Let y = e^mt

Then the characteristic equation is

6m^2 - 5m + 1 = 0

=> (3m - 1)(2m - 1) = 0

=> m = 1/3 or m = ½

So y = Ae^(t/3) + Be^(t/2)

=> y’ = (1/3)Ae^(t/3) + (1/2)Be^(t/2)

y(0) = 4

=> y(0) = 4 = Ae^0 + Be^0 = A + B

So A + B = 4 …………………(1)

y’(0) = 0

=> y’(0) = 0 = (1/3)Ae^0 + (1/2)Be^0 = (1/2)A + (1/2)B

So (1/3)A + (1/2)B = 0 ………...(2)

So we have the system:

A + B = 4 ………………..…(1)

(1/3)A + (1/2)B = 0 ………...(2)

A + B = 4 ………………..…(1)

A + (3/2)B = 0 ……………...(3) = (2) * 3

=> (1/2)B = -4

=> B = -8

But A + B = 4

=> A – 8 = 4

=> A = 12

So y = 12e^(t/3) – 8e^(t/2)