Author Topic: Q5--T0201, T0601  (Read 925 times)

Victor Ivrii

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Q5--T0201, T0601
« on: March 09, 2018, 05:51:10 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 $t \ge  0$.

$$\left\{\begin{aligned}
&x'_1 = 1.25x_1 + 0.75x_2, &&x_1(0) = -2,\\
&x'_2= 0.75x_1 + 1.25x_2, &&x_2(0) = 1.
\end{aligned}\right.$$
« Last Edit: March 09, 2018, 05:56:05 PM by Victor Ivrii »

Junya Zhang

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Re: Q5--T0201, T0601
« Reply #1 on: March 09, 2018, 05:59:39 PM »
a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{4}{3}x_1' - \frac{5}{3}x_1$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{4}{3}x_1'' - \frac{5}{3}x_1'$$
Substitute into the second equation and simplify, we get $$ x_1'' - \frac{5}{2}x_1' + x_1 = 0 $$
which is a second order ODE of $x_1$.


b)
Characteristic equation is $r^2 - \frac{5}{2} r + 1 = (r - \frac{1}{2})(r - 2) = 0$ with roots $r_1 = \frac{1}{2}, r_2 = 2$
General solution for $x_1$ is $x_1 = c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$
Plug in to $x_2 = \frac{4}{3}x_1' - \frac{5}{3}x_1$ get
$$x_2 = -c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$$
So, $$x_1 = c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$$ $$x_2 = -c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$$
Plug in $x_1(0)=-2, x_2(0) = 1$ to get $$-2 = c_1 + c_2 $$  $$1= -c_1 + c_2 $$
Solve the linear system we have
$$c_1 = -\frac{3}{2}, c_2 = -\frac{1}{2}$$
That is, $$x_1 = -\frac{3}{2} e^{\frac{1}{2}t} -\frac{1}{2} e^{2t}$$ $$x_2 = \frac{3}{2} e^{\frac{1}{2}t} -\frac{1}{2} e^{2t}$$

c) See attached picture
Note that as $t \to \infty$, the graph is asymptotic to the line $x_2 = x_1$ in the third quadrant.
« Last Edit: March 09, 2018, 06:15:12 PM by Junya Zhang »