### Author Topic: TUT 5102 QUIZ3  (Read 245 times)

#### yuhan cheng

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##### TUT 5102 QUIZ3
« on: October 11, 2019, 02:00:45 PM »
$y''-2y'-2y=0$

$y''-2y'-2y=0$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation
$r^2-2r-2=0$
We use the quadratic formula which is
$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Hence,
$\left\{ \begin{array}{c} r_1=1+\sqrt3\\ r_2=1-\sqrt3\\ \end{array} \right.$
Since the general solution has the form of
$y=c_1e^{r_1t}+c_2e^{r_2t}$
Therefore, the general solution of the given differential equation is
$y=c_1e^{(1+\sqrt3)t}+c_2e^{(1-\sqrt3)t}$