We first solve the homogeneous equation:

\(y` - y = 0\)

Its characteristic equation is

\(r - 1 = 0 \Rightarrow r = 1\)

Therefore, the solution of the homogeneous equation is

\(y_{H} = Ce^{t}\)

Now we need to find a particular solution. We will try with

\(y_{p}=(At+B)e^{2}t \Rightarrow y'_{p}, =2(At+B)e^{2}t+Ae^{2}t=(2At+(A+2B))e^{2}t\)

Thus,

\(y`_{p} - y_{p} = 8te^{2}t \Rightarrow (2At + (A + 2B))e^{2}t - (At + B)e^{2}t = 8te^{2}t \Rightarrow (At + (A + B))e^{2}t = 8te^{2}t\)

From this we get

\(At + (A + B) = 8t,\)

and the system

\(A = 8\)

\(A + B = 0\)

Therefore,

\(A = 8 B = -A = -8\)

and the particular solution is

\(y_{p} = (8t - 8)e^{2}t\)

This means that the total solution is

\(y = y_{H} + y_{p} = Ce^{t} + (8t - 8)e^{2}t\)

Now we need to find C. We use the fact that y(0) = 1, so

\(1 = y(0) = Ce^{0} + (8 \cdot 0 - 8)e^{2 \cdot 0} = C - 8\)

Therefore, \(C = 9\)

and the solution of the initial equation is

\(y = 9e^{t} + (8t + 8)e^{2}t\)