MAT244-2018S > Quiz-5

Q5--T0301, T0701

(1/1)

Victor Ivrii:
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:
Attached is the solution to TUT 0701

Clearer graph attached on second one

Junya Zhang:
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.

Victor Ivrii:
Guanlin