We can solve vector differential equations by writing separate differential equations for each component. Solve these differential equations, then add them with the appropriate coordinate vectors to the give the answer to the question.
A Cartesian equation may be obtained by elimination of  {jatex options:inline}{/jatex}.
Example: Solve the equation  {jatex options:inline}\frac{d^2 \mathbf{r}}{dt^2}+2 \mathbf{i}=0{/jatex}  with the initial conditions - where  {jatex options:inline}t=0{/jatex},  {jatex options:inline}\mathbf{v}=2 \mathbf{i}+3 \mathbf{j}{/jatex}  m/s,  {jatex options:inline}\mathbf{r}= \mathbf{j}{/jatex}  m
With  {jatex options:inline}\mathbf{r}= x \mathbf{i} + y \mathbf{j}{/jatex}, the equations of the components are
{jatex options:inline}\frac{d^2x}{dt^2}+2=0{/jatex}  (1_
{jatex options:inline}\frac{d^2y}{dt^2}=0{/jatex}  (2)
(1) has solution {jatex options:inline}x=-t^2+At+B{/jatex}.
(2) has solution {jatex options:inline}y=Ct+D{/jatex}.
Then  {jatex options:inline}\mathbf{r} =(-t^2+At+B) \mathbf{i}+ (Ct+D) \mathbf{j}{/jatex}.
Now use the initial conditions.
{jatex options:inline}\mathbf{v}=(-2t+A) \mathbf{i} + (C) \mathbf{j}{/jatex}
{jatex options:inline}\mathbf{v}(0)=2 \mathbf{i}+3 \mathbf{j} \rightarrow 2=(-2(0)+A), \; 3 =C \rightarrow A=2, C=3{/jatex}
{jatex options:inline}\mathbf{r}(0)= \mathbf{j}= (B) \mathbf{i}+ (D) \mathbf{j} \rightarrow B=0, \; D=1{/jatex}.
Then {jatex options:inline}\mathbf{r} =(-t^2-t) \mathbf{i}+ ( 3t+1) \mathbf{j}{/jatex}.
Then  {jatex options:inline}x=-t^2+2t, \; y=3t+1{/jatex}.
From the second equation,  {jatex options:inline}t=\frac{y-1}{3}{/jatex}  so  {jatex options:inline}x=(\frac{y-1}{3})^2+2(\frac{y-1}{3}){/jatex}