# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-5 => Topic started by: Victor Ivrii on November 02, 2018, 03:15:28 PM

Title: Q5 TUT 0301
Post by: Victor Ivrii on November 02, 2018, 03:15:28 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 = 1.25x_1 + 0.75x_2, &&x_1(0) = -2,\\ &x'_2= 0.75x_1 + 1.25x_2, &&x_2(0) = 1. \end{aligned}\right.
Title: Re: Q5 TUT 0301
Post by: Pengyun Li on November 02, 2018, 03:51:40 PM
Isolate $x_2$ in first equation we get $x_2 = \frac{4}{3}x'_1 - \frac{5}{3}x_1$

Differentiate both sides with respect to t we get $x'_2 = \frac{4}{3}x''_1 - \frac{5}{3}x'_1$

Sub into second equation , we get $x''_1 - \frac{5}{2}x'_1 + x_1 = 0$

Characteristic equation is $r^2 - \frac{5}{2} r + 1 = 0$,

hence $r_1 = \frac{1}{2}, r_2 = 2$

General solution for $x_1$ is $x_1 = c_1e^\frac{t}{2}+ c_2 e^{2t}$

Plug into $x_2 = \frac{4}{3}x'_1 - \frac{5}{3}x_1$, we get $x_2 = -c_1 e^\frac{t}{2}+ c_2 e^{2t}$,

So, $x_1 = c_1e^\frac{t}{2}+ c_2e^{2t}$, $x_2 = -c_1 e^\frac{t}{2} + c_2 e^{2t}$

Plug in $x_1(0) = -2, x_2(0) = 1$,

$c_1 + c_2 = -2, -c_1 + c_2 = 1$

hence $c_1 = -\frac{3}{2}, c_2 = -\frac{1}{2}$

Therefore, $x_1 = -\frac{3}{2} e^\frac{t}{2}-\frac{1}{2} e^{2t}$,
$x_2 = \frac{3}{2} e^\frac{t}{2} -\frac{1}{2} e^{2t}$
Title: Re: Q5 TUT 0301
Post by: Guanyao Liang on November 02, 2018, 03:54:00 PM
Answer
Title: Re: Q5 TUT 0301
Post by: Victor Ivrii on November 04, 2018, 07:41:28 PM
Guanyao, no need to post identical solution