Higher order linear equations- ODEs

[Roni1985]

Homework Statement

Verify that the differential operator defined by
L[y] = y⁽ⁿ⁾ + p₁(t)y⁽ⁿ⁻¹⁾ +· · ·+ p_n(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, . . . , y_n are solutions of L[y] = 0, then the linear combination c₁y₁+· · ·+c_ny_n is also a solution of L[y] = 0.

Homework Equations

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

 

[Mark44]

Homework Statement

Verify that the differential operator defined by
L[y] = y⁽ⁿ⁾ + p₁(t)y⁽ⁿ⁻¹⁾ +· · ·+ p_n(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, . . . , y_n are solutions of L[y] = 0, then the linear combination c₁y₁+· · ·+c_ny_n is also a solution of L[y] = 0.

Homework Equations

The Attempt at a Solution

What have you tried?

 

[Roni1985]

What have you tried?

Hello,

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

 

[Mark44]

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

 

[Roni1985]

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

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

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

 

[Mark44]

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