Solution:

$ y(n)-\frac{3}{2} y(n-1)+\frac{1}{2} y(n-2)=2 x(n)+\frac{3}{2} x(n-1)\\ $

Taking Z-transform,

$ \begin{gathered} Y(Z)-\frac{3}{2}\left[Z^{-1} Y(Z)+y(-1)\right]+\frac{1}{2}\left[Z^{-2} Y(Z)+Z^{-1} y(-1)+y(-2)\right] \\\\ =2 X(Z)+\frac{3}{2}\left[Z^{-1} X(Z)+x(-1)\right]\\ \end{gathered} $

$ Y(Z)-\frac{3}{2}\left[Z^{-1} Y(Z)\right]+\frac{1}{2}\left[Z^{-2} Y(Z)+1\right]=X(Z)\left[2+\frac{3}{2} Z^{-1}\right]\\ $

Since x(n) is causal signal $x(-1)=0$

$ \begin{aligned} & x(n)=\left(\frac{1}{4}\right)^n u(n) \Rightarrow X(Z)=\frac{1}{1-\frac{1}{4} Z^{-1}}=\frac{Z}{Z-\frac{1}{4}} \\\\ \therefore & Y(Z)\left[1-\frac{3}{2} Z^{-1}+\frac{1}{2} Z^{-2}\right]=\frac{Z}{Z-\frac{1}{4}}\left[2+\frac{3}{2} \\Z^{-1}\right]-\frac{1}{2}\\ \end{aligned} …