## Related questions with answers

Solve the given system of differential equations.

$x _ { 1 } ^ { \prime } = 2 x _ { 1 } , \quad x _ { 2 } ^ { \prime } = x _ { 2 } - x _ { 3 } , \quad x _ { 3 } ^ { \prime } = x _ { 2 } + x _ { 3 }.$

Solutions

VerifiedWe are given the following system of differential equations

$\begin{align*} x_1'&=2x_1,\\ x_2'&=x_2-x_3,\\ x_3'&=x_2+x_3. \end{align*}$

We need to solve it.

Consider the system of differential equations

$\begin{array}{lll}x'_1=2x_1\\ x'_2=x_2-x_3\\ x'_3=x_2+x_3\end{array}$

.

Since $x'_1=2x_1$ we have that $x(t)=c_1e^{2t}$. Then we are left with the system

$\begin{array}{cc}x'_2=x_2-x_3\\ x'_3=x_2+x_3\end{array}$

which can be written in operator notation

$\begin{array}{cc} (D-1)x_2+x_3=0\\ -x_2+(D-1)x_3=0\end{array}$

.

Applying $(D-1)$ to the both sides of the equation $-x_2+(D-1)x_3=0$ we get $-(D-1)x_2+(D-1)^2x_3=0$.

Adding the equations $(D-1)x_2+x_3=0$ and $-(D-1)x_2+(D-1)^2x_3=0$ gives us the homogeneous equation $(D^2-2D+2)x_3=0$.

The auxiliary polynomial of $(D^2-2D+2)x_3=0$ is $P(r)=r^2-2r+2=0$ which has $r_1=1-i$ and $r_2=1+i$ as roots.

Thus $x_3(t)=c_2e^t\cos t+c_3e^t\sin t$.

Hence since $-x_2+(D-1)x_3=0$ we have that $x_2(t)=(D-1)x_3(t)=c_3e^t\cos t-c_2e^t\sin t$.

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