We first need the complementary solution so the characteristic equation is,

\[{r^3} - 12{r^2} + 48r - 64 = {\left( {r - 4} \right)^3} = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}r = 4\,\,\left( {{\mbox{multiplicity 3}}} \right)\]

We’ve got a single root of multiplicity 3 so the complementary solution is,

\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{4t}} + {c_2}t{{\bf{e}}^{4t}} + {c_3}{t^2}{{\bf{e}}^{4t}}\]

Now, our first guess for a particular solution is,

\[{Y_P} = A + B{{\bf{e}}^{ - 8t}} + C{{\bf{e}}^{4t}}\]

Notice that the last term in our guess is in the complementary solution so we’ll need to add one at least one \(t\) to the third term in our guess. Also notice that multiplying the third term by either \(t\) or \({t^2}\) will result in a new term that is still in the complementary solution and so we’ll need to multiply the third term by \({t^3}\) in order to get a term that is not contained in the complementary solution.

Our final guess is then,

\[{Y_P} = A + B{{\bf{e}}^{ - 8t}} + C{t^3}{{\bf{e}}^{4t}}\]

Now all we need to do is take three derivatives of this, plug this into the differential equation and simplify to get (we’ll leave it to you to verify the work here),

\[ - 64A - 1728B{{\bf{e}}^{ - 8t}} + 6C{{\bf{e}}^{4t}} = 12 - 32{{\bf{e}}^{ - 8t}} + 2{{\bf{e}}^{4t}}\]

Setting coefficients equal and solving gives,

\[\begin{aligned}{t^0} & : & - 64A &= 12\\{{\bf{e}}^{ - 8t}} & : & - 1728B & = - 32\\{{\bf{e}}^{4t}} & : & 6C & = 2\end{aligned}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{array}{*{20}{l}}{A = - \frac{3}{{16}}}\\{B = \frac{1}{{54}}}\\{C = \frac{1}{3}}\end{array}\]

A particular solution is then,

\[{Y_P} = - \frac{3}{{16}} + \frac{1}{{54}}{{\bf{e}}^{ - 8t}} + \frac{1}{3}{t^3}{{\bf{e}}^{4t}}\]

The general solution to this differential equation is then,

\[y\left( t \right) = {c_1}{{\bf{e}}^{4t}} + {c_2}t{{\bf{e}}^{4t}} + {c_3}{t^2}{{\bf{e}}^{4t}} - \frac{3}{{16}} + \frac{1}{{54}}{{\bf{e}}^{ - 8t}} + \frac{1}{3}{t^3}{{\bf{e}}^{4t}}\]