Question

Elementary Differential Equations

Answer 1

what do we use as a general power series?

Answer 2

let\[y=\sum_0a_nx^n\]
define y' and y'' for substitution

Answer 3

y(1) = e(n=1)n*a(n)x(n-1)

Answer 4

y(2) = e(n=2)n*(n-1)*a(n)x(n-2)

Answer 5

good, and im sure youve got the y''

sub in

\[y''+xy' + y = 0\]
\[\sum_2a_nn(n-1)x^{n-2}+x\sum_1a_nnx^{n-1} + \sum_0a_nx^n = 0\]
we need to line up out exponents so that we can then adjust the indexes

Answer 6

i go with the 'lowest' exponent for some reason, so line them up to x^(n-2)

\[\sum_2a_nn(n-1)x^{n-2}+\sum_{1+2}a_{n-2}(n-2)x^{n-2} + \sum_{0+2}a_{n-2}x^{n-2} = 0\]
\[\sum_2a_nn(n-1)x^{n-2}+\sum_{3}a_{n-2}(n-2)x^{n-2} + \sum_{2}a_{n-2}x^{n-2} = 0\]

pull out the odd balls to line up the indexes

Answer 7

\[2a_2+a_{0} +\sum_3a_nn(n-1)x^{n-2}+\sum_{3}a_{n-2}(n-2)x^{n-2} + \sum_{3}a_{n-2}x^{n-2} = 0\]

factor out the powers of x

\[2a_2+a_{0} +\sum_3\left(a_nn(n-1)+a_{n-2}(n-2)x^{n-2} + a_{n-2}\right)x^{n-2} = 0\]

and now we have something to play with to determine a_n

Answer 8

a2 and a0 beocme our arbitrary constants in the end of it all

Answer 9

do we need sth like m = n-2 substitution?

Answer 10

not that im aware of

we know that if the sum of the an parts is 0, and if 2a2 +a0 = 0 then the whole thing is 0

\[\underbrace{2a_2+a_{0}}_{=0} +\underbrace{\sum_3\left(a_nn(n-1)+a_{n-2}(n-2)x^{n-2} + a_{n-2}\right)}_{=0}x^{n-2} = 0\]

Answer 11

i see a stary x power in the middle there ... sneaky little things they are lol

Answer 12

our rule for an is such that:
\[a_nn(n-1)+a_{n-2}(n-1) = 0\]
solve for an

Answer 13

given that n>=3 of course

Answer 14

ann(n−1)+an−2(n−2)=0 ? u mean above

Answer 15

an(n-1) + an-2(n-2) = 0

Answer 16

if we factor an-2, then we have a +1 leaving us with an-2(n-1)

Answer 17

where we got n>3?

Answer 18

from series?

Answer 19

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

since our index starts at n=3, then n>2 is the sufficient condition

Answer 20

ok, i got it, what about initial conditions and convergence radius?

Answer 21

once we have a rule for an im assuming we can determine the radius from that

Answer 22

limit ratio and such

Answer 23

in order of y(0) = 1, then ao = 1

if y'(0) = 0, the a1 = 0

Answer 24

so we can determine all the other an from this

Answer 25

\[y=1+0x+\sum_2a_nx^n~:~y(0)=1\]

\[y'=0+\sum_3a_nnx^{n-1}~:~y'(0)=0\]

so the first 2 coeffs are known

Answer 26

ookay, when we use convergence test?)

Answer 27

once we know the rule for an we can do a limiting ratio if memory serves.

Answer 28

ok, thank you very much, now i have some basic understanding, ur awesome :)

Answer 29

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

Answer 30

youre welcome

Answer 31

a0 = 1
a1 = 0

2a2 + a0 = 0

a2 = -a0/2 = -1/2

a3 = (rule defined)

Answer 32

\[a_3=-\frac{a_{3-2}}{3}=0\]
\[a_5=-\frac{a_{5-2}}{5}=0\]

all odd ns are 0 coeffs

\[a_4=-\frac{a_2}{4}\]
\[a_6=-\frac{a_2}{4.6}\]
\[a_8=-\frac{a_2}{4.6.8}\]

might help to convert these to some k/n! ruleing

Answer 33

its 230am, im going to bed so good luck :)

Answer 34

thank you :D