is of multiplicity two and the second is of multiplicity one (no multiplicity), so the solution of the homogeneous equation is:

 

 Case C: The characteristic equation has complex roots:

In the case that the characteristic equation has roots the complex numbers:

The solution of the characteristic equation will be:

Making a few calculations, this equation can be written in a simpler form:

 

Example:

The characteristic equation is:

The roots of the characteristic equation are:

The solution of the homogeneous equation is:

Also it can be written in the simpler form:

 

Case D: Combination of the above cases

A Homogeneous equation can be combinations of the above three cases. We calculate each case as shown above and then we add the solutions.

 

2nd Stage: The Partial Solution

In order to calculate the partial solution of a differential equation, we examine f(x) at its right side:

Depending on the form of f(x) we have the following four general cases:

Case A: Polynomial of the form:

Case B: Function of the form:

Case C: Function of the form:

Case D: Function of the form:

 

Case A: Polynomial of the form:

The partial solution of the differential equation is:

If, in the left side of the differential equation the factors:

are missing, then the partial solution of the differential equation is:

 

Example 1:

The solution of the homogeneous part is:

To find the partial solution we have to find a solution of the form:

The three derivatives are of the previous polynomial are:

Substituting to the differential equation we have:

The partial solution is:

The general solution of the differential equation is:

 

Example 2:

The characteristic equation of the homogeneous equation is:

The roots of the characteristic equation are:

The solution of the homogeneous part is:

Keeping in mind that the first derivative is missing, to find the partial solution we have to find a solution of the form:

The derivatives of the partial solution polynomial are:

Substituting to the differential equation we have:

So the partial solution is:

The general solution of the differential equation is:

 

Case B: Function of the form:

The partial solution of the differential equation, if q is not a root of the characteristic equation, is:

or:

if q is a root of  the characteristic equation of multiplicity k.

 

Example:

The homogeneous equation is:

The characteristic equation is:

The roots of the characteristic equation are the complex pair:

The solution of the homogeneous equation is:

From the right side of the differential equation, q=1 and it is not a root of the characteristic equation, so for the partial solution we have:

The derivatives of the partial solution are:

Substituting the above equation to the differential equation we have:

 

 So the partial solution is:

The general solution of the differential equation is:

 

Case C: Function of the form:

The partial solution of the differential equation, if the imaginary number mi is not a root of the characteristic equation, is:

 or:

if the imaginary number mi is a root of  the characteristic equation of multiplicity k.

 

Example:

 

The homogeneous equation is:

The characteristic equation is:

The roots of the characteristic equation are the complex pair:

The solution of the homogeneous equation is:

From the right side of the differential equation, mi=i is not a solution of the characteristic equation, so for the partial solution we have:

The derivatives of the partial solution are:

Substituting the above equations to the differential equation we have:

So the partial solution is:

The general solution of the differential equation is:

 

Case D: Function of the form:

This case is a combination of all the above cases. If the complex number q+mi is not a solution of the characteristic equation, the partial solution is of the form:

 

 or:

if the complex number q+mi is a solution of the characteristic equation of multiplicity k.

The polynomials g1 and g2 must be of grade equal to the highest grade of the polynomials f1 and f2.

The procedure of calculations is as shown in the previous cases.