Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz3 => Topic started by: Victorwoshinidie on October 14, 2019, 08:16:43 PM

Questions:y''  2y'  2y = 0
we assume that y = e^(rt)
r^2  2r  = 0
r = b+√b^24ac/2a
Hence, r1 = (1+√33)/4 , r2 = (1√33)/4
y = c1e^r1t +c2e^r2t
y = c1e^(1+√33)/4t +c2e^(1√33)/4 t
c1 + c2 = 0
c1(1+√33)/4 +c2(1√33)/4 = 1
c1 = 2/√33
c2 = 2√33