## anonymous one year ago solve for the two power series solution of the given differential equation: (x-1)y"-xy'+y=0 ; y(0)=-2, y'(0)=6

1. anonymous

for analytic $$y$$ we have $$y=\sum_{n=0}^\infty a_n x^n\\y'=\sum_{n=0}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n\\y''=\sum_{n=0}^\infty n(n+1)a_{n+1}x^{n-1}=\sum_{n=0}^\infty (n+1)(n+2)a_{n+2} x^n$$and $$xy'=x\sum_{n=0}^\infty (n+1)a_{n+1}x^n=\sum_{n=0}^\infty (n+1)a_{n+1}x^{n+1}=\sum_{n=0}^\infty na_nx^n\\xy''=x\sum_{n=0}^\infty (n+1)(n+2)a_{n+2} x^n=\sum_{n=0}^\infty (n+1)(n+2)a_{n+2}x^{n+1}\\\quad\quad=\sum_{n=0}^\infty n(n+1)a_{n+1} x^n$$

2. anonymous

so $$(x-1)y''-xy'+y=0\\\sum_{n=0}^\infty (n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n)=0\\n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}+(1-n)a_n=0\\a_{n+2}=\frac{1-n}{n(n+1)}a_n+\frac{n}{n+2}a_{n+1}$$

3. anonymous

now consider $$y(0)=\sum_{n=0}^\infty a_n (0)^n=a_0\\y'(0)=\sum_{n=0}^\infty (n+1)a_{n+1}(0)^n=(0+1)a_{0+1}=a_1$$

4. anonymous

so $$a_0=-2,a_1=6$$ now evaluate them manually according to the recursion to try and find a pattern

5. anonymous

oops, the recurrence should read: $$a_{n+2}=\frac{1-n}{(n+1)(n+2)}a_n+\frac{n}{n+2}a_{n+1}$$

6. anonymous

so what's the final value for C_1,C_2,C_3,C_4 and so on..?

7. anonymous

anyways $$\begin{array}{c|c|c}n&a_n\\\hline 0&-2\\1&6\\2&-1\\3&-\frac13\\4&-\frac1{12}\\5&\dots \end{array}$$

8. jtvatsim

Oooh... this actually appears to have a very elegant final solution. :)

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