Question

Diff EQ: xy'' + y' + xy = 0, x_0 = 1
find a power series representation.

Answer 1

so I think I remember something like this..
\[y= \sum_{n=0}^{\infty} a_n x^{n} \]
then find y' and y'' from this

Answer 2

is that for x_0 = 0 though?

Answer 3

didn't notice that part
\[y=\sum_{n=0}^{\infty} a_n (x-1)^n\]

Answer 4

ok so i get \[x \sum_{n=2}^{\infty}(n-1)na_{n}(x-1)^{n-2} + \sum_{n=1}^{\infty}na_{n})(x-1)^{n-1} + x \sum_{n=0}^{\infty}a_{n}(x-1)^{n}\]

Answer 5

ok and we want to write as one sum

n-2=0
let k=n-2
then..
\[x \sum_{n-2=0}^{\infty} (n-1)n a_n(x-1)^{n-2} \\ =x \sum_{k=0}^{\infty} (k+2-1)(k+2)a_{k+2}(x-1)^{k} \\ =\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2}(x-1)^k x\]

then we can do something similar for the middle term
n-1=0
let k=n-1
...

Answer 6

Ok let me do that real quick and ill post what i get

Answer 7

wait...

Answer 8

no actually that might work

Answer 9

like all of them will have the (x-1)^k thingy

Answer 10

but there is still that x in the first and third term ...

Answer 11

and then you take out terms to get all the summations to start at the same index?

Answer 12

you cant combine the x and x-1 in any way can you?

Answer 13

i'm rethinking a bit...
okay so we have that x multiple for the first and third
and we have the the whole (x-1) thingy too...
I think I want to write x as (x-1)+1

Answer 14

Question: Is \(x_0=1\) the center of the series, or is it supposed to be an initial condition? The first makes more sense, seeing as \(x=0\) is a singular point.

Answer 15

its the center of the series

Answer 16

Ah got it, didn't notice the second/third comments.

Answer 17

\[y=\sum_{n=0}^{\infty} a_n (x-1)^n \\ y'=\sum_{n=1}^\infty a_n n(x-1)^{n-1} \\ y'' =\sum_{n=2}^{\infty} a_n (n-1)n(x-1)^{n-2} \\ \text{ ... } \\ xy=[(x-1)+1] \sum_{n=0}^{\infty} a_n(x-1)^n =\sum_{n=0}^{\infty} a_n(x-1)^{n+1}+\sum_{n=0}^{\infty} a_n(x-1)^{n} \\ xy''=[(x-1)+1] \sum_{n=2}^{\infty} a_n(n-1)n(x-1)^{n-2} = \\\sum_{n=2}^{\infty} a_n(n-1)n(x-1)^{n-1} +\sum_{n=2}^{\infty}a_n(n-1)n(x-1)^{n-2}\]

Answer 18

\[xy''+y'+xy=0 \\ \text{ gives } \sum_{n=2}^{\infty} a_n(n-1)n(x-1)^{n-1}+\sum_{n=2}^{\infty} a_n(n-1)n(x-1)^{n-2} \\ +\sum_{n=1}^{\infty} a_n n(x-1)^{n-1}+\sum_{n=0}^{\infty} a_n(x-1)^{n+1} +\sum_{n=0}^{\infty}a_n (x-1)^n=0\]
please tell me if you see any type-os in this mess

Answer 19

so far so good

Answer 20

for first term
we want n-1=k
so we have n=k+1
so if n=2 then k=1
---
for second term
we want n-2=k
so we have n=k+2
so if n=2 then k=0
---
for third term
we want n-1=k
so we have n=k+1
so if n=1 then k=0
---
for forth term
we want n+1=k
so we have n=k-1
so if n=0 then k=1
---
for firth term
we want n=k
nothing changes just n to k
or you can just change all the k's to n's later whatever...
\[xy''+y'+xy=0 \\ \text{ gives } \sum_{n=2}^{\infty} a_n(n-1)n(x-1)^{n-1}+\sum_{n=2}^{\infty} a_n(n-1)n(x-1)^{n-2} \\ +\sum_{n=1}^{\infty} a_n n(x-1)^{n-1}+\sum_{n=0}^{\infty} a_n(x-1)^{n+1} +\sum_{n=0}^{\infty}a_n (x-1)^n=0\]
\[\sum_{k=1}^{\infty} a_{k+1}(k)(k+1)(x-1)^{k}+\sum_{k=0}^{\infty} a_{k+2}(k+1)(k+2)(x-1)^{k} \\ +\sum_{k=0}^{\infty} a_{k+1}(k+1)(x-1)^{k}+\sum_{k=1}^{\infty} a_{k-1}(x-1)^{k}+\sum_{k=0}^{\infty} a_k (x-1)^{k}=0\]

Answer 21

now we have the same variable part for each one
that variable part being the (x-1)^k thingy

Answer 22

we are going to need to get the same starting number for each one so we need to expand some of these just a bit

Answer 23

I got\[2a_{2} + a_{1} + a_{0} + \sum_{n=1}^{\infty}[a_{n+1}n(n+1)+a_{n+2}(n+1)(n+2) + a_{n+1}(n+1)+a_{n-1} + a_{n}](x-1)^{n}\]

Answer 24

and it got cut off D:

Answer 25

\[\sum_{k=0}^{\infty} a_{k+2}(k+1)(k+2)(x-1)^{k} =2a_2+\sum_{k=1}^{\infty}a_{k+2}(k+1)(k+2)(x-1)^{k}\]

\[\sum_{k=0}^{\infty} a_{k+1}(k+1)(x-1)^k=a_1+\sum_{k=1}^\infty a_{k+1}(k+1)(x-1)^k \\ \]

\[\sum_{k=0}^{\infty} a_k(x-1)^k=a_0+\sum_{k=1}^{\infty} a_k(x-1)^k\]
so I see how you got those first three terms

Answer 26

let me see if can put it altogether now

Answer 27

\[\sum_{k=1}^{\infty} a_{k+1}(k)(k+1)(x-1)^{k}+\sum_{k=0}^{\infty} a_{k+2}(k+1)(k+2)(x-1)^{k} \\ +\sum_{k=0}^{\infty} a_{k+1}(k+1)(x-1)^{k}+\sum_{k=1}^{\infty} a_{k-1}(x-1)^{k}+\sum_{k=0}^{\infty} a_k (x-1)^{k}=0 \\ \\ \]



\[2a_2+a_1+a_0+\sum_{k=1}^{\infty}a_{k+1}(k)(k+1)(x-1)^{k}+ \\ +\sum_{k=1}^{\infty} a_{k+2}(k+1)(k+2)(x-1)^{k}+\sum_{k=1}^{\infty} a_{k+1}(k+1)(x-1)^k + \\ +\sum_{k=1}^{\infty} a_{k-1}(x-1)^k+\sum_{k=1}^{\infty}a_k(x-1)^k=0\]

Answer 28

\[2a_2+a_1+a_0+ \\ \sum_{k=1}^{\infty} [a_{k+1}k(k+1)+a_{k+2}(k+1)(k+2)+a_{k+1}(k+1)+a_{k-1}+a_k](x-1)^k=0\]

Answer 29

I feel like this problem is so intense

Answer 30

so many type-os or mistakes could happen

Answer 31

this problem sucks lol

Answer 32

For what it's worth, I'm ending up with the same recurrence.

Answer 33

so if k=0 then
\[2a_2+a_1+a_0=0 \\ \]
now the other k integers greater than 0 occur in the series section there...
if k=1,2,3,4,5,... then
\[a_{k+1}k(k+1)+a_{k+2}(k+1)(k+2)+a_{k+1}(k+1)+a_{k-1}+a_k=0\]

that is great to know

Answer 34

\[a_{k-1}+a_k+a_{k+1}[k(k+1)+(k+1)]+a_{k+2}(k+1)(k+2)=0 \\ a_{k-1}+a_k+a_{k+1}(k+1)^2+a_{k+2}(k+1)(k+2)=0 \\ a_{k+2}=\frac{-a_{k-1}-a_k-(k+1)^2a_{k+1}}{(k+1)(k+2)} ; k=1,2,3,...\]

...


\[a_2=\frac{-a_1-a_0}{2} \]


so if k=1...
\[a_3=\frac{-a_0-a_1-4a_2}{2(4)}=\frac{\frac{-a_1-a_0}{2}-2a_2}{4}=\frac{a_2-2a_2}{4}=\frac{-a_2}{4}\]

Answer 35

oops 1+2 is 3 not 4

Answer 36

question

Answer 37

so should be \[a_3=\frac{-a_2}{3}\]

Answer 38

what is the question

Answer 39

the (k+1)^2 how'd you get that

Answer 40

nevermind, figured it out. Sorry lol

Answer 41

I combined like terms..
\[a_{k+1}k(k+1)+a_{k+1}(k+1)=a_{k+1}(k(k+1)+(k+1)) \\ =a_{k+1}(k+1)(k+1)=a_{k+1}(k+1)^2\]

\[a_{k+1}k(k+1)+a_{k+1}(k+1) \\ a_{k+1}(k+1)[k+1] \\ a_{k+1}(k+1)^2 \text{ or you can call it factoring }\]

Answer 42

so we found a pretty form for a_3
\[a_3=\frac{-a_2}{3}\]
...
let's see what happens when we plug in 2 for k...
\[ a_{k+2}=\frac{-a_{k-1}-a_k-(k+1)^2a_{k+1}}{(k+1)(k+2)} ; k=1,2,3,... \\ \]

\[a_4=\frac{-a_{1}-a_2-9a_3}{3(5)} ...\]
we can write a_3 in temrs of a_2...

\[a_4=\frac{-a_1-a_2+3a_2}{3(5)}=\frac{-a_1+2a_2}{3(5)}=\frac{-a_1+2a_2}{15}\]

Answer 43

omg I hate addition lol
\[a_4=\frac{-a_1+2a_2}{3(4)}=\frac{-a_1+2a_2}{12}\]
plug in 3 for k...
\[a_5=\frac{-a_2-a_3-16 a_4}{4(5)}\]

Answer 44

\[a_5=\frac{-a_2+\frac{a_2}{3}-16 \cdot \frac{-a_1+2a_2}{12}}{20} =\frac{-12a_2+4a_2-16(-a_1+2a_2)}{12(20)} \\ a_5=\frac{ -40a_2+16a_1}{12(20)}=\frac{-10a_2+4a_1}{3(20)}=\frac{-5a_2+2a_1}{3(10)}=\frac{-5a_2+2a_1}{30}\]

Answer 45

I'm not seeing a pattern

Answer 46

are you?

Answer 47

not yet, im still catching up though

Answer 48

\[a_3=\frac{-a_2}{3} \\a_4=\frac{-a_1}{12}+\frac{a_2}{6} \\ a_5=\frac{-a_2}{6}+\frac{a_1}{15} \\ y(x)=\sum_{n=0}^{\infty} a_n(x-1)^n \\ y(x)=a_0+a_1(x-1)+a_2(x-1)^2+a_3(x-1)^3+a_4(x-1)^4+a_5(x-1)^5+ \cdots \\ =a_0+a_1(x-1)+a_2(x-1)-\frac{a_2}{3}(x-1)^3+(\frac{-a_1}{12}+\frac{a_2}{6})(x-1)^4+ \\ +(\frac{-a_2}{6}+\frac{a_1}{15})(x-1)^5+ \cdots \]

Answer 49

http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx
I think we should just write this solution like they did in example 3

Answer 50

so you would "combine like terms"
get your a_0's together
your a_1's
and your a_2's

Answer 51

I don't see a more elegant way

Answer 52

but I could just be burnt out on this problem

Answer 53

I dont blame you, i appreciate your help though!

Answer 54

did you have any questions on this question before I leave?

Answer 55

None thus far, I'm understanding everything and getting the same results when i work them out.

Answer 56

like did you see that example 3 on paul's note
our solution is going to look at crappy as that
which is odd
I wanted to find a pattern but I don't see how to do that here

Answer 57

thank you for your help! :)

Answer 58

Yeah i was just going over that, that matches with the solution in the back of the book so i think thats all we can do

Answer 59

really?

Answer 60

that's awesome and kind of a relief giving all the work that this problem took

Answer 61

yep, they didn't get a general formula for the series either

Answer 62

anyways goodnight

Answer 63

goodnight and thank you again! I'd give you many more medals if I could lol

Answer 64

it's cool :)