Finding $C_n=4C_{n-1}+6n-1$ 2.$$\begin{cases}
C_n=4C_{n-1}+6n-1\\
C_0=2\\
\end{cases} $$
$\begin{cases}
C_n=D_n+dn+e\\
D_n=\lambda D_{n-1}\end{cases}$
$D_n+dn+e=4C_{n-1}+6n-1$ 
But $C_{n-1}=D_{n-1}+d(n-1)+e$
So $D_n+dn+e=4(D_{n-1}+d(n-1)+e)+6n-1$
$D_n+dn+e=4D_{n-1}+4dn-4d+4e+6n-1$
But for $\lambda=4$ we get  $4D_{n-1}+dn+e=4D_{n-1}+4dn-4d+4e+6n-1$ and
$0=-dn-e+4dn-4d+4e+6n-1$  or $0=(3d+6)n+3e-4d-1$
So $d=-2,e=1.5$ 
$C_n=D_n+dn+e$ So $C_n=D_n-2n+1.5$ And $C_0=D_0+1.5$ So $2=D_0+1.5$ And $D_0=-0.5$
$D_n=4D_{n-1}$
So
$C_n=D_n+dn+e=4^n-2n+1.5$ which is $O(4^n)$
Is it correct?

No, it's not correct.
$3d+6=0$ and $3e-4d-1=0$ gives $d=-2$ and $e=-\frac73$.
From $D_n=4D_{n-1}$, we get $D_n=D_0 4^n$ with $D_0 = C_0 - 0d - e = 2 + \frac73 = \frac{13}{3}$, that is, $D_n=\frac{13}{3} 4^n$.
This gives $C_n = D_n -2n -\frac73 = \frac{13}{3} 4^n -2n -\frac73 = \frac13 (13 \cdot 4^n - 6 n - 7)$.
Your other mistake was concluding that $D_n=4^n$ instead of $D_n=D_0 4^n$.
