By Eckhard M.S. Hitzer, University of Fukui, Prof. Ryosuke Nagaoka (University of the Air, Japan)

So now I will give you a concrete example of an ordinary homogeneous linear differential equation with constant coefficients here. Coefficients: plus one, minus six, plus eleven, and minus six. And it is of third order, so we have the highest differentiation is three times.

A good way to solve this equation is to transform it into a system of first order equations. And therefore we introduce new variables. We call y (=) y one, and then we call the first differentiation of y (=) y prime (=) y two. And so y two equals y one prime. Differentiating y one, one times we get y two. And we call y three (the second) y double prime and this is the same like y one double prime, y two prime. So we have equations y one prime is y two, y two prime is y three and finally we have the euqation for y three prime is y triple prime.

And this can be written according to the differential equation as six times y double prime, minus 11 (times) y prime plus six (times) y. And now we replace y double prime by y three, y prime by y two and y by y one. So we have this equation y three prime equals this. And then we have the equation y one prime equals y two, and y two prime equals y three here.

And so first, second and third. So these three simultaneous equations can now be written in matrix form and are completely equivalent to the original third order equation. Now they are just first order equations.

And rewriting the equation in matrix form, we have here y one prime, y two prime, y three prime equals the matrix A times y one, y two, y three. And the matrix A must be precisely with the coefficients as they are written here. And a good, the best or easiest strategy for solving this equation is to change coordinates to a system where the matrix A has diagonal form. Now if matrix A has diagonal form, then the entries in the main diagonal of the new form of A will be the eigenvalues of A. Therefore we first will set the characteristic polynomial of A to be zero in order to calculate the eigenvalues.

So the characteristic polynomial is simply (the determinant of) the matrix A minus x times the unity matrix (E) to be zero. And if we develop this matrix (determinant) and we multiply it out, then we see it has this form. And we can factorize the expression on the left hand side into x minus one, times x minus 2, times x minus 3, to be zero. And the values (and) when the polynomial on the left hand side becomes zero are precisely the eigenvalues.

So we have the eigenvalues

x to be equal to one,

x to be equal to two,

x to be equal to three.

Here one, two, three. And this will be the main entries into the diagonal matrix, which is the form,
which we want to transform A to.

And we can transform A into the diagonal form with help of the matrix capital P. And the entries, the columns of this matrix are precisely the eigenvectors of A, which belong to the three eigenvalues, eigenvalue one, eigenvalue two, and eigenvalue three. And here you have the transformation, which we must subject A to, in order to get it into diagonal form, it is the inverse of P times A times P. And here are precisely the three eigenvalues.

Now we can bring our equation into this desired form, where the matrix A becomes diagonal, by
introducing new coordinates. Coordinates z one, z two and z three. And the matrix P times
z one, z two, z three are our old coordinates y one, y two, and y three. And in this new system,
then the three equations (are) now have the most simple form of

d z one by d x is the first eigevalue times z one,

d z two by d x is two, the second eigenvalue times z two,

d z three by d x is the third eigenvalue times z three.

And the general solution of such an equation is very easy. It is again the exponential function with a constant factor in front. And always we have the eigenvalues here in the exponential function. So it is the eigenvalues one, two and three. It is very similiar to the first concrete example, which we had, when k was the constant, and here k has the concrete values of one, two, and three.

From this equation here we see that y one is a linear combination of the three coordinates: z one, z two and z three, with the help of the matrix P. So, because these three constants c one, c two and c three are arbitrary and are to be determined later, any linear combination of the three functions here, of the three exponential functions, will be a solution of the equation, which we are looking for. And by giving - here we have this general solution - by giving conditions on for example on the value of the function at the time t equal zero, on the first derivative, and on the second derivative, we can determine these constants for concrete cases: capital C one, capital C two, and capital C three.

Soli Deo Gloria. Created by
Eckhard Hitzer (Fukui).

Last Modified 16 January 2004

EMS Hitzer and Prof. R. Nagaoka are not responsible for the content of external internet sites.