Author Topic: Quiz3 TUT0702  (Read 7156 times)

Victorwoshinidie

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Quiz3 TUT0702
« 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^2-4ac/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