# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-5 => Topic started by: Victor Ivrii on November 02, 2018, 03:14:18 PM

Title: Q5 TUT 0101
Post by: Victor Ivrii on November 02, 2018, 03:14:18 PM
Transform the given system into a single equation of second order and find the solution $(x_1(t),x_2(t))$, satisfying initial conditions
\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.
Title: Re: Q5 TUT 0101
Post by: Monika Dydynski on November 02, 2018, 03:18:05 PM
a. Transform the given system into a single equation of second order; and
b. Find $x_1$ and $x_2$ satisfying the initial conditions.

\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.

a. Transform the given system into a single equation of second order.

Solving the first equation for $x_2$, we have
$$x_2=-\frac{1}{2}x’_1+\frac{3}{2}x_1\tag{1}$$

Plugging $x_2$ into the second equation gives

$$\left(-\frac{1}{2}x’_1+\frac{3}{2}x_1\right)’=2x_1-2\left(-\frac{1}{2}x’_1+\frac{3}{2}x_1\right)$$
$$-\frac{1}{2}x’’_1+\frac{3}{2}x’_1=2x_1+x’_1-3x_1$$
$$x’’_1-x’_1-2x_1=0.$$

b. Find $x_1$ and $x_2$ satisfying initial conditions.

The characteristic equation is
$$r^2-r-2=0$$
$$(r+1)(r-2)=0 \Rightarrow r_1=-1, r_2=2.$$

The general solution is
$$x_1(t)=c_1e^{-t}+c_2e^{2t}.$$

It follows that the solution for $x_2$ is

$$x_2(t)=-\frac{1}{2}(-c_1e^{-t}+2c_2e^{2t})+\frac{3}{2}(c_1e^{-t}+c_2e^{2t})=2c_1e^{-t}+\frac{1}{2}c_2e^{2t}$$

Satisfying the given initial conditions  $x_1(0)=3$ and $x_2(0)=\frac{1}{2}$, we obtain the following system

\left\{\begin{aligned} &c_1+c_2=3\\ &2c_1+\frac{1}{2}c_2=\frac{1}{2} \end{aligned}\right.

Solving this system, we get

$$\cases{c_1=-\frac{2}{3}\\c_2=\frac{11}{3}}$$

The solutions that satisfy the given initial conditions are

$$x_1(t)=-\frac{2}{3}e^{-t}+\frac{11}{3}e^{2t}$$
$$x_2(t)=-\frac{4}{3}e^{-t}+\frac{11}{6}e^{2t}.$$