Toronto Math Forum

MAT244--2019F => MAT244--Test & Quizzes => Quiz-4 => Topic started by: Yuchen Cong on October 19, 2019, 05:11:25 PM

Title: TUT0602 Quiz4
Post by: Yuchen Cong on October 19, 2019, 05:11:25 PM

Question:
2y''+2y+1=0
Solution:
2r²+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)tcos(1/2)t+C2e^-(1/2)tsin(1/2)t