Question

Use power series to solve the differential equation
\[y''+xy'+y=0\]
\[y=\sum_{n=0}^{\infty} a_n x^n\]
\[y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\]
\[y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}\]
\[xy'=\sum_{n=1}^{\infty}a_nnx^n\]
\[\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]

Answer 1

yep, now what's first?

Answer 2

changing the sums so that they all begin with n=2
changing the last sum from n=0 to n=1 would be
\[a_1+\sum_{n=1}^\infty a_nx^n\]

Answer 3

no we don't strip out the terms yet, we just do an index shift

Answer 4

the first order of business is to get all the x's to be to the nth power through using index shifts

Answer 5

the only series that has x to some power other than n is the first one
how can we change the index of the first series to make the exponent on x be n?

Answer 6

you mean on
\[y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\]

Answer 7

yes, we need to start from a different n to get x^n through an index shift

Answer 8

so I'm shifting the starting point up by 2 so I have to subtract 2 from {n=2-2}
\[-\sum_{n=0}^\infty n(n-1)a_nx^n\] negative to make up for the negative?

Answer 9

nooooo

Answer 10

hold on

Answer 11

\[\sum_{n=0}^\infty n(n+1)a_nx^n\]

Answer 12

nice :)

Answer 13

still not right

Answer 14

because that would give us zero

Answer 15

no but closer

Answer 16

you didn't add 2 to every n

Answer 17

\[\sum_{n=0}^\infty (n+2)(n+1)a_nx^n\]

Answer 18

and I do mean \(every\) n (subscripts included)

Answer 19

again better, but...^

Answer 20

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

Answer 21

can I add them?

Answer 22

now that, I do believe, is correct :)

Answer 23

or multiply

Answer 24

or just leave as is?

Answer 25

add what? oh that? I wouldn't yet

Answer 26

ok

Answer 27

yeah leave as is asd now rewrite what you have

Answer 28

and*

Answer 29

the whole expression please I'm lazy

Answer 30

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

Answer 31

\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]okay next order of business is what you tried to do earlier; get all indices to start at the same n value
the only problem child here is which term?

Answer 32

the middle

Answer 33

so what do we do to fix it?

Answer 34

you have to be a bit clever here...

Answer 35

\[ xa_n +\sum_{n=0}^\infty a_n(n+1)x^n \]

Answer 36

did you try to do an index shift?

Answer 37

it looks like you combined the idea of the index shift with stripping out the first term

Answer 38

oh my...let's try this again

Answer 39

it is a bit tricky so let me give you a hint...

Answer 40

you cannot do another index shift. If you do you will change the exponent on x, which means we are screwed again, so we need to do something else to get this to start from n=0

Answer 41

we can't really strip out a term because we don't have this series defined for n<1, but ask yourself, if it was defined at n=0, what would the first term be?

Answer 42

just have the addition of an a_n up front like in the last example?

Answer 43

no, answer my question above and you will see why...

Answer 44

a 1

Answer 45

times a_n

Answer 46

no, be more careful

Answer 47

what would be the first term of\[\sum_{n=0}^\infty a_n nx^n\]

Answer 48

zero

Answer 49

right, so...

Answer 50

\[0+\sum_{n=1}^\infty a_n nx^n=\sum_{n=0}^\infty a_n nx^n\]so we actually didn't need to change anything; the first term we "stripped out" was a zero anyway! It makes no difference.

Answer 51

so we can go ahead and just change the starting point of that series without further manipulation

Answer 52

always watch for that, it comes in handy

Answer 53

oh ok uhm.....I'll revisit the last example later, because the I'm not convinced that we needed to do a index shift (or was it stripping of a term)

Answer 54

first to get the x^n on all terms we did an index shift (where we add or subtract from each n in the summand)
next to get all the starting points for each series the same, we strip out terms/or do the trick we just did

they are different tricks, and the order in which you do them is imortant

Answer 55

ok

Answer 56

\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=0}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]

Answer 57

it is important for you to try to understand why the index shift requires you to add/subtract each n in the summand while stripping out a term does not

Answer 58

good, so now make it all into one summation...

Answer 59

\[\sum_{n=0}^\infty x^n[(n+2)(n+1)a_{n+2}+a_nn+a_n]=0\]

Answer 60

nice, so what will you set to zero to find the recurrence relation?

Answer 61

(I am going to type minimally now as I am eating dinner at the same time :)

Answer 62

food comes first in my book... :P
Ok let's see here...

Answer 63

I can eat and respond with "yes" or "no" pretty well, no worries :)

Answer 64

\[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=-\frac{a_nn-a_n}{(n+2)(n+1)}\]

Answer 65

si?

Answer 66

si, pero si quieras podrias factor la \(-a_n\)

Answer 67

ok
\[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]

Answer 68

and now comes the tedious part that you can do while I eat
plug in n=0,1,2,3
and check what results from that recurrence relation until you notice a pattern like last time

so just cranks it out; plug in n=0,n=1,n=2
and if a_2 can be written in terms of a_0 or something like last time, do it. That is how we find the pattern.

Answer 69

go ahead and list out the first 5 or so and let's see if we see a relation

Answer 70

\[n=1:a_3=0\]
\[n=2:a_4=-a_2\frac{1}{12}\]
\[n=3:a_5=-a_3\frac{2}{20}\]
\[n=4:a_6=-a_4\frac{3}{30}\]
\[n=5:a_7=-a_5\frac{4}{42}\]

Answer 71

what about n=0 ?
we need to start from n=0 since our series starts from n=0 this time
(last time it started from n=1)

Answer 72

\[n=0:a_2=-a_0\frac{-1}{2}\]

Answer 73

\[n=0:a_2=a_0\frac{1}{2}\]

Answer 74

whatcha eating?

Answer 75

torta de pollo=chicken sandwich on a french sort of bread

Answer 76

con chipotle

Answer 77

nice

Answer 78

oh yeah...the pattern

Answer 79

so you did well in your subs I think, but now you must try to always sub all the way back as far as you can..

Answer 80

yeah, the pattern
write each on as some finite number of unknowns, in this case everything in terms of \(a_0\) and \(a_3\)

Answer 81

each one*

Answer 82

in the previous example you seemed to have simply switched the \[a_{odd}\] to zero and the \[a_{even}\] to \[a_0\]

Answer 83

take you're time...i'm gonna watch a few more mins of my episode

Answer 84

it's not different, it's the same as last time
odd subscript gives zero
even has a pattern reducible to a_0

Answer 85

okay I got a fairly simple answer, let me see if I can confirm it...

Answer 86

stupid wolfram is of no help :/

Answer 87

\[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\]
\[n=1:a_3=0\]
\[n=2:a_4=-a_2\frac{1}{12}=\frac{a_0}{2(3!)}\]
\[n=3:a_5=0\]
\[n=4:a_6=-a_4\frac{3}{30}=\frac{a_0}{5(3!)}\]
\[n=5:a_7=0\]

Answer 88

how did I do?

Answer 89

almost, I actually realized that I made a mistake because I think I see a mistake from you
so you did as well as me I'd say :D

Answer 90

let me try on paper again...

Answer 91

here I think I can demonstrate the benefit of not multiplying out terms and writing things over-explicitly to find patterns...

Answer 92

ok

Answer 93

\[n=0:a_2=\frac{a_0}{2\cdot 1}\]\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}\]now reevaluate the pattern

Answer 94

\[n=0:a_2=\frac{a_0}{2\cdot 1}=\frac{a_0}{2!}\]
\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\]

Answer 95

so try to do the same thing as last time
name all even numbers and make a new series with index k

Answer 96

then represent the coefficients and plug that in for \(a_n\)

Answer 97

n=2k
\[a_n=a_{2k}=\frac{a_0}{k!}\]

Answer 98

is it over k! ?