Toronto Math Forum
MAT2442018S => MAT244Tests => Quiz5 => Topic started by: Victor Ivrii 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 $t \ge 0$.
$$\left\{\begin{aligned}
& x'_1= x_1  2x_2, &&x_1(0) = 1,\\
&x'_2= 3x_1  4x_2, &&x_2(0) = 2.
\end{aligned}\right.$$

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 $x_1 = x_2$.

Also, we can Sketch the graph.
(c) Attached is the graph
Edit: Glad to see Junya add the graph afterwards.