Toronto Math Forum
MAT2442018S => MAT244Tests => Quiz3 => Topic started by: Victor Ivrii on February 10, 2018, 05:16:11 PM

Find the general solution of the given differential equation.
$$
y'' + 3y' + 2y = 0.
$$

Let y = e^{rt},
Substitution of the assumed solution results in the characteristic equation $$r^2+3r+2=0$$
The roots of the equation are r = 2, 1 . Hence the general solution is $y = c_{1}e^{t}+c_{2}e^{2t}$

$$y’’+3y’+2y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2+3r+2=(r+1)(r+2)=0$$
$$\cases{r_1=1\\r_2=2}$$
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
$$y=c_1e^{t}+c_2e^{2t}$$

Meng Wu,
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