Q: Find the general solution of the given differential equation.
$$
y^{\prime\prime}-2y^{\prime}-2y=0
$$
A: We assume that $y=e^rt$, and it follows that $r$ must be a root of the characteristic equation
$$r^2 - 2r -2 = 0$$
Then $r= \frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-2)\pm\sqrt{(-2)^2-4\times2}}{2\times1}= 1\pm\sqrt{3}$
Then we find that $r_1 = 1+\sqrt{3}$ and $r_2 = 1-\sqrt{3}$
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 $$y = c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$
