Give the equation $$y[n]-y[n-1] +\frac14y[n-2]= x[n]$$ With initial conditions $$y[0]=0\\ y[1]= 0$$ Find the step response.
I have been able to get the solution $$y[n] =C_1(\frac12)^n +C_2(\frac12)^n + 1,,$$ I am have problem finding $C_1$ and $C_2$. How do I find $C_1$ and $C_2$?
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.
If your difference equation is:
$$y[n]-y[n-1]+\frac{1}{4}y[n-2]=x[n],$$
then it describes a system with the following transfer function:
$$H(z)=\frac{1}{1-z^{-1}+\frac{1}{4}z^{-2}}$$
If $x[n]=u[n]$ is the unit step function, i.e.:
$$u[n]=\begin{cases} 1, & \text{for } n \ge 0\\ 0, & \text{for } n < 0 \end{cases}$$ then its $z$-transform $\displaystyle U(z)=\frac{1}{1-z^{-1}}$ and the unit step response, in $z$-domain, is:
$$Y_u(z)=H(z)U(z)=\frac{1}{(1-z^{-1}+\frac{1}{4}z^{-2})(1-z^{-1})}=\frac{1}{(1-\frac{1}{2}z^{-1})^2(1-z^{-1})},$$
which you can write as:
$$Y_u(z)=\frac{C_1}{(1-\frac{1}{2}z^{-1})^2}+\frac{C_2}{(1-z^{-1})}$$
where $C_1$ and $C_2$ are constants to be determined. Knowing that ($z$-transform tables):
$$\frac{z^{-1}}{(1-az^{-1})^2}\rightleftharpoons nu[n]a^{n-1}$$
and using the time-shift property you can deduce that:
$$\frac{1}{(1-az^{-1})^2}\rightleftharpoons (n+1)u[n+1]a^{n}.$$
Hence,
$$y_u[n]=C_1(n+1)(\frac{1}{2})^{n}u[n+1]+C_2u[n].$$
All you need to do now is to use the initial conditions to get $C_1$ and $C_2$ ($y_u[0]=0$ and $y_u[1]=0$). You will find two equations and both give you $C_1+C_2=0$ which is not sufficient to find the coefficients. You need then to compute $y_u[2]$ which you will find equal to $1$. Using this, you get in the end:
$$y_u[n]=4(u[n]-(n+1)(\frac{1}{2})^{n}u[n+1])$$
Taking the $z$-transform and applying the initial conditions, $$Y(z)-z^{-1}Y(z)+\frac14z^{-2}Y(z)=X(z)$$ $$Y(z)\left(1-z^{-1}+\frac{1}{4}z^{-2}\right)=X(z)$$ $$\frac{Y(z)}{X(z)}=H(z)=\frac{1}{1-z^{-1}+\frac{1}{4}z^{-2}}=\frac{1}{(1-\frac12z^{-1})^2}=\frac{z^2}{(z-\frac12)^2}$$ Hence, the step response is given by the inverse $z$-transform of $\frac{z^2}{(z-\frac12)^2}\cdot\frac{z}{z-1}$. Can you do it?
Update - Due to a mistake in another answer, I will give a complete answer.
Using partial fraction expansion:
$$\begin{align} F(z)=\frac{S(z)}{z}=\frac{z^2}{(z-\frac12)^2(z-1)}=\frac{A}{(z-\frac12)^2}+\frac{B}{z-\frac12}+\frac{C}{z-1} \end{align}$$ gives $$C=(z-1)F(z)\biggr\vert_{z=1}=\frac{z^2}{(z-\frac12)^2}\biggr\vert_{z=1}=4$$ $$A=(z-\frac12)^2F(z)\biggr\vert_{z=\frac12}=\frac{z^2}{z-1}\biggr\vert_{z=\frac12}=-\frac12$$ $$B=\frac{d}{dz}\left((z-\frac12)^2F(z)\right)\biggr\vert_{z=\frac12}=\frac{2z(z-1)-z^2}{(z-1)^2}\biggr\vert_{z=\frac12}=\frac{z(z-2)}{(z-1)^2}\biggr\vert_{z=\frac12}=-3$$ Now use a table of $z$-transforms and its properties to find the final result:
$$s[n]=\mathcal{Z}^{-1}\left(\frac{Az}{(z-\frac12)^2}+\frac{Bz}{z-\frac12}+\frac{Cz}{z-1}\right)$$
