# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on February 10, 2018, 05:21:03 PM

Title: Q3-T5101
Post by: Victor Ivrii on February 10, 2018, 05:21:03 PM
Find the general solution of the given differential equation.
$$y'' - 2y' - 2y = 0.$$
Title: Re: Q3-T5101
Post by: Darren Zhang on February 10, 2018, 06:04:13 PM
The characteristic equation is $r^2-2r-1=0$, with root of $r = 1 + \sqrt{3}, 1- \sqrt{3}$.
Hence the general solution is  $$y = c_{1}exp(1-\sqrt{3})t+c_{2}exp(1+\sqrt{3})t$$
Title: Re: Q3-T5101
Post by: Meng Wu on February 11, 2018, 09:32:26 AM
$$y''-2y'-2y=0$$
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
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$