Linear Algebra DE

  • #1
129
0

Homework Statement



Let

[tex]V=span(sinx,cosx)[/tex]

be the subspace of Maps(R,R) generated by the functions sin(x) and cos(x), and let

[tex]D:V \to V[/tex]

be the differential operator defined by

[tex]D(y)=y''+y'+y[/tex] for [tex]y E V[/tex].

Show that [tex]Im(D) = V[/tex] and conclude that for every [tex]f E V[/tex], the differential equation

[tex]f=y''+y'+y[/tex]

has a solution [tex]y E V[/tex].

Homework Equations



Not really any, you need Euler's formula to solve the DE though.

The Attempt at a Solution



The differential equation doesn't make any sense to me in that form, so after some research into solving such things (I have never seen one before), I solved

[tex]0=y''+y'+y[/tex]

and obtained

[tex]y(f)= c_1e^{\frac{-f}{2}}sin(\frac{\sqrt{3}}{2}f)+ c_2e^{\frac{-f}{2}}cos(\frac{\sqrt{3}}{2}f)[/tex]

Which is pretty cool, but I'm not entirely sure if that helps me at all. I mean, it looks like a linear combination of things, which is good maybe. Also, can you just make it equal to zero like that?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
You don't want to solve the differential equation. That's just the solution of D(f)=0. It doesn't have much to do with V. You want to show D maps V onto V. Hint: find D(sin(x)) and D(cos(x)).
 
  • #3
Dick
Science Advisor
Homework Helper
26,260
619
You don't want to solve the differential equation. That's just the solution of D(f)=0. It doesn't have much to do with V. You want to show D maps V onto V. Hint: find D(sin(x)) and D(cos(x)).
 
  • #4
129
0
You don't want to solve the differential equation.

OK, thank you. Can I do anything at all with that solution other than say "Look at me, I solved this?"

Hint: find D(sin(x)) and D(cos(x)).

[tex]D(sinx)= (sinx)''+(sinx)'+sinx[/tex]

[tex]D(sinx)= -sinx+cosx+sinx=cosx[/tex]

[tex]D(cosx)= (cosx)''+(cosx)'+cosx[/tex]

[tex]D(cosx)= -cosx-sinx+cosx=-sinx[/tex]

So since V is just linear combinations of sines and cosines, and D maps sine and cosine to more sines and cosines, D maps elements of V to elements of V and Im(D) = V.

Then I can conclude that there is a solution for all elements of V by noting that, since D maps any element of V to another element of V, any f can be made from linear combinations of y.
 
  • #5
9
0
How does [cos(x),-sin(x)] = span[sin(x),cos(x)]?
and also… i think it is a bit quick to say that because of this statement any f can be written from a linear combination of y..is y the function?
 
  • #6
129
0
How does [cos(x),-sin(x)] = span[sin(x),cos(x)]?

Well the span of sin(x) and cos(x) is all linear combinations of sin(x) and cos(x), and

[tex]D(\lambda sinx)=\lambda cosx[/tex]

[tex]D(\lambda cosx)=-\lambda sinx[/tex]

Which is a linear combination of sin(x) and cos(x).

i think it is a bit quick to say that because of this statement any f can be written from a linear combination of y..is y the function?

y is the variable like f(x) = x^2, only instead of x^2 you have a linear combination of derivatives of y.

Keep in mind, I have no idea if any of that is true...I only learned what a DE was this morning on the internet :rofl:
 
Last edited:

Related Threads on Linear Algebra DE

  • Last Post
Replies
5
Views
2K
Replies
9
Views
765
  • Last Post
Replies
6
Views
2K
Replies
3
Views
3K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
5
Views
912
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
6
Views
8K
Replies
1
Views
5K
Top