Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz3 => Topic started by: Victor Ivrii on October 12, 2018, 06:02:55 PM

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

In the attachment.

$$y′′−2y′−2y=0$$
$$r^2−2r−2=0$$
$$r = \frac{−b \pm \sqrt{b^2−4ac}}{2a}$$
$$r_1=1+\sqrt{3}, r_2=1−\sqrt{3}$$
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1\sqrt{3})t}$$

LOL: Yulin, you typed it to produce in the end png. Why not directly into forum?