# Vector Representations of Solutions Consider the linear system of differential equations This system may be rewritten using matrix-notation. Indeed, set ,

then the above system is equivalent to the matricial equation .

Using the matrix product, we get .

The matrix is called the coefficient matrix of the system. Note that the coefficients of the matrix A can be constant or not. The vector function is called the nonhomogeneous term.

Remark: One may think that the equation above is only valid for linear systems of two equations. However, that is not the case. For example, consider the linear system Then, in matricial notation, the system is equivalent to ,

where .

## Equilibrium Points of Homogeneous Linear Systems

Consider the homogeneous linear system The equilibrium points are given by the equations Clearly, x=0 and y=0 give a trivial solution. Hence, the function gives a constant solution to the linear system. We call it the trivial solution. In general, the equilibrium points are the intersection between two lines. Since the two lines intersect, they are the same (if parallel) or the intersection is reduced to one point. So, the set of equilibrium points is the entire line ax+by=0, or the trivial point (0,0). This conclusion is related to the determinant of the matrix coefficient. Indeed, if is not equal to 0 (zero), then we have one equilibrium point (the trivial one).

## The Linearity Principle

This is may be the most important property for linear systems. Consider the homogeneous linear system ,

then

1.
if Y(t) is a solution and k is a constant, then k Y(t) is also a solution;
2.
if and are two solutions, then is also a solution.

This clearly implies that if and are two solutions and and are two arbitrary constants, then is also a solution. This conclusion is also known as the Principle of Superposition.

Clearly, from the Principle of Superposition, we may generate plenty of solutions once two solutions are known. The natural question to ask therefore, is whether we have obtained all the solutions. In order to better appreciate this problem let's consider the following example.

Example: Consider the linear system Show that any solution Y to this system is given as ,

where ,

and and are two constants.

Answer: It is easy to check that indeed and are solutions to the given system. Let Y be any solution. Set .

By the uniqueness and existence theorem, Y is the only solution to the IVP .

Let us find and such that . If this is the case, we should have , which gives ,

which implies Clearly, this gives .

Consider the function .

The linearity principle implies that is a solution. And, since ,

the uniqueness and existence theorem implies that in fact gives the desired conclusion.

Remark: When you look at the above example you will notice that what made the conclusion work is that we were able to solve the algebraic system and this was possible because the two vectors are linearly independent. In fact, the above conclusion is always valid whenever we have a linear independence around.

### Theorem: The General Solution

Suppose and are two solutions to the linear system .

Assume that the vectors and are linearly independent. Then, the solution to the IVP ,

is given by ,

for some constants and . In this case, the two-parameter family ,

where and are arbitrary constants, is called the general solution of the system. Then, the two solutions and are said to be linearly independent.

Example: Consider the undamped harmonic oscillator .

Show that any solution x is given by .

Answer: Consider the associated linear system Set . Note that the second component is just the derivative of the first one. Consider the two vector functions It is easy to check that these two vector functions are in fact solutions to the given system. Also, you may check that the two vectors are linearly independent. Therefore, any solution Y of the system is given by ,

where and are two constants. Using the first component of Y, we see that any solution x(t) of the equation is given by ,

where and are two arbitrary constants.

Clearly, the main problem now is how to find the two linearly independent solutions. This problem will be discussed in the next section and will use the eigenvalue and eigenvector technique. [Differential Equations] [First Order D.E.]
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