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Solve:
$x^{\prime \prime}=y, x(0)=3, x^{\prime}(0)=1$
$y^{\prime \prime}=x, y(0)=1, y^{\prime}(0)=-1$

$x^{\prime \prime}=y \rightarrow x^{\prime \prime}-y=0 \rightarrow D^{2} x-y=0 \cdots (1)$

$y^{\prime \prime}-x=0 \rightarrow-x+y^{\prime \prime}=0 \rightarrow-x+D^{2} y=0 \cdots (2)$ multiply $(2)$ with $D^{2}$

$-D^{2} x+D^{4} y=0 \quad \cdots (2)$

$D^{2} x-y=0$
$-D^{2} x+D^{4} y=0$

$\rightarrow D^{4} y-y=0 \rightarrow\left(D^{4}-1\right) y=0$

$\left(m^{4}-1\right) y=0 \quad y \neq 0 \quad m^{4}-1=0$

$\left(m^{2}+1\right)\left(m^{2}-1\right)=0 \longrightarrow\left(r^{2}-1\right)\left(r^{2}+1\right)=0$

$r_{1}=-1, r_{2}=+1, r_{3}=\pm i$

$y(t)=c_{1} e^{t}+c_{2} e^{-t}+c_{3} \cos (t)+c_{4} \sin (t)$
$y^{\prime}(t)=c_{1} e^{t}-c_{2} e^{-t}-c_3 \sin (t)+c_{4} \cos (t)$
$y^{\prime \prime}(t)=c_{1} e^{t}+c_{2}e^{-t}-c_{3} \cos (t)-c_{4} \sin (t)$

$y(t)=c_{1} e^{t}+c_{2}e^{-t}+c_{3} \cos (t)+c_4 \sin (t)$

$x(t)=c_{1} e^{t}+c_{2}e^{-t}-c_{3} \cos (t)-c_4 \sin (t)$

$3=c_{1}+c_{2}-c_{3}$
$1=c_{1}+c_{2}+c_3$

$x^{\prime}(t)=c_{1} e^{t}-c_{2} e^{-t}+c_{3} \sin (t)-c_4 \cos (t)$

$y^\prime(t)=c_{1} e^{t}-c_2 e^{-t}-c_3 \sin (t)+c_4 \cos (t)$

at $\quad t=0 \rightarrow$

$1=c_{1}-c_{2}-c_{4}$
$-1=c_{1}-c_{2}-c_{3}$

Solving the equations:

$\begin{array}{ll}{c_{1}=1} & {c_{3}=-1} \\ {c_{2}=1} & {c_{4}=-1}\end{array}$

$x(t)=e^{t}+e^{-t}+\cos (t)+\sin (t)$

$y(t)=e^{t}+e^{-t}-\cos (t)-\sin (t)$