Discrepancy when calculating LTI system output using inverse z-Transform I'm given a difference equation, $y[n]-0.4y[n-1]=x[n]$, and asked to find the natural response $y_n[n]$, forced response $y_f[n]$ and complete response $y[n]$ if $x[n]=4 (0.25)^nu[n]$ and $y[0]=0$.

By one approach I've read,
$$
\text{characteristic equation:}\quad z-0.4=0 \quad\therefore y_n[n]=A(0.4)^n \\
\\
y_f[n]=B(0.25)^n \quad\text{substitute back into diffeq:}\\
B(0.25)^n -0.4B(0.25)^{n-1}=B(0.25)^n - \frac{0.4}{0.25}B(0.25)^n= 4(0.25)^nu[n]\\
\therefore B=-\frac{20}{3}, \quad y_f[n]=-\frac{20}{3}(0.25)^nu[n]\\
\therefore y[n] = A(0.4)^n-\frac{20}{3}(0.25)^nu[n]\\
y[0]=0 \rightarrow A=\frac{20}{3} \\
y[n]= \frac{20}{3}(0.4)^n-\frac{20}{3}(0.25)^nu[n]\\
$$ 
Great! Alternatively, I should be able to determine $y[n]$ by taking the inverse z Transform of $Y(z)=H(z)X(z)$, which in this case is (I'm pretty sure)
$$
Y(z)= \frac{z}{z-0.4}\cdot 4 \frac{z}{z-0.25}\\
\begin{align}
\frac{Y(z)}{z} &amp;= \frac{4z}{(z-0.4)(z-0.25)}=\frac{C}{z-0.4}+\frac{D}{z-0.25}\\
&amp;=\frac{32}{3} \frac{1}{z-0.4} - \frac{20}{3}\frac{1}{z-0.25}\\
\end{align}\\
\therefore Y(z)=\frac{1}{3} \left[\frac{32z}{z-0.4} - \frac{20z}{z-0.25}\right]\\
\therefore y[n]= \left[ \frac{32}{3}(0.4)^n - \frac{20}{3}(0.25)^n \right] u[n]
$$
which is almost the same, but not quite.  

Where did I go wrong? Also, in the second approach, how do you take into account the initial condition of $y[0]=0$? And thirdly, in the first approach, is the first term of $y[n]$ multiplied by $u[n]$ or not? If so, how do you show this?

(for anyone wondering, this is not homework for a course, just self study...)
As Matt L pointed out, your initial condition is wrong. y[0] is NOT 0, it's 4. This is obvious from the difference equation. Once you use y[0] = 4, both results come out to be the same.