## Cartesian Equation of Motion From a Vector Differential Equation

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

.
Example: Solve the equation
$\frac{d^2 \mathbf{r}}{dt^2}+2 \mathbf{i}=0$
with the initial conditions - where
$t=0$
,
$\mathbf{v}=2 \mathbf{i}+3 \mathbf{j}$
m/s,
$\mathbf{r}= \mathbf{j}$
m
With
$\mathbf{r}= x \mathbf{i} + y \mathbf{j}$
, the equations of the components are
$\frac{d^2x}{dt^2}+2=0$
(1_
$\frac{d^2y}{dt^2}=0$
(2)
(1) has solution
$x=-t^2+At+B$
.
(2) has solution
$y=Ct+D$
.
Then
$\mathbf{r} =(-t^2+At+B) \mathbf{i}+ (Ct+D) \mathbf{j}$
.
Now use the initial conditions.
$\mathbf{v}=(-2t+A) \mathbf{i} + (C) \mathbf{j}$

$\mathbf{v}(0)=2 \mathbf{i}+3 \mathbf{j} \rightarrow 2=(-2(0)+A), \; 3 =C \rightarrow A=2, C=3$

$\mathbf{r}(0)= \mathbf{j}= (B) \mathbf{i}+ (D) \mathbf{j} \rightarrow B=0, \; D=1$
.
Then
$\mathbf{r} =(-t^2-t) \mathbf{i}+ ( 3t+1) \mathbf{j}$
.
Then
$x=-t^2+2t, \; y=3t+1$
.
From the second equation,
$t=\frac{y-1}{3}$
so
$x=(\frac{y-1}{3})^2+2(\frac{y-1}{3})$