# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on February 10, 2018, 05:16:11 PM

Title: Q3-T0201
Post by: Victor Ivrii on February 10, 2018, 05:16:11 PM
Find the general solution of the given differential equation.
$$y'' + 3y' + 2y = 0.$$
Title: Re: Q3-T0201
Post by: Darren Zhang on February 10, 2018, 05:53:23 PM
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 $y = c_{1}e^{-t}+c_{2}e^{-2t}$
Title: Re: Q3-T0201
Post by: Meng Wu on February 11, 2018, 09:34:16 AM
$$y’’+3y’+2y=0$$
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
$$y=c_1e^{-t}+c_2e^{-2t}$$
Title: Re: Q3-T0201
Post by: Victor Ivrii on February 21, 2018, 07:46:48 AM
Meng Wu,
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