Question

ques

Answer 1

How to prove that
\[\frac{1}{f(D)}e^{ax}=x.\frac{1}{f'(D)}e^{ax}=x.\frac{1}{f'(a)}e^{ax}\]
Where
\[D^n \equiv \frac{d^n}{dx^n}\]
and
\[f(D)=\sum_{i=0}^{n} k_{i}D^{n-i} \space \space \space ; \space \space \space k_{i}=1\]
or
\[f(D)=D^n+\sum_{i=1}^{n}k_{i}D^{n-i}\]
where
\[k_{1},k_{2},k_{3},.......,k_{n}\]
are arbitrary constants
The first part is easy....
\[f(D)e^{ax}=(D^n+\sum_{i=1}^{n}k_{i}D^{n-i})e^{ax}=(a^n e^{ax}+\sum_{i=1}^{n}k_{i}a^{n-i}e^{ax})\]\[f(D)e^{ax}=(a^n+\sum_{i=1}^{n}k_{i}a^{n-i})e^{ax}=f(a)e^{ax}\]\[\implies \frac{1}{f(D)}e^{ax}=\frac{1}{f(a)}e^{ax}\]

Answer 2

sorry I mean
\[k_{0}=1\]

Answer 3

@Nishant_Garg is this by any chance connceted to linear ODE's exponential resoponse formula?? because it looks awfully similar...

Answer 4

Yes, it is the same as finding particular integral or particular solution for a linear ODE of the form
\[f(D)y=e^{ax}\]

Answer 5

you have figured out already : \(\frac{1}{f(D)}e^{ax}=\frac{1}{f(a)}e^{ax}\)

right ?

Answer 6

I think, in ODE context that works when \(f(a)\) doesn't evaluate to \(0\)

Answer 7

Oh, yeah this is the case when f(a)=0
but where does that extra x come from?

Answer 8

if this is from the linear ODE... this link might be useful...I found it too difficult to follow.

http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/linear-operators-linear-time-invariance/MIT18_03SCF11_s17_6text.pdf

Answer 9

\(P(D)e^{ax} = P(a) e^{ax}\)

that is called substitution rule and it seems you have proved it already

Answer 10

but why is there an extra x in
\[x.\frac{1}{f'(D)}e^{ax}\]

Answer 11

familiar with exponential shift rule ?

Answer 12

exponential shift rule :
\[P(D) e^{ax}f(x) = e^{ax}P(D+a)f(x)\]

Answer 13

Nope, don't know about that

Answer 14

is this a question that was given to you/found in a text or are you just trying to figure out something...?

Answer 15

I'm studying ODE's as per my course and one of the methods they require is by using differential operator
They had given the proof for
\[\frac{1}{f(D)}e^{ax}=\frac{1}{f(a)}e^{ax}\]
and I following it easily too but for the second part when the case is f(a)=0, we must use the formula
\[\frac{1}{f(D)}=x.\frac{1}{f'(a)}e^{ax}\]
However there was no proof for this guy so I was curious but I think I only have upto 2nd order ODEs in my syllabus anyway, never heard of the "exponential shift rule" either

Answer 16

ohhh,
you cannot prove those things to be equal..because they are not! those two terms: the one with the extra 'x' and the one without are not equal !

Answer 17

they are two different solutions for two different types of diff equations

Answer 18

namely one where P(D) not equal to zero and other one being P(D)=0 but P' not equal to zero

Answer 19

ok, I think I'll just follow these like a formula then....thx :)

Answer 20

ya... they skipped the proof for a reason...have a look at that link i posted, its hard...

Answer 21

Yep, I'll come back to it when I have free time I guess, for now I only need the results lol

Answer 22

:)

Answer 23

this lecture has proofs for "everything"

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-13-finding-particular-sto-inhomogeneous-odes/

Answer 24

scroll directly to 25minute for exponential shift rule

Answer 25

i have seen it...its more of a verification for the second order ODE...in any case... proving the equality (the posted question) is not possible right?

Answer 26

why not ?
for sure, the solution \(e^{ax}\) is a guess. but we can always prove that particular solution works always by verifying it. that is a legitimate proof. recall that we do this all the time with epsilon delta proofs

Answer 27

what i mean is e^ax/P(a) and xe^ax/P'(a) are not equal. they are two particular solutions for two different ODE's (they differ in the way the constant co-efficients occur). We can prove that they are solutions to their repsective ODE...but their equality cannot be discussed because the P(D) are different for different ODE's

Answer 28

Right, they give two different cases depending on the value of P(a)

Answer 29

well... i was having trouble understanding the general exponential response formula XD.. I considered puttng it up on open study..anyways @ganeshie8 just have a look at that link I posted when you are free....I dont get step (4)...if it makes sense to you, help me!!

Answer 30

@baru
I tried a bit, but i didn't get how they are able to write q(D) in terms of powers of D-a ...
I'll give it a try later... please do let me know if you're able to nail it down :)

Answer 31

exactly my problem...i was hoping it was some binomial expansion related thing or something...nevermind. thanks for trying :)