Question

Need help in this Operator problems.

1. D^2(D-3)^2(D+3)y = e^-3x
2. (D^2 - 1)y = 2e^x
3. (D^3 - D)y = 2x
4. (D^2 + 6D +25)y = 8e^-2x (sinx)

Answer 1

oh. forgot to tell that I need to find the particular solution...

Answer 2

1. \[D^2(D-3)^2(D+3)y_p = e^{-3x}\\\,\\
y_p = \frac1{D^2(D-3)^2(D+3)}e^{-3x}\\\,\\\quad=\]

Answer 3

but sir.. how do you do things in there? I only know D+a and subtracting it to it's exponent. But there's like multiple operator in there..

Answer 4

should I add the exponents of the operators like D^3 * D^2 = D^5??? o.o

Answer 5

\[= \frac1{D^2}\frac1{(D-3)^2}\frac1{(D+3)}e^{-3x}\\\\=\]

Answer 6

so should I go one by one?

Answer 7

can I do this? \[(\int\limits e^{-3x})[3(\int\limits e^{-3x}) ]\]

Answer 8

What operation is D?

Answer 9

what do you mean sir? o.o

Answer 10

Is D a variable, what is it?

Answer 11

D is an operator which means "the derivative" of something.

Answer 12

What is D+a? How do you add differentiation to a constant?

Answer 13

it's under a certain topic.. you don't really add it.. it's just a given equation. o.o

Answer 14

So what is the definition?

Answer 15

err, Sire. I thought you should've known this operator D topic already... o.o

Answer 16

The point is to get you thinking about it.

Answer 17

Treat \(D\) like a variable and expand out the expression into a polynomial.

Answer 18

You can use commutative, associative, even distributive properties on this operator.

Answer 19

...okay. :<
I ain't asking for anything big, I'm looking for someone to guide me.
I really clueless when something I don't know is added.
Like those equations above.
I only know this things like:
\[\frac{ 1 }{ (D-3)^3 } = 12x^{-3}e^{3x}
'cause there's only one equation for D, which is D-3 cube..
._.

Answer 20

This?\[
\frac 1 {(D-3)^3}=12x^{-3}e^{3x}
\]

Answer 21

Yes. o.o

Answer 22

Okay so what do the other equations add which you don't quite get?

Answer 23

those above. ._.

Answer 24

Yes, but what aspect do they add which you don't get? What makes them different?

Answer 25

Have you ever considered, that the fact that they're all factored polynomials means that the advantage of using D operator, is that finding the roots will give insight?

Answer 26

if that's the case.. well.. I've been thinking of that too. I'm just too afraid that I might not get the real answer.

Answer 27

Okay, here is the skinny.
I don't know how to do it, so I learned how to do it.
If you have a polynomial of differential operators \(L(D)\)
Then a solution to the equation: \[
L(D)y=e^{kx}
\]Would be \[
y_1=\frac{e^{kx}}{L(k)}
\]

Answer 28

Let's start with number one.

Answer 29

We have \( e^{-3x}\) so \(k=-3\)

Answer 30

\[
L(D)=D^2(D-3)^2(D+3)
\]And thus: \[
L(k) = L(-3) = (-3)^2(-3-3)^2(-3+3)
\]

Answer 31

Since apparently \(L(k)=0\), it doesn't yeild anything.

Answer 32

err.. so. it's (9)(36) ?

Answer 33

oh. lol. 0. xD

Answer 34

Yeah, this method can't help with this equation.

Answer 35

Actually, I learned some more about this method.
Actually it can still work if \(L(k)=0\)

Answer 36

You have to figure out the multiplicity of the \(0\). In this case it is \(1\).

Answer 37

Since it was caused by \((-3+3)^{\color{red}1}\)

Answer 38

in this case the solution is: \[
\frac{x^ne^{kx}}{L^{(n)}(k)}
\]Where \(n\) is the multiplicity.

Answer 39

So we need the first derivative of \(L(k)\). Can you do that?

Answer 40

Please tell me you know how to take the derivative.

Answer 41

in what equation? o.o

Answer 42

.............. really?

Okay, find t he derivative of \(L(D)\) with respect to \(D\)

\(\color{blue}{\text{Originally Posted by}}\) @wio
\[
L(D)=D^2(D-3)^2(D+3)
\]And thus: \[
L(k) = L(-3) = (-3)^2(-3-3)^2(-3+3)
\]
\(\color{blue}{\text{End of Quote}}\)

Answer 43

so it's uhh..
2D[2(D-3)]D

then.. 2(-3)[2(-3-3)](-3)
(-6)(-12)(-3) = 222 ??

Answer 44

You didn't do the product rule correctly.

Answer 45

It would be easier to just multiply the whole thing out first.

Answer 46

(D−3)2⋅D^2+2⋅(D−3)⋅D^2⋅(D+3)+2⋅(D−3)^2⋅D⋅(D+3)

err.. 0.

Answer 47

Expand it out into a polynomial first, then differentiation. Let's avoid the product rule mess.

Answer 48

okay..

Answer 49

\[
D^2(D-3)^2(D+3)=
27 D^2-9 D^3-3 D^4+D^5
\]

Answer 50

Find the derivative of this polynomial, then put in \(-3\)

Answer 51

-360? o.o

Answer 52

432..

Answer 53

http://www.wolframalpha.com/input/?i=D+%2854-27+D-12+D%5E2%2B5+D%5E3%29+at+-3

Answer 54

I'm getting -108

Answer 55

I used wolfram to do the derivative

Answer 56

so I'd say our solution is: \[
-\frac{xe^{-3x}}{108}
\]

Answer 57

okay..

Answer 58

With the second one, I think the same logic applies.

Answer 59

Well, if you showed me your work I could tell you what is wrong.

Answer 60

well.. I still don't get #1.
I tried solving #2.

Answer 61

Here's what I get:

\[(D^2-1)y = 2e^x \]
factoring out D^2-1
\[(D-1)(D+1)y = 2e^x\]
now,
\[(D+1)yp = 2e^x\]
I'm just setting aside D-1, deriving 2e^x..
\[(D-1)yp = 2e^x +2e^x\]
then,
\[y _p=2e^x+2e^x+2e^x-2e^x\]
\[y_p = 4e^x\]
but from the book it should be xe^x.. dunno why.

Answer 62

\[
(D^2 - 1)y = 2e^x
\]So \[
\frac 12 (D^2 - 1)y = e^x
\]

Answer 63

Now factoring gets \((D-1)(D+1)\) so the mutiplicity of the root is 1.

Answer 64

Doing the derivative of \[
\frac 12 (D^2 - 1)
\]returns \(D\)

Answer 65

So our solution is \[
\frac{xe^x}{1}
\]

Answer 66

err.. how did it became D?

Answer 67

oh. wait. I get it already.

Answer 68

so how did it became xe^x?

Answer 69

Remember it is \(x^n\) where \(n\) is the multiplicity of the root?
Well \(n=1\)

Answer 70

how can I show it properly? o.o

I can only get.. uhh..\[\frac{ D^2-1 }{ 2 }\]

Answer 71

Okay so what is \[
L'(D)
\]In this case?

Answer 72

well.. D is an operator.. I can't derive it properly.. but its on the terms of x, the answer is x

Answer 73

when it becomes just D, it becomes the derivative of e^x which is e^x.. o.o

Answer 74

oh. here. look.

\[\frac{ 1 }{ 2 }(D-1)(D+1)y = e^x\]

Answer 75

well, how do we really get the x from there? o.o

Answer 76

why did D became x? >.>

Answer 77

hello?

Answer 78

Sorry

Answer 79

D didn't become x

Answer 80

\(\color{blue}{\text{Originally Posted by}}\) @wio
in this case the solution is: \[
\frac{x^ne^{kx}}{L^{(n)}(k)}
\]Where \(n\) is the multiplicity.
\(\color{blue}{\text{End of Quote}}\)

Answer 81

Since \(n=1\) we have \(xe^x\)

Answer 82

We also have the first derivative of L

Answer 83

what's the name of that solution?

Answer 84

Exponential input theorem

Answer 85

oh..

Answer 86

\[ \frac1{D^2}\frac1{(D-3)^2}\frac1{(D+3)}\cdot e^{-3x}\]

the inner most bit first ;

\[\frac1{(D+3)}\cdot e^{-3x}=e^{-3x}\frac{1}{(D-3)+3}\cdot1\\
\qquad\qquad\qquad=e^{-3x}\frac1D\\
\qquad\qquad\qquad=e^{-3x}\int dx\]

so

\[= \frac1{D^2}\frac1{(D-3)^2}e^{-3x}\]

Answer 87

note that the particular solution dosent need a consant of integration