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.
Example: solve the equation  {jatex options:inline}\frac{d^2 \mathbf{r}}{dt^2}+9 \mathbf{r}=2t \mathbf{j}{/jatex}  with the initial conditions - where  {jatex options:inline}t=0{/jatex},  {jatex options:inline}\mathbf{v}=2 \mathbf{i}{/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}+9x=0{/jatex}  (1_
{jatex options:inline}\frac{d^2y}{dt^2}+9y=2t{/jatex}  (2)
(1) has solution {jatex options:inline}x=Acos 3t+Bsin3t{/jatex}.
(2) has complementary solution {jatex options:inline}y_c=Ccos 3t+Dsin3t{/jatex}.
We can find a particular solution  {jatex options:inline}y_p{/jatex}  to (2) by assuming  {jatex options:inline}y_p=E+Ft{/jatex}  and substituting into (2).
Differentiating twice gives zero, hence  {jatex options:inline}0+9*(E+Ft)=2t \rightarrow E=0, \; F=\frac{2}{9}\rightarrow y_p=\frac{2}{9}t{/jatex}.
Hence  {jatex options:inline}y=y_c+y_p=Ccos 3t+Dsin3t+ \frac{2}{9}t{/jatex}. Then  {jatex options:inline}\mathbf{r} =(Acos 3t+Bsin3t)+ \mathbf{i}+ (Ccos 3t+Dsin3t+ \frac{2}{9}t) \mathbf{j}{/jatex}.
Now use the initial conditions.
{jatex options:inline}\mathbf{v}=(-3A sin3t+3Bcost) \mathbf{i} + (-3Csin3t+3Dcos3t + \frac{2}{9}) \mathbf{j}{/jatex}
{jatex options:inline}\mathbf{v}(0)=0 \mathbf{i}+ 2 \mathbf{j} =(3B) \mathbf{i} + (3D + \frac{2}{9}) \mathbf{j} \rightarrow B=0, \; D=\frac{2-2/9}{3}=\frac{16}{27}{/jatex}
{jatex options:inline}\mathbf{r}(0)= \mathbf{i}+0 \mathbf{j}= (A) \mathbf{i}+ (C) \mathbf{j} \rightarrow A=1, \; C=0{/jatex}.
Then {jatex options:inline}\mathbf{r} =Acos 3t \mathbf{i}+ ( \frac{16}{27} sin 3t+ \frac{2}{9}t) \mathbf{j}{/jatex}.