Second order and higher order equations occur frequently in science and engineering (like pendulum problem etc.) and hence has its own importance. It has its own flavour also. We devote this section for an elementary introduction.
Here 
 is an interval
contained in 
 and the functions 
 and
 are real valued
continuous functions defined on 
The functions
 and 
 are called the coefficients of
Equation (8.1.1)
and 
 is called the  non-homogeneous term or the  force function.
Equation (8.1.1) is called linear homogeneous
if 
 and non-homogeneous if 
Recall that a second order equation is called nonlinear if it is not linear.
is a second order equation which is nonlinear.
 and We now state an important theorem whose proof is simple and is omitted.
 the function 
 is also a solution of Equation
(8.1.2). It is to be noted here that Theorem 8.1.5 is not an existence theorem. That is, it does not assert the existence of a solution of Equation (8.1.2).
For example, all the solutions of the Equation (8.1.2) form a
solution space. Note that 
 is also a solution of
Equation (8.1.2). Therefore, the solution set of a
Equation (8.1.2) is non-empty. A moments reflection on Theorem
8.1.5 tells us that the solution space of Equation (8.1.2)
forms a real vector space.
The natural question is to inquire about its dimension. This question will be answered in a sequence of results stated below.
We first recall the definition of Linear Dependence and Independence.
The functions
To proceed further and to simplify matters, we assume that 
 in Equation (8.1.2) and that the
function 
 and 
 are continuous on 
In other words, we consider a homogeneous linear equation
The next theorem, given without proof, deals with the existence and
uniqueness of solutions
of Equation (8.1.3) with initial conditions
 for some 
A word of Caution: NOTE THAT THE COEFFICIENT OF
 
An important application of Theorem 8.1.9 is that
the equation (8.1.3) has exactly  
 
 
Use initial condition on  
We now show that any solution of Equation (8.1.3) is a linear 
combination of  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
) IS 
, WE HAVE TO ENSURE THIS
CONDITION. 
 linearly
independent solutions. In other words, the set of all solutions
over 
 forms a real vector space of dimension 
 and 
 be real valued continuous functions on 
 Then
Equation (8.1.3) has exactly two linearly
independent solutions. Moreover, if 
 and 
 are two linearly
independent solutions of Equation (8.1.3), then the solution
space is a linear combination of 
 and 
.
 and 
 be two unique solutions of Equation
(8.1.3) with initial conditions 
 

(8.1.5) 
The unique solutions 
 and 
 exist by virtue of Theorem
8.1.9. We now claim that 
 and 
 are linearly
independent.   Consider the system of linear equations
where 
 and 
 are unknowns. If we can show that the only
solution for the system (8.1.6) is 
,
then the two solutions 
 and 
 will be linearly independent. 
 and 
 to show that the only solution is indeed
. Hence the result follows.
 and 
.
Let 
 be any solution of Equation
(8.1.3) and let 
 and 
 Consider the function 
 defined by
 is a solution of
Equation (8.1.3). Also note that 
 and 
 So, 
 and 
 are two
solution of Equation (8.1.3) with the same
initial conditions. Hence by Picard's Theorem on Existence and
Uniqueness (see Theorem 8.1.9),
 or
 and 
 corresponding to the 
initial conditions 
Consider a 
non-singular matrix 
with 
 Let 
 be a fundamental
system for the differential Equation 8.1.3 and
 Then the rows of the matrix
 also form a fundamental system for
Equation 8.1.3. That is, if 
 is a fundamental
system for Equation 8.1.3 then 
is also a fundamental system whenever 
 is a fundamental system for 
Note that 
 is also a fundamental
system. Here the matrix is 
 and 
 are solutions of
 and 
 are
also solutions of 
 Do 
 and 
form a fundamental set of solutions?
 forms a basis for the solution space of
 find another basis.