Consider the equation : $$ y[n]-\frac{1}{2}y[n-1]=x[n] $$ where $x[n]:=\left(\frac{1}{3}\right)^{n}u[n]$. First, I founded the homogenous solution $y_{h}:=A\frac{1}{2^{n}}$ $$ y_{h}[n]=\frac{1}{2^{n}}y[0] $$ where I insist $A=y[0]$. Furthermore, The particular solution is given of the form $y_{p}[n]:=B\left(\frac{1}{3}\right)^{n}$ I shall determine $B$ as follow : $$ B\left(\frac{1}{3}\right)^{n}u[n]-\frac{1}{2}B\left(\frac{1}{3}\right)^{n-1}u[n-1]=\left(\frac{1}{3}\right)^{n}u[n] $$ \begin{align*} \implies B&=\frac{\displaystyle\left(\frac{1}{3}\right)^{n}u[n]}{\displaystyle\left(\frac{1}{3}\right)^{n}u[n]-\frac{1}{2}\displaystyle\left(\frac{1}{3}\right)^{n-1}u[n-1]}\\ \\ &=\frac{\displaystyle\left(\frac{1}{3}\right)^{n}}{\displaystyle\left(\frac{1}{3}\right)^{n}-\displaystyle\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}}\\ \\ &=-2 \end{align*}
So now that we have $B=-2$, We are given the initial rest condition so what I did was sub $n=0$ in difference equation $$ y[0]-\frac{1}{2}y[-1]=x[0] $$ $$ \implies y[0]=1 $$ $$ **\text{So this means $A=y[0]=1$}** $$ However, the major issue is that : $$ y[0]-\frac{1}{2}y[-1]=x[0] $$ $$ \implies y[0]=1 $$ $$ \implies y[n]:= y_{h}[n]+y_{p}[n] \implies y[0]=y_{h}[0]+y_{p}[0]=1 $$ $$ \implies A\left(\frac{1}{2}\right)^{0} -2\left(\frac{1}{3}\right)^{0}=1 $$ $$ \implies A=3 $$ So I have $A=1$ and $A=3$, an obvious contradiction and I hope someone can help me