Question

I need to determine the indicial equation of the following equation and the exponents at the singularity for each singular point.

I am starting with 0 since its a regular singular point

sin(x)y''+xy'-2y=0

Answer 1

"indicial equation" ?

Answer 2

power series solution

Answer 3

\[y=\sum_{0}a_nx^n\]
\[y'=\sum_{0}a_nnx^{n-1}\]
\[y''=\sum_{0}a_nn(n-1)x^{n-2}\]
\[sin(x)=\sum_0\frac{(-1)^n}{(2n+1)~}x^{2n+1}\]
is that sine right?

Answer 4

might just wanna keep it as sin(x) for simplicity ... unless you like to play with that type of thing

Answer 5

I tried putting sine as series but I don't know how to deal with it

Answer 6

me either at the moment, never had the need to try to work it out :/

Answer 7

I think in this case we put \[y=\Sigma_{n=0}^\infty a_nX^{r+n}\]

Answer 8

lets try it as sin(x) and see where we end up

subbing into the ys into the setup we get

\[sin(x)\sum_0a_nn(n-1)x^{n-2}+x\sum_0a_nnx^{n-1}-2\sum_0 a_nx^n=0\]
\[\sum_0sin(x)a_nn(n-1)x^{n-2}+\sum_0a_nnx^{n}+\sum_0 -2a_nx^n=0\]
\[\sum_0sin(x)a_nn(n-1)x^{n-2}+\sum_0a_n(n-2)x^{n}=0\]
let get the exponents to even up, lowest rules
\[\sum_0sin(x)a_nn(n-1)x^{n-2}+\sum_{0+2}a_{n-2}(n-2-2)x^{n-2}=0\]
\[\sum_0sin(x)a_nn(n-1)x^{n-2}+\sum_{2}a_{n-2}(n-2-2)x^{n-2}=0\]
and just line up our indexes

Answer 9

\[sin(0)a_0~0(0-1)x^{0-2}\\+sin(1)a_1~1(1-1)x^{0-2}\\+\sum_2sin(x)a_nn(n-1)x^{n-2}+\sum_{2}a_{n-2}(n-2-2)x^{n-2}=0\]

\[sin(0)a_0~0(0-1)x^{0-2}\\+sin(1)a_1~1(1-1)x^{1-2}\\+\sum_2sin(x)a_nn(n-1)x^{n-2}+\sum_{2}a_{n-2}(n-4)x^{n-2}=0\]

or does heighest exponent rule?

Answer 10

either way it seems to pan out the same on that

Answer 11

the key is when all a_n parts are zero

Answer 12

\[\sum_2sin(x)a_nn(n-1)x^{n-2}+\sum_{2}a_{n-2}(n-4)x^{n-2}=0~:~a_0=0,a_1=0\]
\[\sum_2[sin(x)a_nn(n-1)+a_{n-2}(n-4)]x^{n-2}=0\]
\[sin(x)a_nn(n-1)+a_{n-2}(n-4)=0\]
\[sin(x)a_nn(n-1)=-a_{n-2}(n-4)\]
\[a_n=-a_{n-2}\frac{(n-4)}{n(n-1)sin(x)}\]

Answer 13

i might wanna solve that for an-2 instead of an

Answer 14

if we put it in the equation we would get:, I did
\[\sum_0 \frac{(-1)^n X^{2n+1}}{(2n+1)!} \sum_{0}(r+n)(r+n-1)a_nX^{r+n-2} +\sum_{0}(r+n)X^{r+n}-\sum_0 2a_n X^{r+n}\]

then change the exponent t
\[\sum_0 \frac{(-1)^n X^{2n+1}}{(2n+1)!} \sum_{-2}(r+n+2)(r+n+1)a_{n+2}X^{r+n} +\sum_{0}(r+n)X^{r+n}-\sum_0 2a_n X^{r+n}\]

but then I don't know what to do
I was thinking of expandind the first few terms at a0
\[[1-\frac{X^3}{3!}..]*[(r(r-1)a_0X^{r-2}+...]+[ra_oX^r+...]-2a_0X^r+...] \] and combine??

Answer 15

because it is't the recurrence relation that we need , we want the indicial equation

Answer 16

\[a_2=a_{0}\frac{4-2}{2(1)sin(x)}\]
\[a_4=a_{2}\frac{4-4}{4(3)sin(x)}\\~~~~~~=a_{0}\frac{4-2}{2(1)sin(x)}\frac{4-4}{4(3)sin(x)}\]
and the recurrence equation unfolds as such

Answer 17

i cant say im all that familiar with your notational setup

Answer 18

we are suppose to get a characteristic equation such as r(r-1) =0 and then get a 2 exponents which we replace r and then find the soulution as we would do normally

Answer 19

this explains it better
http://www.math.mcgill.ca/jakobson/courses/ma261/lecture17.pdf

Answer 20

i can make sense of about half of that :)

Answer 21

the product of sums still gets me

Answer 22

using some simple summations, how would you go about explaining the process? or do you only know how the Frobe is spose to look?