# Differential operator

1. Oct 17, 2005

### asdf1

why is the answer to a differential operator the same as the answer to the original O.D.E. equation?

2. Oct 17, 2005

### HallsofIvy

That question doesn't even makes sense. A "differential operator" is neither an equation not a "problem" and doesn't have an answer!

3. Oct 18, 2005

### dextercioby

I think he means the "fundamental solution" of the differential operator.

Daniel.

4. Oct 18, 2005

### asdf1

for example:
(D^2-d-2)y=0
if you solve D, which is D=2,-1
it's the same as solving the equation~

5. Oct 19, 2005

### dextercioby

In the light of your latest discoveries, your question still doesn't make too much sense.

Daniel.

6. Oct 19, 2005

### HallsofIvy

Actually, I would argue that that doesn't make sense either- an operator is not an equation. An equation may have a solution, but not the operator!

Okay, I can understand that, although your terminology is still odd!
I presume you are referring to the differential equation:
(D2- D- 2)y= 0 where "D" is the differential operator d/dx (or d/dt). D2- D- 2 would also be a linear differential operator.

However, solving the equation D2- D- 2= 0 to get D= 2 or D= -1 is what mathematicians call "abuse of terminology". If you intend D to be "d/dx", it clearly doesn't make sense to turn around and say that D= 2!
It is, though, convenient shorthand and we do it all the time. It is convenient shorthand for the "characteristic equation". If we were to look for a solution of the form $y= e^{\lambda x}$, putting that into the equation would give $\lambda^2 e^{\lambda x}- \lambda e^{\lambda x}- 2e^{\lambda x}= 0$ or $\left(\lambda^2- \lambda- 2\right)e^{\lambda x}= 0$. Since $y= e^{\lambda x}$ is never 0, we must have $\lambda^2- \lambda- 2= 0$, the characteristic equation.
That "looks like" the original equation, especially in operator notation (which is the main reason for using it) because of that very nice property of exponentials:
$$\frac{d^n(e^{\lambda x}}{dx^n}= \lambda^n e^{\lambda x}$$

Caution!! This is only true for a very limited (though important) class of differential equations: linear equations with constant coefficients.

Last edited by a moderator: Oct 19, 2005
7. Oct 19, 2005

### asdf1

sorry, i didn't make myself clear~
for the example above, usually the normal way is to suppose that
y=ce^(namda)x
and find namda
but the answer to namda is the same as finding the answer to the differential operator~

8. Oct 21, 2005

### HallsofIvy

Yes, that was what I just said. Oh, by the way, the Greek letter is "lambda", not "nambda".