Q3-T5101

[Victor Ivrii]

Find the general solution of the given differential equation.
$$
y'' - 2y' - 2y = 0.
$$

[Darren Zhang]

The characteristic equation is $r^2-2r-1=0$, with root of $r = 1 + \sqrt{3}, 1- \sqrt{3}$.
Hence the general solution is  $$y = c_{1}exp(1-\sqrt{3})t+c_{2}exp(1+\sqrt{3})t$$

[Meng Wu]

$$y''-2y'-2y=0$$
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
$$r={-b\pm \sqrt{b^2-4ac}\over 2a}$$
Hence,
$$\cases{r_1={1+\sqrt{3}}\\r_2=1-\sqrt{3}}$$
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
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$