Question

ok

Answer 1

@UnkleRhaukus

Answer 2

Shifting the index gives you the same series as \(\sum_{n=0}^\infty C_{n+2}x^{n+2}\)... it only discards a finite number of terms.

Answer 3

I would keep it as \(\sum_{n=2}^\infty C_n x^n\), personally; you then find \((x-1)y'=(x-1)\sum_{n=2}^\infty nC_nx^{n-1}=\sum_{n=2}^\infty C_n(nx^n-x^{n-1})\).

Answer 4

The initial conditions won't work though...

Answer 5

Not sure what the problem is, but if you know the first two terms isn't it just \[
a_0+a_1+\sum_{n=2}^\infty a_n
\]

Answer 6

\(y(2)=\) one?, or eleven?

Answer 7

without working out any of the mathing yet, im curious as to why n=0 is a fail?

Answer 8

\[y'' + (x-1)y' + y = 0\]
\[\sum_0a_nn(n-1)x^{n-2} + (x-1)\sum_0a_nnx^{n-1} + \sum_0a_nx^n = 0\]
\[\sum_0a_nn(n-1)x^{n-2} + \sum_0a_nnx^{n}-\sum_0a_nnx^{n-1} + \sum_0a_nx^n = 0\]
\[\sum_0a_nn(n-1)x^{n-2}-\sum_0a_nnx^{n-1} + \sum_0a_n(1+n)x^n = 0\]
\[\sum_0a_nn(n-1)x^{n-2}-\sum_{0+1}a_{n-1}(n-1)x^{n-1-1} + \sum_{0+2}a_{n-2}(1+n-2)x^{n-2 }= 0\]
\[\sum_0a_nn(n-1)x^{n-2}-\sum_{1}a_{n-1}(n-1)x^{n-2} + \sum_{2}a_{n-2}(n-1)x^{n-2 }= 0\]
\[a_00(-1)x^{-2}+a_11(0)x^{-1}+\sum_2a_nn(n-1)x^{n-2}\\
-a_{0}(0)x^{-1}-\sum_{2}a_{n-1}(n-1)x^{n-2}\\
+ \sum_{2}a_{n-2}(n-1)x^{n-2 }= 0\]
a series is identically zero when all the coefficients equal zero; therefore, let a0 and a1 be unknown constants that will help to define the rule; in the meantime, lets combine the summations

\[\sum_2a_nn(n-1)x^{n-2}+\sum_{2}-a_{n-1}(n-1)x^{n-2} + \sum_{2}a_{n-2}(n-1)x^{n-2 }= 0\]
\[\sum_2[a_nn(n-1)-a_{n-1}(n-1)+a_{n-2}(n-1)]x^{n-2}= 0\]
\[a_nn(n-1)-a_{n-1}(n-1)+a_{n-2}(n-1)=0\]
\[(n-1)(a_nn-a_{n-1}+a_{n-2})=0\]
seeing the n>=2 .... this tells me that the a parts will have to equate to zero
\[a_nn-a_{n-1}+a_{n-2}=0\]
\[a_n=\frac{a_{n-1}-a_{n-2}}{n}\]

Answer 9

pretty sure there might be some typos along the way :)