Author Topic: Q5--T0401, T0901  (Read 1920 times)

Victor Ivrii

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Q5--T0401, T0901
« on: March 09, 2018, 05:49:00 PM »
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.
« Last Edit: March 09, 2018, 05:52:10 PM by Victor Ivrii »

Darren Zhang

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Re: Q5--T0401, T0901
« Reply #1 on: March 09, 2018, 05:57:49 PM »
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.
« Last Edit: March 09, 2018, 06:00:43 PM by Junjie Zhang »

Junya Zhang

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Re: Q5--T0401, T0901
« Reply #2 on: March 09, 2018, 06:19:35 PM »
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.
« Last Edit: March 09, 2018, 06:25:33 PM by Junya Zhang »