x 1 ′ − 2 x 1 − 3 x 2 = 0 x_1'− 2x_1 − 3x_2 = 0 x 1 ′ − 2 x 1 − 3 x 2 = 0
x 2 ′ + x 1 + x 2 = 0 x_2' + x_1 + x_2 = 0 x 2 ′ + x 1 + x 2 = 0
x 1 ( 0 ) = 3 , x 2 ( 0 ) = 0 x_1(0) = \sqrt3,x_2(0) = 0 x 1 ( 0 ) = 3 , x 2 ( 0 ) = 0
from second equation:
x 1 = − x 2 ′ − x 2 x_1=-x_2'-x_2 x 1 = − x 2 ′ − x 2
x 1 ′ = − x 2 ′ ′ − x 2 ′ x_1'=-x_2''-x_2' x 1 ′ = − x 2 ′′ − x 2 ′
then:
− x 2 ′ ′ − x 2 ′ − 2 ( − x 2 ′ − x 2 ) − 3 x 2 = 0 -x_2''-x_2'-2(-x_2'-x_2)-3x_2=0 − x 2 ′′ − x 2 ′ − 2 ( − x 2 ′ − x 2 ) − 3 x 2 = 0
x 2 ′ ′ − x 2 ′ + x 2 = 0 x_2''-x_2'+x_2=0 x 2 ′′ − x 2 ′ + x 2 = 0
characteristic equation:
k 2 − k + 1 = 0 k^2-k+1=0 k 2 − k + 1 = 0
k = 1 ± 1 − 4 2 = 1 / 2 ± i 3 / 2 k=\frac{1\pm \sqrt{1-4}}{2}=1/2\pm i\sqrt 3/2 k = 2 1 ± 1 − 4 = 1/2 ± i 3 /2
x 2 = e t / 2 ( c 1 c o s 3 2 t + c 2 s i n 3 2 t ) x_2=e^{t/2}(c_1cos\frac{\sqrt 3}{2}t+c_2sin\frac{\sqrt 3}{2}t) x 2 = e t /2 ( c 1 cos 2 3 t + c 2 s in 2 3 t )
x 2 ( 0 ) = c 1 = 0 x_2(0)=c_1=0 x 2 ( 0 ) = c 1 = 0
x 2 ′ = e t / 2 2 c 2 s i n 3 2 t + e t / 2 3 2 c 2 c o s 3 2 t 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 2 ′ = 2 e t /2 c 2 s in 2 3 t + 2 e t /2 3 c 2 cos 2 3 t
x 1 = − e t / 2 2 c 2 s i n 3 2 t − e t / 2 3 2 c 2 c o s 3 2 t − e t / 2 c 2 s i n 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 = − 2 e t /2 c 2 s in 2 3 t − 2 e t /2 3 c 2 cos 2 3 t − e t /2 c 2 s in 2 3 t
x 1 ( 0 ) = − 3 2 c 2 = 3 x_1(0)=-\frac{\sqrt{3}}{2}c_2=\sqrt 3 x 1 ( 0 ) = − 2 3 c 2 = 3
c 2 = − 2 c_2=-2 c 2 = − 2
then:
x 2 ( t ) = − 2 e t / 2 s i n 3 2 t x_2(t)=-2e^{t/2}sin\frac{\sqrt 3}{2}t x 2 ( t ) = − 2 e t /2 s in 2 3 t
Answer: (c)
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