Consider the differential equation $$ \frac{d \textbf{r}}{dt}=(t^3+4t)\,\vec i + t \vec j, \qquad \textbf{r}(0)= \vec i+ \vec j $$
I got
$$\textbf{r}(t) = (\frac{t^4}{4}+ 2t^2) \vec i + \frac{t^2}{2} \vec j.$$
What's the next step? What to do with $\textbf{r}(0)=\vec i+ \vec j$ ?
It seems that this is a problem in the 3d physical space. If this is the case, you cannot in general neglect the $\vec k$ component. The general solution would then be
$$r(t) = \left(\frac 14 t^4 + 2t^2 + c_1\right) \vec i + \left(\frac 12 t^2+ c_2\right) \vec j + c_3 \vec k$$
The initial condition $r(0)= \vec i + \vec j$ means that when $t=0$ the $\vec i$ and $\vec j$ components evaluate to 1 and the $\vec k$ component evaluates to zero. Taking this into consideration, you know that $c_1=c_2 = 1$ and $c_3=0$.