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
