Question

Frobenius Series

Answer 1

xy''-5y'+7xy = 0, find the indicial equation, in terms of r,

Answer 2

I manage to determine the p(x) and q(x) which is -5 and 7x^2 respectively. But what confused me is I do not know how shift the terms in the series.

Answer 3

\[y = \sum_{m=0}^{\infty} a_mx^(m+r) \]
\[y' = \sum_{m=0}^{\infty} (m+r)a_mx^(m+r-1)\]
\[y'' = \sum_{m=0}^{\infty} (m+r)(m+r-1)a_mx^(m+r-2)\]
I subs. into the equation but then i just got lost.

Answer 4

First off, your starting points on y' and y" are m=1 and m=2 respectively.
Then, put everything into the original one
But for xy", you need to multiple x into y" to get
\[xy"= \sum_{m=2}^\infty (m+r)(m+r-1)a_m x^{m+r-1}\]
\[-5y =\sum_{m=1}^\infty -5(m+r)a_m x^{m+r-1}\]
\[7xy=\sum_{m=0}^\infty 7a_mx^{m+r+1}\]
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\[xy"-5y+7xy=\sum_{m=2}^\infty (m+r)(m+r-1)a_m x^{m+r-1} \\+\sum_{m=1}^\infty -5(m+r)a_m x^{m+r-1}+\sum_{m=0}^\infty 7a_mx^{m+r+1}\]

Answer 5

Now, you need change the starting points. For the first one, y", to get m =0 you need subtract 2, right? Hence, whenever you see m, you take m -2
so
\[xy"=\sum_{m=0}^\infty (m-2+r)(m-2+r-1)a_{m-2}x^{m-2+r-1}\\ =\sum_{m=0}^\infty (m+r-2)(m+r-3)a_{m-2}x^{m+r-3}\]
Do the same with y'
after getting the same starting point m=0, you can combine all in 1 summation.

Answer 6

So after i did as you said in one summation, now i am just looking at a long equation which confuses me even more.

Answer 7

\[\sum_{m=0}^{\infty} [(m+r-2)(m+r-3)a _{m-2}x^{m+r-3}-5(m+r-1)a_{m-1}x^{m+r-2}+7a_mx^{m+r-1} \]