Question

how to solve power series expansion of xy''-2y'+xy=0 when x0 = 1

Answer 1

Let \( y = \sum\limits_{n=0}^\infty a_nx^{n+k}\) and find the values of \( a_n's \) interms of \( x_0\) and \( x_1 \)

Answer 2

you get something like ... from Maple.
\[ y(x) = C_1x^2(1-(1/8)*x^2+(1/192)*x^4+O(x^6))+\\C_2*(\ln(x)*(x^2-(1/8)*x^4+O(x^6))-2+(3/32)*x^4+O(x^6)) \]

Answer 3

this is a big question, requires a lot to work out

Answer 4

solution is something like
\[ y(x) = C_1\;x\;BesselJ(1, x)+C_2 \;x\;BesselY(1, x) \]
You might wanna check
http://en.wikipedia.org/wiki/Bessel_polynomials
http://mathworld.wolfram.com/BesselPolynomial.html

Answer 5

I did that part and what I get is \[x* \sum_{k=0}^{\infty}a _{k}x ^{k}k(k-1)(x-1)^{k-2}-2* \sum_{k=0}^{\infty}a _{k}x ^{k}k(x-1)^{k-1}+x*\sum_{k=0}^{\infty}a _{k}(x-1)^{k}\]

how do i get rid of the x terms?

Answer 6

I dont think you get rid of the x, but you need to plug something into x

Answer 7

I tried shifting the indexes over and the x terms don't go away so I can't get the recurrence relation

Answer 8

\[y = \sum\limits_{n=0}^\infty a_nx^{n+k} \\
y' = \sum\limits_{n=0}^\infty a_n (n+k)x^{n+k-1} \\
y'' = \sum\limits_{n=0}^\infty a_n (n+k)(n+k-1)x^{n+k-2} \\\]
put these stuff on DE
\[ x \sum\limits_{n=0}^\infty a_n (n+k)(n+k-1)x^{n+k-2} - \sum\limits_{n=0}^\infty a_n (n+k)x^{n+k-1} + x \sum\limits_{n=0}^\infty a_nx^{n+k} =0\]

Answer 9

\[ \sum\limits_{n=0}^\infty a_n (n+k)(n+k-1)x^{n+k-1} - \sum\limits_{n=0}^\infty a_n (n+k)x^{n+k-1} + \sum\limits_{n=0}^\infty a_nx^{n+k+1} =0 \]
since first and second term has same power of x
\[ \sum\limits_{n=0}^\infty a_n(n+k) ((n+k-1) -1 )x^{n+k-1} + \sum\limits_{n=0}^\infty a_nx^{n+k+1} =0 \]

Answer 10

I see now, wow thanks for the help! Been working on this problem for 3 hours!

Answer 11

\[ \sum\limits_{n=0}^\infty a_n(n+k) (n+k-2 )x^{n+k-1} + \sum\limits_{n=0}^\infty a_nx^{n+k+1} =0 \\
a_0 (0+k)(0 + k -2) x^{k-1}+ a_1 (1+k)(1 + k -2)x^k + \\
\sum\limits_{n=0}^\infty a_{n+2}((n+2)+k) ((n+2)+k-2 )x^{(n+2)+k-1} + \sum\limits_{n=0}^\infty a_nx^{n+k+1} =0 \]
now they have same power. you can get the recursion relation from here.