• Support PF! Buy your school textbooks, materials and every day products Here!

How to prove this theorem about linear ODE's?

  • Thread starter AdrianZ
  • Start date
  • #1
319
0

Homework Statement


Suppose that we have the following linear ODE:
[tex]y^{(n)} + a_{n-1}y^{(n-1)} + a_{n+2}y^{(n-2)} + ... + a_2y'' + a_1y' + a_0y = 0[/tex]
Prove that if we have (n-1) linearly independent {y1,...,yn-1} solutions then we can find yn.

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:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,808
933
I'm not sure what you mean by "find [itex]y_n[/itex]". 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
319
0
I'm not sure what you mean by "find [itex]y_n[/itex]". 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!
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.

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.
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.
 

Related Threads on How to prove this theorem about linear ODE's?

  • Last Post
Replies
3
Views
639
  • Last Post
Replies
1
Views
823
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
2
Views
802
  • Last Post
Replies
2
Views
648
  • Last Post
Replies
0
Views
858
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
956
  • Last Post
Replies
10
Views
2K
Top