I have this diff equation: $\frac{d^2 h(y)}{dy^2}-2y\frac{dh(y)}{dy}+(e-1)h(y)=0$
I am having trouble understanding the method of solution given by the textbook:
This is only part of the solution but i having trouble understating the step that led to (9.27).
- How are we able to change the $s->s+1$ for just one term and not the whole term.
- Why is $s=0$ in that term and not $s=1$
- Is there a general rule for relabelling the dummy summation index that you could list
The name of summing index is always dummy this means that $$\sum_{s=0}^{\infty} 2s(2s-1) a_s y^{2s-2}= \sum_{t=0}^{\infty} 2t(2t-1) a_t y^{2t-2}$$ Now let the first term in (9.26) be $$F=\sum_{t=0}^{\infty} 2t(2t-1) a_t ~y^{2t-2}$$ $$F= 0 a_0 y^{-2}+ \sum_{t=1}^{\infty} 2t(2t-1) a_t~ y^{2t-2}$$ Now put $t=s+1 \implies t-1=s $ to get $$F=\sum_{s+1=1}^{\infty} 2(s+1)(2s+1) a_{s+1}~ y^{2s}=\sum_{s=0}^{\infty} 2(s+1)(2s+1) a_{s+1}~ y^{2s}$$
The main motivation was to have similar power of $y$ as $y^{2s}$in both terms of 9.26. Now replace the first term of 9.26 as above and get 9.27. Yhis is usually done in the series solution of ODEs.