# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: yueyangyu on October 11, 2019, 02:00:02 PM

Title: quiz3 tut0502
Post by: yueyangyu on October 11, 2019, 02:00:02 PM
Find the general solution of the given differential equation.
$$y''-2y'-2y=0$$

Assume that $$y=e^{rt}$$ and it follows that r must be a root of characteristic equation
$$r^2-2r-2=0$$
$$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Then,
$$r_1=1+\sqrt{3} \quad r_2=1-\sqrt{3}$$

Therefore, the general solution of the given differential equation is:
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$