Author Topic: Q3-T0201  (Read 859 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2466
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Q3-T0201
« on: February 10, 2018, 05:16:11 PM »
Find the general solution of the given differential equation.
$$
y'' + 3y' + 2y = 0.
$$

Junjie Zhang

  • Full Member
  • ***
  • Posts: 19
  • Karma: 12
    • View Profile
Re: Q3-T0201
« Reply #1 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 $y = c_{1}e^{-t}+c_{2}e^{-2t}$

Meng Wu

  • Elder Member
  • *****
  • Posts: 91
  • Karma: 36
  • MAT3342018F
    • View Profile
Re: Q3-T0201
« Reply #2 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
$$y=c_1e^{-t}+c_2e^{-2t}$$

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2466
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Q3-T0201
« Reply #3 on: February 21, 2018, 07:46:48 AM »
Meng Wu,
Warning: stop spamming and flooding