Answer to Question #285965 in Differential Equations for emperoremhyr

Question

Given the initial value problem    x 0 1 − 2x1 − 3x2 = 0 x 0 2 + x1 + x2 = 0 x1(0) = √ 3, x2(0) = 0 , find x2(t). (a) x2(t) = e t 2 sin √ 3 2 t (b) x2(t) = e t  cos 1 2 t + sin 1 2 t  (c) x2(t) = −2e t 2 sin √ 3 2 t (d) x2(t) = 2e t 2 cos √ 3 2 t + sin √ 3 2 t ! (e) x2(t) = 2e t 2 sin √ 3 2 t 

Answer 1

$x_1'− 2x_1 − 3x_2 = 0$

$x_2' + x_1 + x_2 = 0$

$x_1(0) = \sqrt3,x_2(0) = 0$

from second equation:

$x_1=-x_2'-x_2$

$x_1'=-x_2''-x_2'$

then:

$-x_2''-x_2'-2(-x_2'-x_2)-3x_2=0$

$x_2''-x_2'+x_2=0$

characteristic equation:

$k^2-k+1=0$

$k=\frac{1\pm \sqrt{1-4}}{2}=1/2\pm i\sqrt 3/2$

$x_2=e^{t/2}(c_1cos\frac{\sqrt 3}{2}t+c_2sin\frac{\sqrt 3}{2}t)$

$x_2(0)=c_1=0$

$x_2'=\frac{e^{t/2}}{2}c_2sin\frac{\sqrt 3}{2}t+\frac{e^{t/2}\sqrt 3}{2}c_2cos\frac{\sqrt 3}{2}t$

$x_1=-\frac{e^{t/2}}{2}c_2sin\frac{\sqrt 3}{2}t-\frac{e^{t/2}\sqrt 3}{2}c_2cos\frac{\sqrt 3}{2}t-e^{t/2}c_2sin\frac{\sqrt 3}{2}t$

$x_1(0)=-\frac{\sqrt{3}}{2}c_2=\sqrt 3$

$c_2=-2$

then:

$x_2(t)=-2e^{t/2}sin\frac{\sqrt 3}{2}t$

Answer: (c)