Second-Order Linear Ordinary Differential Equations


This section contains information about

General Form of a Linear Second-Order ODE

A linear second-order ODE has the form:

displaymath53

On any interval where S(t) is not equal to 0, the above equation
can be divided by S(t) to yield

displaymath55

The equation is called homogeneous if f(t)=0. Otherwise, it is called
nonhomogeneous.

Existence and Uniqueness

A second-order differential equation is accompanied by initial conditions
or boundary conditions. Initial conditions are in the form y(t_0)=y_0 and
y'(t_0)=y'_0. Boundary conditions might be of the form: y(t_0)=a and y(t_1)=b.

For the initial value problem, the existence and uniqueness theorem states
that if p(t), q(t) and f(t) are continuous on some interval (a,b) containing t_0,
then there exists a unique solution y(t) to the ode in the whole interval (a,b).

Procedure for Solving Linear Second-Order ODE

The procedure for solving linear second-order ode has two steps

(1)  Find the general solution of the homogeneous problem:
displaymath57

According to the theory for linear differential equations, the general
solution of the homogeneous problem is

displaymath59

where C_1 and C_2 are constants and y_1 and y_2 are any two
linearly independent solutions to the homogeneous equation.

(2) Find a particular solution of the nonhomogeneous problem:
displaymath61

The particular solution is any solution of the nonhomogeneous
problem and is denoted y_p(t).

The general solution of the full nonhomogeneous problem is

displaymath63

The key point to note is that all possible solutions to a linear second-order
ode can be obtained from two linearly independent solutions to the
homogeneous problem and any particular solution.

Here is an example. Consider the ode

displaymath65

The homogeneous equation is

displaymath67

It can be shown that y_1=exp(-t) and y_2=exp(-2t) are solutions to the
homogeneous equation. Plug these expressions into the ode and verify!

A particular solution of the nonhomogeneous equation is exp(t). Hence,
the general solution of the ode is

displaymath69

where C_1 and C_2 are constants.

Linear Dependence

Two functions are linearly independent if they are not multiples of each other.
For example, exp(-t) and exp(-2t) are linear independent. On the other
hand, t+3 and 7t+21 are linearly dependent since the latter is 7 times the former.

Techniques for Solving Homogeneous Linear Second-Order ODE

Certain classes of homogeneous linear second-order ode can be solved
analytically. We will consider two classes:

(1) Constant-coefficient ode:
displaymath71

p(t) and q(t) are constants.

(2) Euler-Cauchy ode:
displaymath73

p and q are constants.

Techniques for Determining a Particular Solution

There are two principal techniques for determining a particular solution:

(1) The Method of Undetermined Coefficients
(2) Variation of Parameters


[ODE Home] [1st-Order Home] [2nd-Order Home] [Laplace Transform Home] [Notation] [References]

Copyright © 1996 Department of Mathematics, Oregon State University

If you have questions or comments, don't hestitate to contact us.