MAT244-2018S > Quiz-3

Q3-T5101

(1/1)

**Victor Ivrii**:

Find the general solution of the given differential equation.

$$

y'' - 2y' - 2y = 0.

$$

**Darren Zhang**:

The characteristic equation is $r^2-2r-1=0$, with root of $r = 1 + \sqrt{3}, 1- \sqrt{3}$.

Hence the general solution is $$y = c_{1}exp(1-\sqrt{3})t+c_{2}exp(1+\sqrt{3})t$$

**Meng Wu**:

$$y''-2y'-2y=0$$

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

$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$

Navigation

[0] Message Index

Go to full version