How did you get "t1/2" part? Thankx though!
2 Answers

General solution= Ke^(2t)+Ke^tt1/2

Note that the above answer is wrong as there should be two arbitrary constants.
Find the complementary function by solving the auxiliary equation:
y''  y'  2 = 0
m² + m  2 = 0
(m + 2)(m  1) = 0
m + 2 = 0 OR m  1 = 0
m = 2 OR m = 1
yᶜ = C₁℮ᵗ + C₂℮^(2t)
yᶜ = C₁℮ᵗ + C₂ / ℮^(2t)
Find the particular integral by comparing coefficients:
yᵖ = At + B
yᵖ' = A
yᵖ'' = 0
yᵖ''  yᵖ'  2yᵖ = 2t
A  2(At + B) = 2t
A  2At  2B = 2t
2At + (A  2B) = 2t
2A = 2
A = 1
A  2B = 0
2B = A
B = A / 2
B = ½
yᵖ = t  ½
Find the general solution by combining these two parts:
y = yᶜ + yᵖ
y = C₁℮ᵗ + C₂ / ℮^(2t)  t  ½