Answer
\begin{array}{l}
\quad\left(\frac{1}{3} t^{3}+1\right) \mathrm{i}+\left(t^{2}+1\right) \mathrm{j}
\end{array}
Work Step by Step
We find:
\begin{array}{l}
\int y^{\prime}(t) d t=\frac{1}{3} t^{3} \mathrm{i}+t^{2} \mathrm{j}+\mathrm{C} =\mathrm{y}(t)\\
\mathrm{y}(0)=\mathrm{i}+\mathrm{j} =\mathrm{C}\\
\left(\frac{1}{3} t^{3}+1\right) \mathrm{i}+\left(t^{2}+1\right) \mathrm{j} =\mathrm{y}(t)\\
\end{array}
