# Properties of the D operator

1. Jan 30, 2012

### DryRun

The problem statement, all variables and given/known data

The attempt at a solution
I have been trying to understand how to obtain the R.H.S. of each property from its L.H.S. but i can't find how, although i know that it's somehow related to differentiating the L.H.S. I am having a hard time to prove these properties, starting with the first one.

2. Jan 30, 2012

### I like Serena

What does the L function stand for?

If we disregard the L, or consider it the identity, you've got the differentiation rules for a couple of standard functions, combined with the application of the chain rule.

3. Jan 30, 2012

### DryRun

In my notes, L(D) is a function of the D operator.
Symbolically, a differential equation can be written in the form: L(D)y=f(x)

4. Jan 30, 2012

### I like Serena

The question remains what kind of function.
It doesn't seem to do anything useful.

Your differential equation L(D)y=f(x) would be the same as y'=f(x).
Or with other notations: $D_x y=f(x)$, or ${dy \over dx}=f(x)$.

5. Jan 30, 2012

### lanedance

is L just any function?
$$L(D)y(x) = f(x)$$

for example when $L(D)=D^2 +1$ you have
$$L(D)e^ax = D(De^{ax}) +e^{ax} = (aDe^{ax})+e^{ax}=(a^2+1)e^{ax}=L(a)e^{ax}$$

6. Jan 30, 2012

### DryRun

There's not much more explanation about the function L in my notes.

The Particular Integral, $y_p=\frac{1}{L(D)}f(x)$ and then there's a whole table of Inverse Operator Techniques. For example, $y_p=\frac{1}{L(D)}ke^{ax}$ gives $\frac{ke^{ax}}{L(a)}$, $L(a)\not=0$

It seems to me like the function L simply retains the value that has to be substituted into the function for D. For example: $L(D)=5D^2+3D+1$ where D=a=2 would give something like $L(2)=5(2)^2+3(2)+1$ but i don't know what kind of function it is.

Last edited: Jan 30, 2012
7. Jan 30, 2012

### micromass

There must be something given for L??

Is L linear?? A polynomial?? Given by a power series?? Continuous??