Question

Solve the DE:
If G = D(xD-1) and H = (xD-1)(xD+2)
Determine GH and HG.

Answer 1

@hartnn

Answer 2

what does D stand for in this context. the derivative ?

Answer 3

yes :)

Answer 4

D stands for Differential Operator

Answer 5

i might need some theory, maybe you could post me the name of your book or something

Answer 6

we don't have a book.. just handouts...
the topic is about Differential Operator

Answer 7

openstudy says errors

Answer 8

Here, @perl http://prntscr.com/4wnm9o

Answer 9

one example

Answer 10

http://prntscr.com/4wnncd

Answer 11

@amistre64 pls. help me ..

Answer 12

its better if you rewrite your original expression

Answer 13

These are linear operators, and they don't commute in general.

For example, does multiplying by x and taking the derivative as two operators commute? Well, let's try to do them on some random function y and see what happens:

First, the operators x with a hat just means multiply by x. D with a hat means take the derivative.

\[\LARGE \hat x (\hat D(y))=\hat x \frac{dy}{dx}=x*\frac{dy}{dx}\]
Now let's reverse the order \[\LARGE \hat D ( \hat x(y))= \hat D(x *y)=1*y+x \frac{dy}{dx}\] Since \[\LARGE x y' \ne y + x y'\] then we say that these operators don't commute since order of operation mattes! This is one of the main rules that makes operator algebra different than normal algebra. So now when you look at G and H you have to be careful when you multiply them together than you keep the multiplication straight, otherwise it will be wrong. You should expect GH to not be equal to HG, so if you get the same answer for both you might be doing something wrong.

Give it your best shot, and I'll help you finish it. =)

Answer 14

One last note, if I slightly change how I wrote it so it looks more like what you were give it might be more clear to you. So if I strip away all the stuff we were talking about and just leave the operators, it will look like this: \[\LARGE xD \ne Dx\]

Answer 15

i'll post my sol'n later... I hope you'll help me understand this.. thanks a lot!!!!!! later..

Answer 16

\[G=D(xD-1) \]\[H=(xD-1)(xD+2)\]
\[GHy=D(xD-1)[(xD-1)(xD+2)(y)]\]

I AM STOCK,..

how can I multiply (xD-1)(xD+2)(y)??

Answer 17

@Kainui

Answer 18

Just distribute like you normally would. As an example:

(A+B)(A-B)= A^2+BA-AB-B^2

The middle two terms don't cancel because we can't say BA=AB like we usually do in commutative algebra.

Answer 19

\[x^2 +2xD - xD - 2\] ?

Answer 20

\[xD(xD+2) - xD - 2 \]

??
@ Kainui

Answer 21

@Kainui

Answer 22

@amistre64 ?

Answer 23

im not familiar with the D notation, ive seen it around, but i just dont have the practice needed to make it set in my head.

Answer 24

Dy means, dy/dx ? or something else

Answer 25

got a sideways picture lol yeah, Dy means dy/dx

Answer 26

G = D(xD-1)
H = (xD-1)(xD+2)

Determine GH and HG.

is that G(Hy) and H(Gy)

Answer 27

yes.. :)

Answer 28

I wonder xD-1 means
(or xD+2)
I know D means d/dx
but I don't understand what are we taken the derivative of since there is nothing to follow the D.

Answer 29

Gy = Dy (x Dy-1)
Hy = (x Dy-1)(x Dy+2)

Dy = y' sooo

Gy = y' (x y'-1)
Hy = (xy'-1)(xy'+2)
----------------------

G(Hy) = D (x D-1) (xy'-1)(xy'+2)

G(Hy) = (xD^2 - D) (xy'-1)(xy'+2)

G(Hy) = xD^2[(xy'-1)(xy'+2)] - D[(xy'-1)(xy'+2)]

Answer 30

G(Hy) is essentially: x times the second derivative of Hy, minus the first derivatrive of Hy

Answer 31

\[D[(xy'-1)(xy'+2)] \]
\[(xy'-1)'(xy'+2) + (xy'-1)(xy'+2)'\]
\[(x'y'+xy''-1) (xy'+2) + (xy'-1)(x'y'+xy''+2)\]
and the second derivative is just one more after that

Answer 32

pfft, -1 to 0 and +2 to 0

Answer 33

\[(x'y'+xy'') (xy'+2) + (xy'-1)(x'y'+xy'')\]
\[(y'+xy'') [(xy'+2) + (xy'-1)]\]
\[D(Hy)=(y'+xy'') (2xy'+1)\]

Answer 34

so, does this make sense?

Answer 35

why we must derive Hy?

Answer 36

G(Hy) is essentially: x times the second derivative of Hy, minus the first derivatrive of Hy

Answer 37

use Dy instead of y' pls? im confused......

Answer 38

youll have to do the modifications, im using what makes the most sense to me ...

Answer 39

okay.. ill try... sorry :)

Answer 40

just change the y' to Dy, y'' to D^2y etc ...

but as far as processing it, im gonna have to use the prime notation to make sure i dont mess up

Answer 41

Gy= Dy(xDy-1)
why not
Gy= Dy(xDy-y) ? :)

Answer 42

hmmm, let me redo that .....

G = D(xD-1)

G = xD^2 - D

Gy = xD^2y - Dy

Gy = D(xDy - y)

D is not a variable, its an operator. we need to treat it like an operator then.

I most likely misthought Gy to start with ... lets chk up my Hy :)

Answer 43

H = (xD-1)(xD+2)

H = x^2 D^2 +xD -2

Hy = x^2 D^2y +xDy -2y

Hy = xD(x Dy +y) -2y

Answer 44

Gy = D(xDy - y)

does this simplify to:

GHy = D(xDHy - Hy) ??

Answer 45

D(xD-1)
product rule?
=(xD^2 + D - D)
=xD^2
?

Answer 46

\[D(xD-1)\]
\[\frac d{dx}(x\frac d{dx}-1)\]
\[x\frac {d^2}{dx^2}-\frac d{dx}\]

Answer 47

assuming of course that im catching on to this stuff ... correct me as you see fit

Answer 48

explain why you want to use a product rule

Answer 49

x'D + xD' ?

Answer 50

d/dx is not a variable (or a function), its an operator. it only has a definition when it actually operates on something

d/dx of y = dy/dx

d/dx of x^2 = 2x dx/dx = 2x

d/dx of ax^2 = da/dx x^2 + a 2x dx/dx

i see why you are wanting to apply it to x * d/dx but im just not sure if thats applicable. can you show me a rule or thrm for it?

Answer 51

uhhmm that's what my prof. taught us..... :/

Answer 52

D (xD-1) y

D(x Dy - y)

Dx Dy + xD^2y - Dy

Dx = 1 assuming D = d/dx

Dx Dy + xD^2y - Dy

Dy + xD^2y - Dy

xD^2y = xy''

if we use the y to define the operator as a function, then i can see it as that

Answer 53

(xD-1) Dy

xD^2y -Dy = xy'' - y' so commutability is not gaurenteed with out constant stuff to me

Answer 54

D(xy' - y)

(x'y + xy'' - y')

yeah, im going to need alot more practice at this then i currently have to be of any use in these kinds of things :/

Answer 55

:) me too haha
it is late already... need to sleep
BIG THANKS TO YOU!!!
hope I can finish answering this tomorrow..
night! :)

Answer 56

good luck, if i can review this today i will and if a come up with anything profound .... ill try to let you know :)

Answer 57

Nonononono we are not taking any derivatives, we are purely doing operator algebra! We are never once taking an actual derivative. The only reason I showed derivatives was to justify that operator algebra is not commutative. Everything else is basically the same, and if you decide to, you can operate on functions with them later.

So I guess I'll post the answer since this has gone on too long.

Answer 58

If G = D(xD-1) and H = (xD-1)(xD+2)
Determine GH and HG.

So let's distribute out G and H separately
\[\large G=D(xD-1)=D(xD)-D=D+xD^2-D=xD^2\] See how I did the "product rule" here? We can test it by running a function through it to make sure:
\[\large D(xD-1)y=D(xy'-y)=y'+xy''-y'=xy''=xD^2y\]
In fact, we can use "running a function through" as our method of simplifying it and just "factor out" the differentiation operator like I did in the last step. Remember when I said earlier that \[\LARGE xD \ne Dx\] you should be incredibly cautious and go up and reread my explanation above, as this is key, especially since xD shows up here.

So now, we need to simplify H. I'll save that for the next post.

Answer 59

Ok you should have attempted the next one on your own so that you can make sure you're doing it correctly. This is also for @amistre64 so you can learn too! =D

\[\large H=(xD-1)(xD+2)=xD(xD)+xD2-xD-2\] since differentiation is a linear operator, multiplying by a scalar actually commutes with everything. So you can safely say xD2-xD=xD. Just as a quick reminder that we're talking about derivatives, here's an example of this: \[\LARGE \frac{d}{dx}(3x^2)=3 \frac{d}{dx}(x^2)=6x\] See how we can slide scalars around without consequences? Coolio. Now notice we have the same D(xD) we had last time, but now we have an x in there. This is just multiplying afterwards, still nothing crazy going on, it's only derivatives that should give you some hesitation. So now we have as our final result for H:
\[\LARGE H=x^2D^2+2xD-2\]

Now we can multiply them together the same route we've been going.

Like I said earlier, you can just take an arbitrary function y, find Hy, then find GHy and after you have simplified it factor out the D's from the y's and throw the y's away to get the GH operator. Rinse and repeat for HG.

Answer 60

\[G=xD^2 \]
\[Hy=x^2D^2y+2xDy-2y \]
\[GHy= xD^2 (x^2D^2y+2xDy-2y)\]
\[=xD(x^2D^3y+2xD^2y+2xD^2y+2Dy-2Dy)\]
\[=xD(x^2D^3y+4xD^2y)\]
\[=x(x^2D^4y+2xD^3y+4xD^3y+4D^2y)\]
\[GHy=x^3D^4y+6x^2D^3y+4xD^2y \]
\[GH=x^3D^4+6x^2D^3+4xD^2\]

???

Answer 61

\[HGy=(x^2D^2+2xD-2)(xD^2y)\]
\[=x^2D^2(xD^2y) + 2xD(xD^2y)-2xD^2y\]
\[=x^2D(xD^3y+D^2y)+2x(xD^3y+D^2y)-2xD^2y\]
\[=x^2(xD^4y+D^3y+D^3y)+2x^2D^3y+2xD^2y-2xD^2y\]
\[=x^3D^4y+2x^2D^3y+2x^2D^3y \]
\[HGy=x^3D^4y+4x^2D^3y \]
\[HG=x^3D^4+4x^2D^3\]
???

Answer 62

did i do it right? @Kainui

Answer 63

@abdullah1995 hi can you help me? :)

Answer 64

I am sorry but either i don't understand what's going on or the math is too advanced for me :c

Answer 65

it's alright.. :) hope you'll learn from this too :D

Answer 66

indeed. what university year is this question ?

Answer 67

damn lol. i just started university like 2 months ago for electrical and computer engineering

Answer 68

@Kainui ??

Answer 69

hmm... THANK YOU VERY MUCH @Kainui !!!