### Author Topic: Q3 TUT 0301  (Read 2975 times)

#### Victor Ivrii

• Elder Member
• Posts: 2607
• Karma: 0
##### Q3 TUT 0301
« on: October 12, 2018, 06:04:24 PM »
Find a differential equation whose general solution is
$$y=c_1e^{-t/2}+c_2e^{-2t}.$$

#### Yunqi(Yuki) Huang

• Jr. Member
• Posts: 11
• Karma: 5
##### Re: Q3 TUT 0301
« Reply #1 on: October 12, 2018, 06:16:38 PM »
in the attchement

#### Keyue Xie

• Newbie
• Posts: 2
• Karma: 1
##### Re: Q3 TUT 0301
« Reply #2 on: October 12, 2018, 06:34:41 PM »
$$y=c_1e^{\frac{-1}2t} + c_2e^{(-2t)}$$
$$r_1=\frac{-1}2, r_2=-2$$
$$(r+\frac{1}{2})(r+2) = 0$$
$$r^2+ 2r+\frac{1}2r +1 = 0$$
Therefore
$$r^2+\frac{5}{2}r+1 = 0$$
$$2r^2 + 5r +2 = 0$$
$$2y''(t) + 5y'(t) + 2y(t) = 0$$
« Last Edit: October 12, 2018, 06:55:18 PM by Keyue Xie »

#### Yunqi(Yuki) Huang

• Jr. Member
• Posts: 11
• Karma: 5
##### Re: Q3 TUT 0301
« Reply #3 on: October 12, 2018, 06:49:55 PM »
$$r=-\frac{1}{2}or-2$$
$$(r+\frac{1}{2})(r+2) = 0$$
$$r^2+ 2r+\frac{1}2r +1 = 0$$
$$r^2+\frac{5}{2}r+1 = 0$$
$$2r^2 + 5r +2 = 0$$
Therefore
$$2y'' + 5y' + 2y = 0$$