Toronto Math Forum
MAT2442018S => MAT244Tests => Quiz3 => Topic started by: Victor Ivrii on February 10, 2018, 05:19:20 PM

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

The characteristic equation is $r^22r2=0$, with root of $r = 1+\sqrt{3}, 1\sqrt{3}$.
Hence, the general solution should be $y = c_{1}exp(1\sqrt{3})t+c_{2}(1+\sqrt{3})t$

$$y''2y'2y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^22r2=0$$
We use the quadratic formula which is
$$r={b\pm \sqrt{b^24ac}\over 2a}$$
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
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1\sqrt{3})t}$$