Question #285965

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 


1
Expert's answer
2022-01-11T00:18:04-0500

x12x13x2=0x_1'− 2x_1 − 3x_2 = 0

x2+x1+x2=0x_2' + x_1 + x_2 = 0

x1(0)=3,x2(0)=0x_1(0) = \sqrt3,x_2(0) = 0


from second equation:

x1=x2x2x_1=-x_2'-x_2

x1=x2x2x_1'=-x_2''-x_2'


then:

x2x22(x2x2)3x2=0-x_2''-x_2'-2(-x_2'-x_2)-3x_2=0

x2x2+x2=0x_2''-x_2'+x_2=0

characteristic equation:

k2k+1=0k^2-k+1=0

k=1±142=1/2±i3/2k=\frac{1\pm \sqrt{1-4}}{2}=1/2\pm i\sqrt 3/2

x2=et/2(c1cos32t+c2sin32t)x_2=e^{t/2}(c_1cos\frac{\sqrt 3}{2}t+c_2sin\frac{\sqrt 3}{2}t)

x2(0)=c1=0x_2(0)=c_1=0

x2=et/22c2sin32t+et/232c2cos32tx_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


x1=et/22c2sin32tet/232c2cos32tet/2c2sin32tx_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

x1(0)=32c2=3x_1(0)=-\frac{\sqrt{3}}{2}c_2=\sqrt 3

c2=2c_2=-2


then:

x2(t)=2et/2sin32tx_2(t)=-2e^{t/2}sin\frac{\sqrt 3}{2}t


Answer: (c)


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