Homework Help: Higher order linear equations- ODEs

1. Feb 24, 2010

Roni1985

1. The problem statement, all variables and given/known data

Verify that the differential operator defined by
L[y] = y(n) + p1(t)y(n−1) +· · ·+ pn(t)y

is a linear differential operator. That is, show that
L[c1y1+ c2 y2] = c1L[y1] + c2L[y2],

where y1 and y2 are n times differentiable functions and c1 and c2 are arbitrary constants.
Hence, show that if y1, y2, . . . , yn are solutions of L[y] = 0, then the linear combination c1y1+· · ·+cnyn is also a solution of L[y] = 0.

2. Relevant equations

3. The attempt at a solution

I think I don't understand what I need to find.
What's the question here ?
First I need to verify that it's a linear differential operator.
Next, I need to show that
L[c1y1+ c2 y2] = c1L[y1] + c2L[y2]

and then, there is another part ?
does this question have three parts ?

I don't really understand how to approach this question.

Would appreciate any help.

Thanks,
Roni.

EDIT:

ohh, to show that it's a linear differential operator, I just need to show that this one is true:
L[c1y1+ c2 y2] = c1L[y1] + c2L[y2]

correct ?

now I'm trying to understand the second part of the question

Last edited: Feb 24, 2010
2. Feb 24, 2010

Staff: Mentor

Re: higher order linear equations

What have you tried?

3. Feb 24, 2010

Roni1985

Re: higher order linear equations

Hello,

I edited my post....
I am still trying to understand the question...

4. Feb 24, 2010

Staff: Mentor

Re: higher order linear equations

There are two parts.
Show that L is a linear differential operator.
Show that if y1, y2, ..., yn are solutions of L[y] = 0, then any linear combination of the yi's is also a solution of L[y] = 0. IOW, show that L[c1y1 + c2y2 + ... + cnyn] = 0.

They're both pretty straightforward, especially if you keep in mind some fo the basic principles of differentiation; namely, (f + g)'(x) = f'(x)+ g'(x) and (cf)'(x) = cf'(x).

5. Feb 24, 2010

Roni1985

Re: higher order linear equations

oh right, I just had to read the definition :\

thank you very much for your help, I appreciate it.

6. Feb 24, 2010

Staff: Mentor

Sure, you're welcome! A thank you goes a long way!