Question

Perform the multiplication:

Answer 1

(xD - 1)D

Answer 2

Note: this is differential equations NOT algebra

Answer 3

0

Answer 4

lol

Answer 5

is it rdrr ?

Answer 6

i dont know. how did you do it?

Answer 7

rdrr is a mnuemonic for har dee har har

Answer 8

lol. seriously do you know how to do this?

Answer 9

im not sure what you mean by multiply differential

Answer 10

oh :( it's operations on higher order linear DE thingy

Answer 11

\[\frac{dy}{dx}\frac{dy}{dx}=(\frac{dy}{dx})^2\]

Answer 12

seems beyond my current knowledge :/

Answer 13

aww :( all i know is it's not equal to D(xD - 1)..other than that im stuck

Answer 14

This is a TROLL-QUESTION!!!!!! did you notice that ' xD ' there?? :O

Answer 15

loll

Answer 16

D means differential operator :p x is a variable

Answer 17

(xD-1)(D) =\(xD^2-D\) .. good question :D

Answer 18

lol it's not that simple :p

Answer 19

D is a differential operator that is : \(D = \frac{d}{dx}\)..

xD can be written as :

\[x \frac{d}{dx} \implies = \frac{dx}{dx} = 1\]

Taken the x inside..

So,

(xD - 1)D

(1 - 1)D

= 0

I think so..

Answer 20

so what would be D(xD - 1)?

Answer 21

Solve inner one first..

xD you can write it as Dx..

Dx = 1

So,

\[\frac{d}{dx}(1 - 1) = 0\]

Answer 22

D(xD - 1) shouldnt be equal to (xD - 1)D

Answer 23

so which one is wrong in your solution?

Answer 24

See, I think (xD - 1)D is wrong actually.

This should be D(xD - 1)

See the difference:

\[(x \frac{d}{dx} - 1) \frac{d}{dx}\]

Does it make sense??

Or,

\[\frac{d}{dx}(x \frac{d}{dx} - 1)\]

This also does not make sense but as variable x can be taken inside so,

\[\frac{d}{dx}( \frac{dx}{dx} - 1)\]
This makes sense.

Answer 25

uhmm what's your point?

Answer 26

i really dont see what you mean

Answer 27

D(xD - 1) is right in my point..

Answer 28

so (xD - 1)D is?

Answer 29

See, you can take (xD - 1) inside;

So it will become:

D(xD - 1)

Answer 30

\[D(xD - 1) \ne (xD - 1)D\]
according to that rule thingy... i dont get what you mean @waterineyes

Answer 31

Call @UnkleRhaukus , he will give some opinions on this..

Answer 32

D and xD - 1 do not commute

Answer 33

D(xD - 1) = D + xD^2
(xD-1)D = xD^2 - D

Answer 34

hmm so it becomes \[x \frac{d^2 y}{dx^2} - \frac{dy}{dx}?\]

Answer 35

if D is defined as d/dx then ... that is the last one. all right gotta go for a while.

Answer 36

aww..i need to know how you got it :(

Answer 37

D(xD - 1) = D(xD) - D(1) = D(xD) <-- chain rule =>D(x) D + x DD = D + x D^2

(xD - 1)D = xD(D) - 1(D) = xD^2 - D

Answer 38

D(1) = 0

Answer 39

OHHHH i see

Answer 40

\[\qquad(x\text D - 1)\text D=x\text D^2-\text D\]
\[\left( x\frac{\text d}{\text dx} -1\right)\frac{\text d}{\text dx}=\left( x\frac{\text d^2}{\text dx^2} -\frac{\text d}{\text dx}\right)\]

Answer 41

how do you make the parentheses big? using array?

Answer 42

\left( \right)