# How to prove this theorem about linear ODE's?

1. Oct 29, 2011

### AdrianZ

1. The problem statement, all variables and given/known data
Suppose that we have the following linear ODE:
$$y^{(n)} + a_{n-1}y^{(n-1)} + a_{n+2}y^{(n-2)} + ... + a_2y'' + a_1y' + a_0y = 0$$
Prove that if we have (n-1) linearly independent {y1,...,yn-1} solutions then we can find yn.

3. The attempt at a solution
well, this is my idea:
Imagine that C(a,b) is the linear space of all the functions that are continuous in the interval [a,b]. Let's define an inner product on this space as the following:
<f,g> = ∫abf(x)g(x)dx
Now, Imagine I have n-1 linearly independent functions y1,...,yn-1 then the problem is transformed into finding a function yn such that yn is perpendicular to all the other yi's (i=1,...,n-1).

well, it means I should show that there exists a function that for any yi I have:
abynyidx

Here is where I'm stuck. What should I do?

Last edited: Oct 29, 2011
2. Oct 29, 2011

### HallsofIvy

I'm not sure what you mean by "find $y_n$". Certainly it is true that if you know n-1 (independent) solutions to an nth order linear d.e., you can reduce to a first order equation for the nth independent solution. But, of course, you may not be able to solve that first order equation!

I see many problems with your proposed method. For example, you talk about "functions continous on [a, b]". Where did "a" and "b" come from? There is no such interval given in the problem nor is there any reason to think a solution would be defined only on such an interval.

3. Oct 29, 2011

### AdrianZ

well, would you tell me how I can reduce the equation to a first order equation for the nth independent solution? Would that new ODE be linear? If yes, then why I may not able to solve that first order ODE?
Anyways, your idea seems nice. Please tell me how I can reduce my equation to a first ODE if I have (n-1) independent solutions.

well, you're right but a and b are not my problem right now to be honest because the inner product I've defined stays well-defined even If I define <f,g> = ∫f(x)g(x)dx. Am I right?
My idea was naive. I thought we have a nth order linear ODE and we know we have (n-1) vectors (linearly independent solutions), so If I find a vector (in this case a continuous function) that is perpendicular to all of them, then it surely is linearly independent and I'll have a basis for the linear space of its solutions. That was my idea.

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