## Solving a Vector Differential Equation With Initial Conditions

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
$\frac{d^2 \mathbf{r}}{dt^2}+9 \mathbf{r}=2t \mathbf{j}$
with the initial conditions - where
$t=0$
,
$\mathbf{v}=2 \mathbf{i}$
m/s,
$\mathbf{r}= \mathbf{j}$
m
With
$\mathbf{r}= x \mathbf{i} + y \mathbf{j}$
, the equations of the components are
$\frac{d^2x}{dt^2}+9x=0$
(1_
$\frac{d^2y}{dt^2}+9y=2t$
(2)
(1) has solution
$x=Acos 3t+Bsin3t$
.
(2) has complementary solution
$y_c=Ccos 3t+Dsin3t$
.
We can find a particular solution
$y_p$
to (2) by assuming
$y_p=E+Ft$
and substituting into (2).
Differentiating twice gives zero, hence
$0+9*(E+Ft)=2t \rightarrow E=0, \; F=\frac{2}{9}\rightarrow y_p=\frac{2}{9}t$
.
Hence
$y=y_c+y_p=Ccos 3t+Dsin3t+ \frac{2}{9}t$
. Then
$\mathbf{r} =(Acos 3t+Bsin3t)+ \mathbf{i}+ (Ccos 3t+Dsin3t+ \frac{2}{9}t) \mathbf{j}$
.
Now use the initial conditions.
$\mathbf{v}=(-3A sin3t+3Bcost) \mathbf{i} + (-3Csin3t+3Dcos3t + \frac{2}{9}) \mathbf{j}$

$\mathbf{v}(0)=0 \mathbf{i}+ 2 \mathbf{j} =(3B) \mathbf{i} + (3D + \frac{2}{9}) \mathbf{j} \rightarrow B=0, \; D=\frac{2-2/9}{3}=\frac{16}{27}$

$\mathbf{r}(0)= \mathbf{i}+0 \mathbf{j}= (A) \mathbf{i}+ (C) \mathbf{j} \rightarrow A=1, \; C=0$
.
Then
$\mathbf{r} =Acos 3t \mathbf{i}+ ( \frac{16}{27} sin 3t+ \frac{2}{9}t) \mathbf{j}$
.

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