### Author Topic: 3.1 #2  (Read 1256 times)

#### Hyunmin Jung

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##### 3.1 #2
« on: September 30, 2014, 02:32:14 AM »
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*}
« Last Edit: September 30, 2014, 06:02:46 PM by Victor Ivrii »