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This module contains information on solving linear constant coefficient differential equations.


The approach to solving linear constant coefficient ordinary differential equations is to find the general form of all possible solutions to the equation and then apply a number of conditions to find the appropriate solution. The two main types of problems are initial value problems, which involve constraints on the solution and its derivatives at a single point, and boundary value problems, which involve constraints on the solution or its derivatives at several points.

The number of initial conditions needed for an N th order differential equation, which is the order of the highest order derivative, is N , and a unique solution is always guaranteed if these are supplied. Boundary value problems can be slightly more complicated and will not necessarily have a unique solution or even a solution at all for a given set of conditions. Thus, this module will focus exclusively on initial value problems.

Solving linear constant coefficient ordinary differential equations

Consider some linear constant coefficient ordinary differential equation given by A x ( t ) = f ( t ) , where A is a differential operator of the form

A = a n d n d t n + a n - 1 d n - 1 d t n - 1 + . . . + a 1 d d t + a 0 .

Let x h ( t ) and x p ( t ) be two functions such that A x h ( t ) = 0 and A x p ( t ) = f ( t ) . By the linearity of A , note that A ( x h ( t ) + x p ( t ) ) = 0 + f ( t ) = f ( t ) . Thus, the form of the general solution x g ( t ) to any linear constant coefficient ordinary differential equation is the sum of a homogeneous solution x h ( t ) to the equation A x = 0 and a particular solution x p ( t ) that is specific to the forcing function f ( t ) .

We wish to determine the forms of the homogeneous and nonhomogeneous solutions in full generality in order to avoid incorrectly restricting the form of the solution before applying any conditions. Otherwise, a valid set of initial or boundary conditions might appear to have no corresponding solution trajectory. The following discussion shows how to accomplish this for linear constant coefficient ordinary differential equations.

Finding the homogeneous solution

In order to find the homogeneous solution to A x ( t ) = f ( t ) , consider the differential equation A x ( t ) = 0 . We know that the solutions have the form c e λ t for some complex constants c , λ . Since A c e λ t = 0 for a solution, it follows that

a n d n d t n + a n - 1 d n - 1 d t n - 1 + . . . + a 1 d d t + a 0 e λ t = 0 ,

so it also follows that

a n λ n + a n - 1 λ n - 1 . . . + a 1 λ + a 0 = 0 .

Therefore, the parameters of the solution exponents are the roots of the above polynomial, called the characteristic polynomial.

For equations of order two or more, there will be several roots. If all of the roots are distinct, then the the general form of the homogeneous solution is simply

x h ( t ) = c 1 e λ 1 t + . . . + c n e λ n t .

If a root has multiplicity that is greater than one, the repeated solutions must be multiplied by each powers of t from 0 to one less than the root multiplicity (in order to ensure linearly independent solutions). For instance, if λ 1 had multiplicity 2 and λ 2 had multiplicity 3, the homogeneous solution would be

Questions & Answers

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Preparation and Applications of Nanomaterial for Drug Delivery
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Source:  OpenStax, Signals and systems. OpenStax CNX. Aug 14, 2014 Download for free at http://legacy.cnx.org/content/col10064/1.15
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