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} $

$ \begin{gathered} Y(Z)=\frac{Z}{Z-\frac{1}{4}} \frac{\left(2+\frac{3}{2} Z^{-1}\right)}{\left(1-\frac{3}{2} Z^{-1}+\frac{1}{2} Z^{-2}\right)}-\frac{\frac{1}{2}}{1-\frac{3}{2} Z^{-1}+\frac{1}{2} Z^{-2}} \\\\ Y(Z)=\frac{Z}{Z-\frac{1}{4}}\left(\frac{2 Z^2+\frac{3}{2} Z}{Z^2-\frac{3}{2} Z+\frac{1}{2}}\right)-\frac{\frac{1}{2} Z^2}{Z^2-\frac{3}{2} Z+\frac{1}{2}}\\ \end{gathered} $

$ \begin{aligned} & \frac{Y(Z)}{Z}=\frac{\left(2 Z^2+\frac{3}{2} Z\right)}{\left(Z-\frac{1}{4}\right)(Z-1)\left(Z-\frac{1}{2}\right)}-\frac{\frac{1}{2} Z}{(Z-1)\left(Z-\frac{1}{2}\right)}=\frac{\left(2 Z^2+\frac{3}{2} Z\right)-\frac{1}{2} Z\left(Z-\frac{1}\\{4}\right)}{\left(Z-\frac{1}{4}\right)(Z-1)\left(Z-\frac{1}{2}\right)} \\\\ & =\frac{2 Z^2+\frac{3}{2} Z-\frac{1}{2} Z^2+\frac{1}{8} Z}{\left(Z-\frac{1}{4}\right)(Z-1)\left(Z-\frac{1}\\{2}\right)}=\frac{\frac{3}{2} Z^2+\frac{13}{8} Z}{\left(Z-\frac{1}{4}\right)(Z-1)\left(Z-\frac{1}{2}\right)} \\\ & =\frac{A}{\left(Z-\frac{1}{4}\right)}+\frac{B}{(Z-1)}+\frac{C}{\left(Z-\frac{1}{2}\right)} \\\\ & \end{aligned} $

$ \frac{3}{2} Z^2+\frac{13}{8} Z=A(Z-1)\left(Z-\frac{1}{2}\right)+B\left(Z-\frac{1}{4}\right)\left(Z-\frac{1}{2}\right)+C\left(Z-\frac{1}{4}\right)(Z-1)\\ $

$\operatorname{At} Z=\frac{1}{4}: A=\frac{8}{3}$

At $Z=1: B=\frac{25}{3}$

At $Z=\frac{1}{2}: C=\frac{-19}{2}$

$ \begin{gathered} \therefore \frac{Y(Z)}{Z}=\frac{\frac{8}{3}}{Z-\frac{1}{4}}+\frac{\frac{25}{3}}{Z-1}-\frac{\frac{19}{2}}{Z-\frac{1}{2}} \\\\ Y(Z)=\frac{8}{3} \frac{Z}{Z-\frac{1}{4}}+\frac{25}{3} \frac{Z}{Z-1}-\frac{19}{2} \frac{Z}{Z-\frac{1}{2}}\\ \end{gathered} $

Applying inverse Z transform,

$ y(n)=\frac{8}{3}\left(\frac{1}{4}\right)^n u(n)+\frac{25}{3} u(n)-\frac{19}{2}\left(\frac{1}{2}\right)^n u(n)\\ $