Question

Perform the multiplication

Answer 1

\[(\text D - x)(\text D + x)\]

Answer 2

Note: this is differential equations NOT algebra

Answer 3

mm.. And D are differential operators?

Answer 4

yup

Answer 5

Do they have a concrete expression?

Answer 6

\(D^2 + Dx - xD - x^2\)

Answer 7

Dx = 1 in this case..

Answer 8

Why?

Answer 9

why is D^2 not operating on anything ?!?!

Answer 10

D is given as:
\[\frac{d}{dx}\]

So,

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

Answer 11

it'd be \[\frac{d}{dx}[1]\]

Answer 12

And what does that mean x*d/dx?

Answer 13

sir can't it be\[(d^2-x^2)\]

Answer 14

@lgbasallote sir.

Answer 15

@jiteshmeghwal9 it is not algebra lgba has told it already..

Answer 16

i wants to understand differential equations:)

Answer 17

im thinking it's (D^2 - x^2) too lol but i dont think that's what happens in d.e.

Answer 18

Yes..

Answer 19

Even I too need to understand DE.. @jiteshmeghwal9

Answer 20

can anybody teach me differential equations????????
@lgbasallote sir,plz

Answer 21

\(((D+x)(D-x))y = (D+x)((D-x)y) = (D+x)(Dy - xy) = D^{2}y - D(xy) + xD(y) - x^{2}y\)

\(D(xy) = xD(y) - y\text{ So, the expression boils down to}\)

\((D^{2}-x^{2}+1)y\)

Answer 22

that would make sense Fool as it doesn't make sense for a differential operator to be by itself

Answer 23

Dx would be \[\frac{dx}{dt}\] @rafabc02

Answer 24

Oh! So D=d/dt and not d/dx, right?

Answer 25

That \(y\) is necessary because \(D\) is an operator and it needs to be operated on something. And to see the result of multiplication of operators, you need to operate it on a variable.

Answer 26

@jiteshmeghwal9 im still learning this thing lol

@FoolAroundMath your explanation got cut

Answer 27

ok i got it..but can i also say \[( \text D + x)(\text D - x)y \implies (\text Dy + xy)(\text D - x)\]
or no?

Answer 28

\((D-x)(D+x)y = (D-x)(Dy + xy) = (D^{2}(y) + D(xy) - xD(y) - x^{2}y\)

\(D(xy) = xD(y) + y \)

\(\text{Final answer: }(D^{2}-x^{2}+1)y\)

Answer 29

@rafabc02 Dx is the derivative of the function x in respects to what x=f(?)... I was thinking of system of equations when you have two functions

Answer 30

parameters i guess you could say

Answer 31

I think that FoolingAroundMath used that D=d/dx, because of this:
D(xy)=xD(y)+y
There he used the product rule, right?

Answer 32

And he used that D(x)=1

Answer 33

could someone answer that question i asked :S lol

Answer 34

Yes. I assumed that \(D = d/dx \) and then used product rule.

Answer 35

I think that you can't do that, @lgbasallote

Answer 36

okay thanks :D

Answer 37

Assume???

D is defined for \(\frac{d}{dx}\) always..

Answer 38

@lgbasallote I suggest you that you should confirm the answer from @UnkleRhaukus , he is master in DE he can surely give his point in this..

Answer 39

okay i'll wiat

Answer 40

which needs confirmation?

Answer 41

idk,. @waterineyes ?

Answer 42

Yes I need confirmation..

Answer 43

@UnkleRhaukus please perform the multiplication for this:

\[(D - x)(D+x)\]

Answer 44

i am no master of DE's.. yet
but i get this
\[(D - x)(D+x)=D^2+Dx-xD-x^2\]\[\qquad\qquad\qquad\quad=D^2+1-xD-x^2\]

\[(D - x)(D+x)y=D^2y+Dxy-xDy-x^2y\]\[\qquad\qquad\qquad\quad=D^2y+xDy+y-xDy-x^2y\]\[\qquad\qquad\qquad\quad=D^2y+y-x^2y\]

Answer 45

I also got the same..
Thanks Uncle Rocks always Rocks...