Linear difference equation
As shown in other answers, one way to solve such problems is using the $\mathcal{Z}$-transform. However, your approach of finding a general and a particular solution also works, but you've got a few things wrong. Since this looks like homework, I'll just try to give you a few essential hints to enable you to solve the problem yourself (and to actually learn something).

You seem to have found the characteristic polynomial and its solutions:

$$\lambda^2-\lambda+\frac14\lambda=\left(\lambda-\frac12\right)^2=0\tag{1}$$

The characteristic polynomial has a double root at $\lambda=\frac12$, which must be taken into account when forming the general solution. That's where you've gone wrong. You treated the double root as two distinct single roots, which doesn't work. The solution of the homogeneous difference equation ($x[n]=0$) has the form

$$y_h[n]=C_1\left(\frac12\right)^n+C_2n\left(\frac12\right)^n,\qquad n\ge 0\tag{2}$$

A particular solution (taking into account the excitation $x[n]=u[n]$) can be found by inspection:

$$y_p[n]=C_3\tag{3}$$

First determine $C_3$ by simply plugging the particular solution $(3)$ into the difference equation. The final solution is then given by

$$y[n]=y_h[n]+y_p[n]\tag{4}$$

The constants $C_1$ and $C_2$ must now be chosen such that the given initial conditions are satisfied.