Question

http://math.stackexchange.com/questions/864114/find-a-solution-to-a-linear-system-using-the-d-operator-and-method-of-eliminatio

Answer 1

@thomaster this is from differential equations ... using the D operator and method of elimination. The link has my attempt and LATEX.

Answer 2

@ganeshie8

Answer 3

say \(\large p(D) = D^2 + 2D+1\)

\[\large (D^2+2D+1)x_1 = e^{\alpha t}\]

\[\large Y_p = \dfrac{ e^{at}}{p(\alpha )}\]

Answer 4

you can plugin \(\alpha = 0\) right ?

Answer 5

I would tweak this step

\[ (D+7) \left((D−5)x_1+6x_2−1=0\right) \]
to
\[ (D+7)(D-5) x_1 + 6(D+7) x_2 - (D+7) 1 = 0 \]

although we it looks like "multiplying", we are "applying an operator"
so, in particular, (D+7) 1 means D 1 + 7*1
D 1 means take the derivative of 1 with respect to t. we get 0, and the result
\[ (D+7)(D-5) x_1 + 6(D+7) x_2 - 7 = 0 \]

you show a stand-alone D (which we must interpret as D applied to 1 i.e. D 1)
but it might be more clear to be more explicit about what we are doing.

For the particular solution, we try the ersatz solution xp = A in the differential equation
\[ (D^2 +2D + 1) A = 1 \]
or
\[ D^2 A +2D A + A= 1 \]

A is a constant, so D applied to A yields 0. we get
\[ A = 1\]
i.e. the particular solution is
\[ x_p = 1 \]

Answer 6

@phi what is the ersatz solution? Are we doing something that's similar to the method of undetermined coefficients here? Because if Y_p = A then the first and second derivative would be 0

Answer 7

but where do I plug in the A when I'm doing the equation? I mean
Y_p = A
Y'_p and Y''_p = 0

0+2(0) + 1 = 1 ??!?!!??!

Answer 8

unless?
0+2(0) + A(1) = 1
A = 1?!?!?!

Answer 9

ugh this one step... man :(... I know the rest... what is going on here :(

Answer 10

@perl

Answer 11

shoot, I would have to read up on this,do you have any books online for this

Answer 12

^ nooooooooooooooooooo :(.

Answer 13

this one step oh man T____T! @ganeshie8 ahhhhhhhhhhh save me

Answer 14

yes 1 satisfies the DE, right ?

Answer 15

I think \(Y_p = 1\)

Answer 16

\(\large (D^2+2D+1)x_1 = 1 = e^{0 t} \)

\(\large Y_p = \dfrac{e^{\alpha t}}{P(\alpha) } = \dfrac{e^{0t}}{0^2 + 2(0) + 1} = 1 \)

Answer 17

.. where alpha = 0 and then subsitute to get 1? I'm confused x.x

Answer 18

Recall the substitution rule :

\(\large D^n e^{at} = a^n e^{at}\)

Answer 19

:/ but I don't see any e^alpha t ' s in my problem... only the polynomial one

Answer 20

your problem is a special case when a = 0

Answer 21

\[
1 = 1^t = (e^{0})^{t} = e^{0t}
\]

Answer 22

Though, are you even familiar with this method of solving?

Answer 23

maybe not... I'm trying to self study xD

Answer 24

but I have read over and over and followed the examples... and did the earlier problems... I got the right answers.. it was this particular problem that was nuts

Answer 25

\(\large D^n e^{at} = a^n e^{at} \)

since \(D\) is a linear operator,

\(\large P(D) e^{at} = P(a) e^{at} \)

do we agree on this ?

Answer 26

P(D) is just a polynomial in D

Answer 27

yeah

Answer 28

then its easy for you to find a particular solution for below DE :

\(P(D) x_1 = e^{at}\)

Answer 29

Keep in mind, your original DE is a special case of this : \(P(D) x_1 = 1\)

Answer 30

lets find the particular solution of
\(P(D) x_1 = e^{at}\)

Answer 31

which is \[(D^2+2D+1) x_1 = 1\] for the special case right??

Answer 32

yess

Answer 33

let me pull up my notes..

Answer 34

Theorem :
For the DE : \(P(D) x_1 = e^{at}\) ,

\(Y_p = \dfrac{e^{at}}{P(a)}\)



lets prove this..

Answer 35

more proving D:,,,,ahhhhhhhhhh

Answer 36

its more of like verifying, than a proof -
just plugin \(Y_p\) into the DE and see if it satisfies

Answer 37

so we need to plug in that monster into the equation?

Answer 38

Proof :
\(P(D) x_1 = e^{at}\)

plugin \(Y_p\) into the equation :

\(P(D) Y_p = e^{at}\)

Answer 39

\(P(D) \dfrac{e^{at}}{P(a)} = e^{at}\)

Answer 40

but earlier, we agreed on this : \(P(D) e^{at} = P(a) e^{at}\)

Answer 41

\(P(D) \dfrac{e^{at}}{P(a)} = e^{at}\)

becomes


\(P(a) \dfrac{e^{at}}{P(a)} = e^{at}\)

Answer 42

Clearly left and right balances out

Answer 43

thats the end of proof.

Answer 44

sigh if only proofs were that easy... i mean I know some but ... can't get the ideas flowing

Answer 45

its called exponential input theorem,
and most proofs in DEs are this easy... all the nasty proofs are pushed to analysis, so that DE stays light and easy :)

Answer 46

uh huh .__.

Answer 47

still hate proofs.. after I took the course...hate it even more x.x

Answer 48

well I know 5.5 out of 8 chapters at least...just trying to figure out what I'm not getting x.x

Answer 49

lol proofs won't come in exams i guess... you only get specific, clear, nice looking problems

Answer 50

this prof knows how to teach DEs efficiently :
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-13-finding-particular-sto-inhomogeneous-odes/

Answer 51

@ganeshie8
thankyou for that link, i will be watching videos (easier than reading a 300 page textbook ;)

Answer 52

oops, I meant ansatz solution (see http://en.wikipedia.org/wiki/Ansatz)
basically, you guess the answer! and then verify it
\[ (D^2+2D+1)x_1 = 1 \]
In the above we assume the particular solution is \(x_1=A\) , where A is an unknown constant, and test:
\[ (D^2+2D+1)A = 1 \\ D^2 A +2D A + A = 1\]
as you know, D applied to A is 0, and we find
\[ 0+0+A= 1 \\ A=1\]
we conclude \(x_p=A=1\)
and add that to the homogenous solution, which you found to be

\[ x_h= c_1 e^{-t}+c_2te^{-t} \]
so the total solution is
\[ x= x_h+x_p=
c_1 e^{-t}+c_2te^{-t}+1\]

Answer 53

In case it is not clear, if we have
(D+1) A
this means
\[ \left(\frac{d}{dt} + 1\right) A \\ \frac{d}{dt}A + A \\ \frac{dA}{dt} + A\]

when A is a constant, this simplifies to just A

Treating the "operators" as algebraic polynomials might be confusing?