Question

Hi everyone! When using series expansion to solve the DE y'+y=0, the recursion formula is C_(k+1)=(-C_k)/(k+1) and when only solving for "y" the answer is
y=C_zero[1-x+(1/2)x^2-(1/6)x^3+...]
If C_1=-C_zero, and C_2=C_0/2 and C_3=-C_0/6 can anyone tell me "mechanically" exactly how and why the answer is written this way? I'm just confused about the C_zero and C_1 mostly...I posted the work as well! Thanks! :o)

Answer 1

Basically I'm just a little confused about the C_1 and C_0

Answer 2

There must be some initial condition ?

Answer 3

no initial condition

Answer 4

for example...bare with me...

Answer 5

Okay, we may assume \(y(0) = c_0\) as the initial condition

Answer 6

The goal here is to find the power series of \(y\) : \[\sum\limits_{n=0}^{\infty}\color{blue}{c_n}x^n\]

the effort reduces to finding the coefficients \(\color{blue}{c_n}\)

Answer 7

when I am plugging going through each k=0,1,2,3,4 etc and plugging each number into the recursion formula in order to get values for C0,C1,C2,etc, I have written down...
C1=-C0
C2=CO/2
C3=-CO/6
etc
when I go to map each constant with the corresponding C0+C1X+C2X^2+C3X^3 ETC, i can't seem to get what they got
does that make any sense?

Answer 8

I don't need help understanding how to do all the problem, just why c1 and c0 are written the way they are...thats all

Answer 9

you said @rational "we may assume y(0)=c0 as the initial condition"
why can we assume that?

Answer 10

I must have missed that part in my notes :o/

Answer 11

Lets see, the recurrence relation you have is
\[c_{n+1}=\dfrac{-c_n}{n+1}\]

Answer 12

plugin \(n=0\) and get
\[c_{0+1}=\dfrac{-c_0}{0+1}\implies c_1=-c_0\]

Answer 13

yes, that's correct for the recursion

Answer 14

yep, had that!

Answer 15

C2=CO/2
C3=-CO/6

Answer 16

again, what exactly is your question ?
your c3 is not matching with their c3 is it?

Answer 17

one second so I can check and clarify my question...just a sec

Answer 18

no...my c3 is fine...I just need someone to hold my hand I guess and show me how exactly"mechanically" to write out the solution

Answer 19

once you have first few coefficeints just plug them in the expansion of power series of y :

\[y=\sum\limits_{n=0}^{\infty}\color{blue}{c_n}x^n=\color{blue}{c_0}+\color{blue}{c_1}x+\color{blue}{c_2}x^2+\color{blue}{c_3}x^3+\cdots\]

Answer 20

okay, well my co would be -c1 then
how do I do that?

Answer 21

express all the coefficients in terms of \(c_0\)

Answer 22

It seems you have :
\(c_0 = c_0\)

\(c_1=-c_0\)

\(c_2=\frac{1}{2}c_0\)

\(c_3=-\frac{1}{6}c_0\)

\(c_4=\frac{1}{24}c_0\)

right?

Answer 23

we both agree that I am trying to write the answer as:
y=C_zero[1-x+(1/2)x^2-(1/6)x^3+...] right?

since c1=-co, then I guess since y=c0+c1x+c2x^2+c3x^3
I just replace c1 with -co, then factor it out?

Answer 24

first plugin c1, c2, c3, c4... values into the series expansion

Answer 25

\[\begin{align}y=\sum\limits_{n=0}^{\infty}\color{blue}{c_n}x^n&=\color{blue}{c_0}+\color{blue}{c_1}x+\color{blue}{c_2}x^2+\color{blue}{c_3}x^3+\cdots\\~\\&=?\end{align}\]

Answer 26

yeah I actually understand that part, I'm just confused on the c1 and co for some silly reason...
re-read my last post and tell me if that is correct"as far as the c1 part goes"

Answer 27

replace c1 by -c0

Answer 28

you're correct about that
and we need to do just that for all other coefficients too

Answer 29

so Y=C0-C0X+BLA+BLA+BLA right? :o)

Answer 30

bla bla doesn't look mathematical,
but yes its as simple as that

Answer 31

lol...omg...something so simple as just replace the c1 with -co
I feel pretty silly now ugh! X0/

Answer 32

Haha also that textbook solution wants you notice 6 = 3! and 24 = 4!

Answer 33

I think now that I am looking at it differently is that what was confusing me was the fact that I everything else had a numbered coefficient but really, there is a Co there if you don't factor it out of everything

Answer 34

yes they skipped a step at second line from last

Answer 35

why those crazy solution writers skipping steps...grrr

Answer 36

anyhooo...I think I get it now! Thanks sooo much! :o)

Answer 37

Hey rational, does it show you getting a medal?
I just clicked on it 3 times but it isn't working?

Answer 38

sounds good! you should check out paul site on series solutions if you haven't already, its pretty good

Answer 39

I have it on another tab already! lol
I have like 47 tabs open! :o)

Answer 40

okay, you have your medal now...I will close the question...thanks again! :o)

Answer 41

Okay! Enjoy...

Answer 42

:o)