If $\frac{d^2\vec{r}}{dt^2}=6t\hat{i}-24t^2\hat{j}+4\sin t \hat{k}$, then prove that $\vec{r}=(t^3-t+2)\hat{i}+(1-2t^4)\hat{j}+(t-4\sin t)\hat{k}$ Problem : If $\dfrac{d^2\vec{r}}{dt^2}=6t\hat{i}-24t^2\hat{j}+4\sin t \hat{k}$ and if $\vec{r}=2\hat{i}+\hat{j}$ and $\dfrac{d\vec{r}}{dt}=-\hat{i}-3\hat{k}$ when $t=0$, then show that $\vec{r}=(t^3-t+2)\hat{i}+(1-2t^4)\hat{j}+(t-4\sin t)\hat{k}$.  
Try: We have
\begin{align*}
\dfrac{d^2\vec{r}}{dt^2}=  6t\hat{i}-24t^2\hat{j}+4\sin t \hat{k}\implies &  d\left(\dfrac{d\vec{r}}{dt}\right)=(6t\hat{i}-24t^2\hat{j}+4\sin t \hat{k})\, dt \\
\implies &  \dfrac{d\vec{r}}{dt}=3t^2\hat{i}-8t^3\hat{j}-4\cos t \hat{k}+ \vec{C}\\
\implies &  \dfrac{d\vec{r}}{dt}=(3t^2-1)\hat{i}-8t^3\hat{j}-(4\cos t+7) \hat{k}\\
\implies &  \vec{r}=(t^3-t)\hat{i}-2t^4\hat{j}-(4\sin t+7t) \hat{k}+\vec{D}\\
\implies &  \vec{r}=(t^3-t)\hat{i}-2t^4\hat{j}-(4\sin t+7t) \hat{k} \quad [\because \vec{D}=\vec 0]
\end{align*} 
which is different from the original result. But the given result is true. My derived result is also satisfies the given diff equn.  Where is the problem?
 A: It seems you made $2$ mistakes. First, with
$$\dfrac{d\vec{r}}{dt}=3t^2\hat{i}-8t^3\hat{j}-4\cos t \hat{k}+ \vec{C} \tag{1}\label{eq1A}$$
From $\dfrac{d\vec{r}}{dt}=-\hat{i}-3\hat{k}$ when $t=0$, you get
$$\begin{equation}\begin{aligned}
-4\hat{k} + \vec{C} & = -\hat{i}-3\hat{k} \\
\vec{C} & = -\hat{i}+\hat{k}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus, you would get
$$\dfrac{d\vec{r}}{dt}=(3t^2-1)\hat{i}-8t^3\hat{j}-(4\cos t-1) \hat{k} \tag{3}\label{eq3A}$$
It seems you subtracting $4\hat{k}$ on the right instead of adding it. Anyway, this is basically what the answer also states.
Near the end, using the correct derivative expression, you would get
$$\vec{r}=(t^3-t)\hat{i}-2t^4\hat{j}-(4\sin t-t) \hat{k}+\vec{D} \tag{4}\label{eq4A}$$
Using the initial condition $\vec{r}=2\hat{i}+\hat{j}$ when $t = 0$ gives
$$2\hat{i}+\hat{j} = \vec{D} \tag{5}\label{eq5A}$$
I don't know why you think $\vec{D} = \vec 0$. Using \eqref{eq5A} in \eqref{eq4A}, you would get
$$\begin{equation}\begin{aligned}
\vec{r} & =(t^3-t+2)\hat{i}-(2t^4-1)\hat{j}-(4\sin t-t) \hat{k} \\
& = (t^3-t+2)\hat{i}+(1-2t^4)\hat{j}+(t-4\sin t)\hat{k}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
This matches what you're asked to show.
A: It looks like you might have an error in the following line:
$\dfrac{d\vec{r}}{dt}=(3t^2-1)\hat{i}-8t^3\hat{j}-(4\cos t+7)\hat{k}$
I believe you should be subtracting 1 in the $\hat{k}$ term instead of adding 7, to match the initial condition given for $t=0$.  I believe it should read:
$\dfrac{d\vec{r}}{dt}=(3t^2-1)\hat{i}-8t^3\hat{j}-(4\cos t-1)\hat{k}$
I hope this helps.
