# Homework Help: Differential Eq. Proof

1. May 24, 2005

How would I prove that $(D-r)^m$ annihilates $x^ke^{rx}$ where $k=0,1,...,m-1$.

At first I tried to do it as a computation by expanding the whole the thing out but that turned out to be a nightmare. After that I went down a different path that seems promising but I can't seem to figure it out. Here's what I have so far:

$$(D-r)(x^ke^{rx})=D(kx^{k}e^{rx})-rkx^{k}e^{rx} =kx^{k-1}e^{rx}$$
and
$$(D-r)^m=(D-r)^{m-1}(D-r)$$

I think I'm on the right track here but I'm seriously stuck. I just need a little nudge in the right direction. I appreciate any help.

John

2. May 24, 2005

### Hurkyl

Staff Emeritus
induction...

(Too bad there isn't a whisper font)

3. May 25, 2005

Unfortunately we never learned induction and the instructor knows this...don't ask me how we did series without it.

I did a little reading and from what I gather you prove the case for n=1, then assume the theorem is true for n-1 and doing some algebraic manipulation you can find that the two are equal to each other.

We're trying to prove that
$$(D-r)^n(x^{n-1}e^{rx})=0$$

The first step is easy to do. (n=1)
$$(D-r)^1(x^0e^{rx})=D(e^{rx})-re^{rx}=0$$

Then we assume that
$$(D-r)^{n-1}(x^{n-2}e^{rx})=0$$

Then since
$$(D-r)^{n}(x^{n-1}e^{rx})=(D-r)^{n}(x^{n-2}xe^{rx})=(D-r)^{n-1}x^{n-2}(D-r)xe^{rx}$$
and
$$(D-r)xe^{rx}=e^{rx}$$
so
$$(D-r)^{n-1}x^{n-2}(D-r)xe^{rx}=(D-r)^{n-1}x^{n-2}e^{rx}$$
which we assumed was zero.

Is this right? I understand the process but I have a question. How do I know that my assumption is a good assumption? Is there another step I have to do to complete the proof?

BTW, is there another proof without using induction? The instructor gave us some hints and that's what led me to what I have in my original post but I'm not sure what to do beyond that.

Thanks again for the help.

4. May 25, 2005

### Hurkyl

Staff Emeritus
$$(D-r)^{n}(x^{n-1}e^{rx})=(D-r)^{n}(x^{n-2}xe^{rx})=(D-r)^{n-1}x^{n-2}(D-r)xe^{rx}$$

This isn't right -- you can't split the operand up like that.

At least, unless you've managed to prove:

$$(D-r)x^{n-1}e^{rx} = x^{n-2} (D-r)xe^{rx}$$

Also, your original work is wrong too.

$$(D-r)(x^ke^{rx})=D(kx^{k}e^{rx})-rkx^{k}e^{rx}$$

Where'd all those k's come from?

5. May 27, 2005

I think I was a little too liberal with copy and pasting in the original.

$$(D-r)(x^ke^{rx})=D(x^ke^{rx})-rx^ke^{rx}=kx^{k-1}e^{rx}$$

I'm not sure why I can't split up the operand. I tried to get it to work but I couldn't so I started down a different path along my original line of thinking.

$$(D-r)^{m-1}(D-r)x^ke^{rx}$$
and since
$$(D-r)(x^ke^{rx})=kx^{k-1}e^{rx}$$
we can substitute
$$(D-r)^{m-1}kx^{k-1}e^{rx}$$
and if you keep going
$$(D-r)^{m-2}kx^k(k-1)e^{rx}$$
and if you substite m-1 for k
$$(D-1)^{m-2}(m-1)x^{m-1}(m-2)e^{rx}$$
Now this does annihilate the function for any m but I'm not sure how to actually prove it.

I'm just no good with proofs...I appreciate the help, at least it feels like I'm making some progress.

6. May 27, 2005

### Hurkyl

Staff Emeritus
But you can't give a reason why you can either.

Here's a trivial example:

D2(x2) = D(x) D(x)

Do you know the basic form of an induction proof?

7. May 27, 2005

No I never learned induction so I was trying my best with what I had. And it should have been x^(m-2) in my previous post. For splitting the operand up, I just factored a (D-r) out and made an assumption, which may have been false, that (D-r) is commutative. Plus it just seemed to work out nicely I spent a lot of time trying to prove that you could split it up from your suggestion that I prove
$$(D-r)x^{n-1}e^{rx} = x^{n-2} (D-r)xe^{rx}$$
but I haven't had any luck.

I'm sorry, it's probably frustrating seeing me fumbling around with this problem. Maybe if I had learned induction...

8. May 27, 2005

### Hurkyl

Staff Emeritus
The basic strategy is this:

(base case) You prove your theorem for the smallest case.

(inductive step) Then, you prove the general case by assuming it's true for the previous cases.

This proves it's true for every case, because if there was a counterexample, there would be a smallest counterexample. But if there was a smallest counterexample, the inductive step would prove it was true for that case!

(Of course, this is assuming the cases are parametrized by positive integers, or something sufficiently similar)

9. Jun 1, 2005

I've been working on this for the past couple of days and I'm just going around in circles.

I'm trying to prove
$$Q(m)=(D-r)^m x^{m-1} e^{rx}=0$$
So using induction I prove this for m=1
$$Q(1)=(D-r)^1 x^0 e^{rx}=D(e^{rx})-re^{rx}=0$$
Then I assume that the following is true
$$Q(m-1)=(D-r)^{m-1} x^{m-2} e^{rx}=0$$
Then prove that Q(m) is true
$$Q(m)=(D-r)^{m} x^{m-1} e^{rx}=0$$

So I tried this...
$$Q(m)=(D-r)^{m-1} (D-r) x^{m-1} e^{rx}=(D-r)^{m-1} (m-1) x^{m-2} e^{rx}$$
but that doesn't get me anywhere.

This is basically what I did the first time but I'm just spinning my wheels after this. What do I do now?

10. Jun 2, 2005

### Hurkyl

Staff Emeritus
Well, can you see any way to turn

$$(D-r)^{m-1} (m-1) x^{m-2} e^{rx}$$

into something with

$$(D-r)^{m-1} x^{m-2} e^{rx}=0$$

in it?

11. Jun 3, 2005

So I could just do this
$$(D-r)^{m-1} (m-1) x^{m-2} e^{rx}=(m-1)(D-r)^{m-1} x^{m-2} e^{rx}$$

Then I just say
$$(D-r)^{m-1} x^{m-2} e^{rx}=0$$
so
$$(m-1)(D-r)^{m-1} x^{m-2} e^{rx}=0$$
right?

Is that all I have to show? The original problem used $k=0,1,...,m-1$. Does this prove it for all the other cases too?

12. Jun 3, 2005

### Hurkyl

Staff Emeritus
That looks good to me!

13. Jun 3, 2005