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

• Screwdriver
In summary, the conversation discusses a subspace of Maps(R,R) generated by the functions sin(x) and cos(x), and a differential operator D:V \to V defined by D(y)=y''+y'+y for y E V. The goal is to 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. Through finding D(sin(x)) and D(cos(x)), it is shown that D maps elements of V to elements of V, and therefore Im(D) = V. This allows for the conclusion that any f can be written from a linear combination of y, where y is a variable representing a linear combination of

## 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?

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)).

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)).

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.

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?

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:

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