Solving Linear Algebra DE: Im(D)=V, f=y''+y'+y

[Screwdriver]

Homework Statement

Let

$$V=span(sinx,cosx)$$

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

$$D:V \to V$$

be the differential operator defined by

$$D(y)=y''+y'+y$$ for $$y E V$$.

Show that $$Im(D) = V$$ and conclude that for every $$f E V$$, the differential equation

$$f=y''+y'+y$$

has a solution $$y E V$$.

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

$$0=y''+y'+y$$

and obtained

$$y(f)= c_1e^{\frac{-f}{2}}sin(\frac{\sqrt{3}}{2}f)+ c_2e^{\frac{-f}{2}}cos(\frac{\sqrt{3}}{2}f)$$

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?

 

[Dick]

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)).

 

[Dick]

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)).

 

[Screwdriver]

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)).

$$D(sinx)= (sinx)''+(sinx)'+sinx$$

$$D(sinx)= -sinx+cosx+sinx=cosx$$

$$D(cosx)= (cosx)''+(cosx)'+cosx$$

$$D(cosx)= -cosx-sinx+cosx=-sinx$$

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.

 

[knotman2]

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?

 

[Screwdriver]

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

$$D(\lambda sinx)=\lambda cosx$$

$$D(\lambda cosx)=-\lambda sinx$$

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: