Author Topic: TUT0502 QUIZ3  (Read 257 times)

Wusijia

  • Newbie
  • *
  • Posts: 4
  • Karma: 0
    • View Profile
TUT0502 QUIZ3
« on: October 11, 2019, 02:00:02 PM »
\begin{equation*}
y''-2y'-2y=0
\end{equation*}
\begin{equation*}
r^2-2r-2=0
\end{equation*}
\begin{equation*}
r_1=\frac{2+\sqrt{4+8}}{2}=1+\sqrt{3}
\end{equation*}
\begin{equation*}
r_2=\frac{2-\sqrt{4+8}}{2}=1-\sqrt{3}
\end{equation*}
\begin{equation*}
y=C_1e^{(1+\sqrt{3})t}+C_2e^{(1-\sqrt{3})t)}
\end{equation*}