Toronto Math Forum
MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: maoyafei on October 13, 2019, 11:51:38 PM
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Find the general solution of the given differential equation
y''+3y'+2y=0
assume y=e^(rt), y'=re^(rt), y''=(r^2)e^(rt)
r^2+3r+2
(r+1)(r+2)=0
r1=-1, r2=-2
y=C1e^(r1t)+C2e^(r2t)
Ans: y=C1e^(-t) + C2e^(-2t)