Question

not sure how I am supposed to arrive to an answer

Answer 1

Integrating both sides once gives

$\dfrac{\mathrm d\mathbf r}{\mathrm dt}=2e^t\,\mathbf i+3e^{-t}\,\mathbf j+4e^{2t}\,\mathbf k+\mathbf c$

where $\mathbf c$ is an arbitrary constant vector. Use the initial condition to find its value:

$\dfrac{\mathrm d\mathbf r}{\mathrm dt}(0)=-\mathbf i+7\,\mathbf j=(2+c_1)\,\mathbf i+(3+c_2)\,\mathbf j+(4+c_3)\,\mathbf k$

$\implies\mathbf c_1=-3\,\mathbf i+4\,\mathbf j-4\,\mathbf k$

Integrate again:

$\mathbf r(t)=2e^t\,\mathbf i-3e^{-t}\,\mathbf j+2e^{2t}\,\mathbf k+\mathbf c_1t+\mathbf c_2$

where $\mathbf c_2$ is another arbitrary vector of constants. Use the other initial condition to determine its components:

$\mathbf r(0)=6\,\mathbf i+\mathbf j+3\,\mathbf k=(2+c_1)\,\mathbf i+(-3+c_2)\,\mathbf j+(2+c_3)\,\mathbf k$

$\implies\mathbf c_2=4\,\mathbf i+4\,\mathbf j+\mathbf k$

Then the particular solution to this ODE is

$\boxed{\mathbf r(t)=(2e^t-3t+4)\,\mathbf i+(-3e^{-t}+4t+4)\,\mathbf j+(2e^{2t}-4t+1)\,\mathbf k}$