Solve the vector differential equation $\mathbf{r}^{\prime} (t)=2t \mathbf{i}+e^{t}\mathbf{j}+e^{-t}\mathbf{k}$ with the initial condition $\mathbf{r}(0)=\mathbf{i}-\mathbf{j}+\mathbf{k}$.

Solution The general solution to the differential equation is $$\begin{eqnarray*} \mathbf{r}(t)=\int \mathbf{r}^{\prime} (t)dt&=&\int (2t\mathbf{i}+e^{t}\mathbf{j }+e^{-t}\mathbf{k})dt\\ &=&\left(\int 2tdt\right) \mathbf{i}+\left(\int e^{t}~dt\right) \mathbf{j}+\left( \int e^{-t}dt\right) \mathbf{k}\\ &=&(t^{2}+c_{1})\mathbf{i}+(e^{t}+c_{2})\mathbf{j}+(-e^{-t}+c_{3}) \mathbf{k} \end{eqnarray*}$$

Now we use the initial condition $\mathbf{r}(0)=\mathbf{i}-\mathbf{j}+ \mathbf{k}$. $$\begin{equation*} \mathbf{r}( 0)\;=\;c_{1}\mathbf{i}+(1+c_{2})\mathbf{j}+(-1+c_{3}) \mathbf{k=i}-\mathbf{j}+\mathbf{k} \end{equation*}$$ from which we find $$\begin{align*} \begin{array}{rcl@{\qquad}rcl} c_{1}=1\qquad 1+c_{2}&=-1&-1+c_{3}=1\\ c_{2}&=-2&\qquad c_{3}=2 \end{array} \end{align*}$$

797

The vector function $\mathbf{r=r}( t)$ is $$\begin{equation*} \mathbf{r}(t)=(t^{2}+1)\mathbf{i}+(e^{t}-2)\mathbf{j}+(2-e^{-t})\mathbf{k} \notag \end{equation*}$$