Author Topic: Q5--T0501, T5101  (Read 984 times)

Victor Ivrii

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Q5--T0501, T5101
« on: March 09, 2018, 05:57:36 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 $t \ge  0$.

$$\left\{\begin{aligned}
& x'_1= x_1 - 2x_2, &&x_1(0) = -1,\\
&x'_2= 3x_1 - 4x_2, &&x_2(0) = 2.
\end{aligned}\right.$$

Junya Zhang

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Re: Q5--T0501, T5101
« Reply #1 on: March 09, 2018, 05:59:13 PM »
a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{1}{2}x_1 - \frac{1}{2}x_1'$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{1}{2}x_1' - \frac{1}{2}x_1''$$
Substitute into the second equation and simplify, we get $$ x_1'' + 3 x_1' + 2x_1 = 0 $$
which is a second order ODE of $x_1$.


b)
Characteristic equation is $r^2 +3 r + 2 = (r + 2)(r + 1) = 0$ 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 $x_2 = \frac{1}{2}x_1 - \frac{1}{2}x_1'$ get
$$x_2 = \frac{3}{2}c_1 e^{-2t} + c_2 e^{-t}$$
So, $$x_1 = c_1 e^{-2t} + c_2 e^{-t}$$ $$x_2 = \frac{3}{2}c_1 e^{-2t} + c_2 e^{-t}$$
Plug in $x_1(0)=-1, x_2(0) = 2$ to get $$-1 = c_1 + c_2 $$  $$2= \frac{3}{2}c_1 + c_2 $$
Solve the linear system we have
$$c_1 = 6, c_2 = -7$$
That is, $$x_1 = 6 e^{-2t} -7 e^{-t}$$ $$x_2 = 9 e^{-2t}  -7 e^{-t}$$

c) See attached image
Note that as $t\to \infty$, the graph approaches the origin in the third quadrant tangent to the line $x_1 = x_2$.
« Last Edit: March 09, 2018, 06:08:41 PM by Junya Zhang »

Junjie Zhang

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Re: Q5--T0501, T5101
« Reply #2 on: March 09, 2018, 06:02:12 PM »
Also, we can Sketch the graph.
(c) Attached is the graph
Edit: Glad to see Junya add the graph afterwards.
« Last Edit: March 09, 2018, 06:18:56 PM by Junjie Zhang »