Solution: We begin by finding the general solution of the homogeneous equation \[ y'' - 3y' + 2y = 0. \] Assume a solution of the form \(y = e^{rx}\). Then \[ r^2e^{rx} - 3re^{rx} + 2e^{rx} = 0, \] or equivalently, after dividing by \(e^{rx}\) (which is never zero), \[ r^2 - 3r + 2 = 0. \] This factors as \[ (r-1)(r-2)=0, \] so that the characteristic roots are \(r=1\) and \(r=2\). Therefore, the general solution is \[ y(x)= C_1e^{x}+C_2e^{2x}. \] Now we use the initial conditions to find \(C_1\) and \(C_2\). For \(x=0\): \[ y(0)= C_1+C_2 = 3. \] Differentiating, we have \[ y'(x)= C_1e^{x}+2C_2e^{2x}. \] Thus, at \(x=0\): \[ y'(0)= C_1+2C_2 = 1. \] Subtracting the first equation from the second, we get: \[ (C_1+2C_2)-(C_1+C_2)= 1-3 \quad \Longrightarrow \quad C_2 = -2. \] Then, \(C_1= 3-C_2= 3-(-2)= 5\). Thus, the solution is \[ \textcolor{blue}{\boxed{y(x)= 5e^{x}-2e^{2x}.}} \]