Solution of Linear Ordinary Differential Equations (LODE) with Constant Coefficients ( A Short Review )

Standard Form of nth Order LODE:


where the ais are constant coefficients such that an is not zero, and f(t) is the forcing function. If f(t)=0 then equation (1) is said to be homogenous.


The complete solution to equation (1) is a sum of two parts: the complementary solution xc, and the particular solution xp,

x (t) = xc (t) + xp (t)                                                                          (2)

The complementary part, xc (t) is the solution to the equation when f(t) is replaced by 0, i.e. made homogenous. It contains n arbitrary constants, which are evaluated later to fit boundary or initial conditions.

The particular part, xp (t), is a solution which solves the original equation, i.e. when f(t) is put back into the equation. It does not have any arbitrary constant and must be linearly independent from the complementary solution.

The procedure involves the following steps:

  1. Replace f(t) by zero and solve for xc (t)
  2. If f(t) is zero to begin with, then xp (t)=0. Otherwise, find the particular solution that satisfies the original differential equation but is linearly independent from xc (t).
  3. Evaluate the arbitrary constants to satisfy the boundary conditions.

Complementary Solution:

First, assume that the complementary solution has the form:

xc = exp( s t )                                                                              (3)

where s is a parameter yet to be determined. Then the kth derivative of xc (t) is given by


Thus after substitution of (3) into (1), while setting f(t)=0,


here p(s) is a polynomial function of s of degree n given by

Since exp(st) is not equal to 0 for arbitrary values of t, equation (5) can be satisfied only if p(s)=0, i.e. we need to solve for the roots of p(s) defined by (6).


  1. p(s) is called the characteristic polynomial.
  2. p(s)=0 is called the characteristic equation.
  3. The roots of p(s)=0 are called the characteristic roots or eigenvalues

Let the roots of p(s) be {s1, s2, , sn}, then the complementary solution is going to be a sum of terms of exp(skt) multiplied by arbitrary coefficients and/or appropriate powers of t depending on how many times the root is repeated.

For each real root sA that may have been repeated kA times in the set, include the following terms to the complementary solution,

where A0, , AkA are arbitrary coefficients.

For each pair of complex roots sB= a + ib and sC = a - ib that may have been repeated kB=kC times in the set, include the following terms to the complementary solution,

where B0 , , BkB , C0 , , CkC are arbitrary coefficients.

Particular Solution (using the method of undertermined coefficients)

Depending on the form of f(t), assume a similar form for xp (t) but containing unknown coefficients. (If a term in the assumed xp (t) matches any term of xc (t), multiply these terms by t m where m is large enough to keep it independent xc (t).

These assumed terms are then substituted into the differential equation. The following steps are then implemented:

  1. The resulting algebraic equation is rearranged by collecting coefficients.
  2. Coefficients from both sides of the equation are matched.
  3. The unknown parameters are then evaluated by solving the resulting simultaneous equations.
Choices for xp (t)
A ( a constant )
A t m ( m > 0 )
D0 + D1 t + + Dm tm
A t m exp( r t )
(D0 + D1 t + + Dm tm) exp( r t )
A t m exp( r t ) cos ( q t ) or
A t m exp( r t ) sin ( q t )
(D0 + D1 t + + Dm tm) exp( r t ) cos ( q t) +
(E0 + E1 t + + Em tm) exp( r t ) sin ( q t )

Click here for an example.

This page is maintained by Tomas B. Co ( Last revised 11/30/1999.

     Tomas B. Co
     Associate Professor
     Department of Chemical Engineering
     Michigan Technological University
     1400 Townsend Avenue
     Houghton, MI 49931-1295

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