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?

We start by solving the homogenous part: letting $x^n=C_n$ gives the characteristic equation $$x^n=4x^{n-1}\Rightarrow x=4,$$ which gives $$C_n=A(4^n).$$ Now, for the the inhomogenous part, let us guess that the solution is of the form $$p(n)=d_0+d_1n,$$ substituting into the recurrence gives $$d_0+d_1n=4(d_0+d_1(n-1))+6n-1,$$ which, by comparing coefficients gives $$d_0=-\frac{7}{3},\quad d_1=-2.$$ The final solution is obtained by adding the homogenous and inhomogenous parts; and thus $$C_n=A(4^n)-\frac{7}{3}-2n.$$ Finally, we may find $A$ by imposing the initial condition $$2=A-\frac{7}{3}\Rightarrow A=\frac{13}{3}.$$ This therefore gives $$C_n=\frac{13}{3}(4^n)-\frac{7}{3}-2n.$$
