Q5--T0401, T0901

[Victor Ivrii]

a.  Transform the given system into a single equation of second order.

b. Find $x_1$ and $x_2$ that also satisfy the given initial conditions.

c.  Sketch the graph of the solution in the $(x_1,x_2)$-plane for $t \ge  0$.


$$\left\{\begin{aligned}
&x'_1= 2x_2, &&x_1(0) = 3,\\
&x'_2= -2x_1, &&x_2(0) = 4.
\end{aligned}\right.$$

[Darren Zhang]

Solving the first equation for B , we have $x_{2} = x_{1}'/2$. Substitution into the second equation results in $$x_{1}''/2 = -2x_{1}$$
The resulting equation is $x_1'+4x_1 = 0$, with general soln. $$x_{1}(t) = c_{1}cos2t+c_{2}sin2t$$
With $x_2$ given in terms of x_{1}, it follows that $$x_2(t) = -c_{1}sin2t+c_{2}cos2t$$.
Imposing the specific conditions, we have $c_1=3,c_2 = 4$.
Therefore, $x_1(t) = 3cos2t+4sin2t$ and $x_2(t)=-3sin2t+4cos2t$
Attached is the graph.

[Junya Zhang]

Junjie's solution is correct, but here's a more detailed version

a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{1}{2}x_1'$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{1}{2}x_1''$$
Substitute into the second equation and simplify, we get $$ x_1'' + 4 x_1 = 0 $$
which is a second order ODE of $x_1$.


b)
Characteristic equation is $r^2 +4 = 0$ with roots $r_1 = 2i, r_2 = -2i$
General solution for $x_1$ is $x_1 = c_1 \cos{2t} + c_2 \sin{2t}$
Plug in to $x_2 = \frac{1}{2}x_1'$ get
$$x_2 = -c_1 \sin{2t} + c_2 \cos{2t}$$
So, $$x_1 = c_1 \cos{2t} + c_2 \sin{2t}$$ $$x_2 = -c_1 \sin{2t} + c_2 \cos{2t}$$
Plug in $x_1(0)=3, x_2(0) = 4$ to get
$$c_1 = 3, c_2 = 4$$
That is, $$x_1 = 3 \cos{2t} + 4 \sin{2t}$$ $$x_2 = -3 \sin{2t} + 4 \cos{2t}$$

c) See attached graph.
Note that the graph is a circle center at origin with radius 5, and as $t\to \infty$, it is moving clockwise.