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, \[\mathbf{r}= \mathbf{j}\]
mWith
\[\mathbf{r}= x \mathbf{i} + y \mathbf{j}\]
, 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
\[\mathbf{r} =Acos 3t \mathbf{i}+ ( \frac{16}{27} sin 3t+ \frac{2}{9}t) \mathbf{j}\]
.