# Something about homogenous DE with constant co-efficients:

1. Nov 13, 2007

### O.J.

I know the solution to them is done by taking it for granted that the solution MUST be of the exponential form e^mx. But in the special case where we get two REPEATED roots in which case we multiply one of the solution by x to get an independent solution of the form x e^mx.

This is what's getting me all twisted up, I aked my professor and he didn't really remove the vagueness surroudning this. Our method of solution is based on the assumption that all solutions must be of the form e^mx. But the new solution we got which is x e^mx contradicts our assumptions. That solution is not a pure exponential; it is of a different form.

I know it is foudn by the method of reduction of order, btw, and i still find the result contradictive to what we assumed. Please clarify.

2. Nov 13, 2007

### HallsofIvy

No, you do not know that- it isn't true! Looking for a solution of the form y= erx leads you to the characteristic equation but you should shortly learn that there is no reason why a solution MUST be of that form. Fortunately, we can then see what to do in that case but looking for a solution of the form y= erx is only a "stopgap". Indeed, it is not just xemx that is not of that form. Strictly speaking, in terms of real solutions, cos(mx) or sin(mx) are not of that form. And certainly there exist homogeneous differential equation with constant coefficients that have only polynomials as solution.

Again, the point of view "the solution MUST be of this form" is incorrect. It is more a viewpoint of "suppose I try to find a solution of this form, what happens?"

3. Nov 16, 2007

### O.J.

so you're saying, that finding one solution can lead to other solutions using reduction of order?

4. Nov 16, 2007

### HallsofIvy

I am saying that trying something as a solution can lead to the unexpected solutions! Trying something is always better than sitting and staring at the equation.

5. Nov 16, 2007

### O.J.

Ok, thats nice. It is the viewpoint of 'trying' you say. Suppose you 'try' a form of solution of a DE and it works and you write the general solution based on it. How would you be sure there ARENT other solutions in OTHER forms after you stated the general solution which should encompass all solutions (independent ones).

6. Nov 17, 2007

### HallsofIvy

Because, in addition to "trying" specific things, we also learn the theory. We know that the set of all solutions to an nth order, linear, homogenoeous differential equation form a vector space of dimension n. If we are able, by any means at all, to find n independent solutions, then we know any solution can be written as a linear combination of the n independent solutions.

7. Nov 17, 2007

### O.J.

ok now im lost. whats this theory you speak of? they haven't taught it to us.

8. Nov 18, 2007

### HallsofIvy

No person should ever take differential equations without having take Linear Algebra as a prerequisite! It is the whole theory underlying "linear differential equations". Your questions are "theory" questions and apparently your differential equations course was a "cookbook" course where you just learn to solve specific problems. A pox on such things!