Answer
$x_1(t)=5\sin t\\
x_2(t)=\cos t-2\sin t$
Work Step by Step
$x_1'=2x_1+5x_2$
$x_2'=-x_1-2x_2$
$x_1(0)=0\\
x_2(0)=1$
We begin by rewriting the system in operator form as $(D−2)x_1 - 5x_2=0$ (1)
$x_1+(D + 2)x_2 =0 $ (2)
To eliminate $x_2$ between these two equations, we first operate on equation (2) with $-\frac{1}{4}(D-2)$ to obtain:
$-(D−2)x_1 -(D-2)(D+2)x_2=0$
Adding equation (1) to this equation eliminates $x_2$ and yields
$(-D^2+1)x_2 =0\\
(D^2+1)x_2=0$.
This constant coefficient differential equation has auxiliary polynomial
$P(r)=r^2+1$
Consequently, $x_2(t)=c_1\cos t+c_2\sin t$
Since $x_1(0)=0\\
x_2(0)=1$
Substitute:
$-2c_1-c_2=0\\
c_1=1$
Solve the system of equation
$c_1=1\\
c_2=-2$
Hence,
$x_1(t)=5\sin t\\
x_2(t)=\cos t-2\sin t$
