Finding the general solution of the given differential equation
$$y" + 3y' + 2y = 0 $$
assuming $y = e^{rt}$ is a solution (I have not fully understand why $y = e^{rt}$ is a solution) to the derive characteristic equation. Can I get some clear explanation?
We are looking the solution in this form (intuition)--V.I.
I understand that it satisfies equation such as $y" - y = 0$) Nope unless $r=\pm 1$
\begin{gather*}
y" = r^2e^{rt},\\
y' = re^{rt},\\
y = e^{rt.}
\end{gather*}
Characteristic Equation $r^2+3r+2 = 0$.
Roots are $r_1=-1 and $r_2=-2$
\begin{gather*}
y_1 = e^{-t},\\
y_2 = e^{-2t},\\
y = c_2e^{-t}+c_2*e^{-2t}
\end{gather*}
