Answer
$f(t)=2t^{3}+\cos t+C_{1}t^{2}+Dt+E$
Work Step by Step
$\left[\begin{array}{ll}
\text{}f''' & \text{particular antiderivative,}f''\\
12=12\cdot t^{0} & 12\cdot\dfrac{t^{0+1}}{0+1}=12t\\
\sin t & -\cos t
\end{array}\right]$
$f''(t)=12t-\cos t+C$
$\left[\begin{array}{ll}
\text{}f'' & \text{particular antiderivative,}f'\\
12t=12t^{1} & 12\cdot\dfrac{t^{1+1}}{1+1}=6t^{2}\\
-\cos t & -\sin t\\
C & Ct
\end{array}\right]$
$f'(t)=6t^{2}-\sin t+Ct+D$
$\left[\begin{array}{ll}
\text{}f' & \text{particular antiderivative,}f\\
6t^{2} & 6\cdot\dfrac{t^{2+1}}{2+1}=2t^{3}\\
-\sin t & \cos t\\
Ct & C\cdot\frac{t^{1+1}}{1+1}=\frac{Ct^{2}}{2}=C_{1}t^{2}\\
D & Dt
\end{array}\right]$
$f(t)=2t^{3}+\cos t+C_{1}t^{2}+Dt+E$
