Author Topic: Q5--T0101, T0801  (Read 1133 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2467
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Q5--T0101, T0801
« on: March 09, 2018, 05:46:05 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 $t \ge  0$.

$$\left\{\begin{aligned}
& x'_1= -0.5x_1 + 2x_2, &&x_1(0) = -2,\\
&x'_2= -2x_1 - 0.5x_2, &&x_2(0) = 2
\end{aligned}\right.$$
« Last Edit: March 09, 2018, 05:53:12 PM by Victor Ivrii »

Junjie Zhang

  • Full Member
  • ***
  • Posts: 19
  • Karma: 12
    • View Profile
Re: Q5--T0101, T0801
« Reply #1 on: March 09, 2018, 05:53:39 PM »
Solving the first equation for B , we obtain $x_2 = x_1'/2+x_1/4$. Substitution into the second equation results in $$x_1''/2+x_1'/4 = -2x_1-(x_1'/2+x_1'/4)/2$$.
Rearranging the terms, the single differential equation for $x_1$ is $$x_1''+x_1'+\frac{17}{4}x_1=0$$
The general soln is $$x_1(t) = e^{-t/2}(c_1cos2t+c_2sin2t)$$. With $x_2$ given in terms of $x_1$, it has
$$x_2(t) = e^{-t/2}(-c_1cos2t+c_2sin2t)$$.
Imposing the specified initial conditions, we obtain $c_1 = -2, c_2 = 2$. Hence,
$$x_1(t) = e^{-t/2}(-2cos2t+2sin2t)$$.
$$x_2(t) = e^{-t/2}(2cos2t+2sin2t)$$.
Attached is the graph.
« Last Edit: March 09, 2018, 06:09:30 PM by Junjie Zhang »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2467
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Q5--T0101, T0801
« Reply #2 on: March 10, 2018, 08:45:11 AM »
You should write \cos (and so on)