The theory of linear differential equations is closely related to the theory of systems of linear equations in linear algebra. As a result of this theory, we will be able to say a lot about the structure of the solutions. Specifically, our main results will be the following:
Example: One solution of the non-homogeneous differential equation is . Two linearly independent solutions of the homogeneous equation are and . We're not concerned right now with how to obtain these solutions, but it's easy to check that they are solutions. The conclusion will be that the general solution of is .
To make the differential equation look
more like linear algebra, we define a linear operator:
Theorem: Suppose is one solution of the equation . Then
the solutions of consist of all functions of the form
where is
a solution of the homogeneous equation .
The solutions of the homogeneous equation form a vector space.
Proof:
If is any solution of the homogeneous equation, then
. Conversely if is any solution of
, then
so
with a solution of the homogeneous equation.
To show the solutions of the homogenous equation form a vector space,
we need to show that
if and are any solutions of and and are
constants, then
is also a solution. But this is easy
since
As initial conditions for a second order equation, we need to specify
both the value of and its derivative at some point . We then
have the following
Existence and Uniqueness Theorem for second order linear equations:
Theorem:
Suppose , and are continuous on some interval
. Given any with
and any real numbers and
, there is exactly one solution of the differential equation
with initial conditions ,
,
defined for .
(We will not prove this)
Example: Consider again . We know that is a solution for any constants and . We can satisfy any initial conditions at using a function of this form: and so we would take and . By the uniqueness part of the Existence and Uniqueness Theorem, this gives us the only solution satisfying those initial conditions. Since every solution must satisfy some initial conditions at , we can conclude that all solutions are of the form .
In our example, the solutions of the homogeneous equation are all functions of the form . These form a vector space. It is in fact a two-dimensional vector space: every one of these functions is written in exactly one way as a linear combination of the two functions and which form a basis of the space. They are known as a fundamental set of solutions. Recall from linear algebra:
Definition:
A set of vectors
is linearly independent
if the only way to write a linear combination
is with all the scalars .
Note that in our case, the vectors are functions and the scalars are constants; ``'' means for all . For a set of two functions, linear independence is easy to check: they are linearly independent unless one is a constant multiple of the other.
Definition:
A basis of the vector space of solutions of a second order
homogeneous linear equation is called a
fundamental set of solutions of the
equation.
Theorem:
The vector space of solutions of a second order homogeneous linear equation is
two-dimensional. Thus a fundamental set of solutions of the equation
consists of two linearly independent solutions.
Proof: This is a consequence of the Existence and Uniqueness Theorem. Take any in the interval where and are defined. Let be the solution of the equation with initial conditions , . Let be the solution with initial condition , . These are linearly independent (any constant multiple of has , and any constant multiple of has ). To satisfy any initial conditions at , say , , we can take . Thus any initial conditions at can be obtained with a linear combination of and . Since every solution satisfies some initial conditions at , every solution is a linear combination of and . So is a basis of the space of solutions. Since this basis has two elements, the space is two-dimensional.
The two results we stated at the beginning are now clear. For a homogeneous
equation, we need to find a fundamental set of solutions, which consists of
two linearly independent solutions. This constitutes a basis of the space of
solutions of the homogeneous equation. For a non-homogeneous equation,
we need one solution of it and a basis of solutions,
again consisting of two linearly
independent solutions, of the
homogeneous
equation.
Now we have the general structure of the solutions.
What we have to do next is find those solutions.