- #1
Dustinsfl
- 2,281
- 5
$\alpha_nr^n + \beta_nr^{-n}$
We know that
\begin{alignat*}{3}
\alpha_na^n+\beta_na^{-n} & = & A_n\\
\alpha_nb^n+\beta_nb^{-n} & = & 0
\end{alignat*}
So I ended up with
$$
\alpha_n = \frac{-A_n}{b^{2n}/a^n-a^n}
$$
and
$$
\beta_n = \frac{A_n}{1/a^n-a^n/b^{2n}}
$$
When I plug them in, I obtain
$$
\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}/a^n-r^{2n}}{b^{2n}-a^{2n}}
$$
However, the solution is supposed to be
$$
\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}
$$
I cannot find my error.
We know that
\begin{alignat*}{3}
\alpha_na^n+\beta_na^{-n} & = & A_n\\
\alpha_nb^n+\beta_nb^{-n} & = & 0
\end{alignat*}
So I ended up with
$$
\alpha_n = \frac{-A_n}{b^{2n}/a^n-a^n}
$$
and
$$
\beta_n = \frac{A_n}{1/a^n-a^n/b^{2n}}
$$
When I plug them in, I obtain
$$
\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}/a^n-r^{2n}}{b^{2n}-a^{2n}}
$$
However, the solution is supposed to be
$$
\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}
$$
I cannot find my error.