Toronto Math Forum
Welcome,
Guest
. Please
login
or
register
.
1 Hour
1 Day
1 Week
1 Month
Forever
Login with username, password and session length
News:
Home
Help
Search
Calendar
Login
Register
Toronto Math Forum
»
MAT244--2019F
»
MAT244--Test & Quizzes
»
Quiz-4
»
TUT0602 Quiz4
« previous
next »
Print
Pages: [
1
]
Author
Topic: TUT0602 Quiz4 (Read 6542 times)
Yuchen Cong
Newbie
Posts: 3
Karma: 0
TUT0602 Quiz4
«
on:
October 19, 2019, 05:11:25 PM »
Question:
2y''+2y+1=0
Solution:
2r
2
+2r+1=0
r=[-2+-((-2)
2
-(4)(2)(1))
(1/2)
]/2(2)
r=[-2+-(-4)
(1/2)
]/4
r=-(1/2)+-(1/2)i
λ=-(1/2), µ=1/2
y(t)=C1e
-(1/2)t
cos(1/2)t+C2e
-(1/2)t
sin(1/2)t
Logged
Print
Pages: [
1
]
« previous
next »
Toronto Math Forum
»
MAT244--2019F
»
MAT244--Test & Quizzes
»
Quiz-4
»
TUT0602 Quiz4