### Author Topic: Q5--T0301, T0701  (Read 1572 times)

#### Victor Ivrii ##### Q5--T0301, T0701
« on: March 09, 2018, 05:54:45 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 = 3x_1 - 2x_2, &&x_1(0) = 3,\\ &x'_2= 2x_1 - 2x_2, &&x_2(0) = \frac{1}{2}. \end{aligned}\right.

#### Guanlin Song

• Newbie
• • Posts: 2
• Karma: 1 ##### TUT0701
« Reply #1 on: March 09, 2018, 05:56:40 PM »
Attached is the solution to TUT 0701

Clearer graph attached on second one
« Last Edit: March 09, 2018, 06:16:02 PM by Guanlin Song »

#### Junya Zhang

• Full Member
•   • Posts: 27
• Karma: 29 ##### Re: Q5--T0301, T0701
« Reply #2 on: March 09, 2018, 06:01:19 PM »
Guanlin's solution is correct! But here's the typed solution a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{3}{2}x_1 - \frac{1}{2}x_1'$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{3}{2}x_1' - \frac{1}{2}x_1''$$
Substitute into the second equation we get $$\frac{3}{2}x_1' - \frac{1}{2}x_1'' = 2x_1 - 2(\frac{3}{2}x_1 - \frac{1}{2}x_1')$$
Simplify the expression we get $x_1'' - x_1' -2x_1 = 0$,which is a second order ODE of $x_1$.

b)
Characteristic equation is $r^2 - r - 2 = (r-2)(r+1) = 0$ with roots $r_1 = 2, r_2 = -1$
General solution for $x_1$ is $x_1 = c_1 e^{2t} + c_2 e^{-t}$
Plug in to $x_2 = \frac{3}{2}x_1 - \frac{1}{2}x_1'$ get
$$x_2 = \frac{3}{2} ( c_1 e^{2t} + c_2 e^{-t}) - \frac{1}{2} (2c_1 e^{2t} - c_2 e^{-t}) = \frac{1}{2}c_1e^{2t} + 2c_2e^{-t}$$
So, $$x_1 = c_1 e^{2t} + c_2 e^{-t}$$ $$x_2 = \frac{1}{2}c_1e^{2t} + 2c_2e^{-t}$$
Plug in $x_1(0)=3, x_2(0) = \frac{1}{2}$ to get $$3 = c_1 + c_2$$  $$\frac{1}{2} = \frac{1}{2}c_1 + 2c_2$$
Solve the linear system we have
$$c_1 = \frac{11}{3}, c_2 = -\frac{2}{3}$$
That is, $$x_1 = \frac{11}{3} e^{2t} -\frac{2}{3} e^{-t}$$ $$x_2 = \frac{11}{6} e^{2t} -\frac{4}{3} e^{-t}$$

c) see attached graph
Note that as $t \to \infty$, the graph is asymptotic to the line $x_2 = \frac{1}{2} x_1$ in the first quadrant.
« Last Edit: March 09, 2018, 06:27:30 PM by Junya Zhang »

#### Victor Ivrii ##### Re: Q5--T0301, T0701
« Reply #3 on: March 10, 2018, 08:47:31 AM »
Guanlin