Question #285963

Which of the following is the solution of the initial value problem y (5) + 4y (4) + 4y (3) = 0 , y(0)= 0 , y0 (0)= − 1 2 , y(2)(0)= 5 , y(3)(0)=−14 , y (4)(0)= 36 ? (a) y = −1 − t 4 + t 2 2 + e −2t (b) y = 1 4 − t + t 2 4 − 1 4 e −2t − te−2t (c) y = − 1 4 + t + 1 4 e −2t − te−2t (d) y = −1 + t 2 − t 2 4 + (1 + t 2 )e 2t (e) y = 1 2 + t − 1 2 e 2t + t 2 e 2t


1
Expert's answer
2022-01-10T15:38:41-0500
y(5)+4y(4)+4y(3)=0y^{(5)}+4y^{(4)}+4y^{(3)}=0

Characteristic (auxiliary) equation


r5+4r4+4r3=0r^5+4r^4+4r^3=0

r3(r+2)2=0r^3(r+2)^2=0

r1=r2=r3=0,r4=r5=2r_1=r_2=r_3=0, r_4 = r_5=-2

The general solution of the homogeneous differential equation is


y=c1+c2t+c3t2+c4e2t+c5te2ty=c_1+c_2t+c_3t^2+c_4e^{-2t}+c_5te^{-2t}


y=c2+2c3t2c4e2t+c5e2t2c5te2ty'=c_2+2c_3t-2c_4e^{-2t}+c_5e^{-2t}-2c_5te^{-2t}

y=2c3+4c4e2t4c5e2t+4c5te2ty''=2c_3+4c_4e^{-2t}-4c_5e^{-2t}+4c_5te^{-2t}

y(3)=8c4e2t+12c5e2t8c5te2ty^{(3)}=-8c_4e^{-2t}+12c_5e^{-2t}-8c_5te^{-2t}

y(4)=16c4e2t32c5e2t+16c5te2ty^{(4)}=16c_4e^{-2t}-32c_5e^{-2t}+16c_5te^{-2t}

y(4)(0)=36=>16c432c5=36y^{(4)}(0)=36=>16c_4-32c_5=36

=>4c48c5=9=>4c_4-8c_5=9

y(3)(0)=14=>8c4+12c5=14y^{(3)}(0)=-14=>-8c_4+12c_5=-14

=>4c46c5=7=>4c_4-6c_5=7

c4=14,c5=1c_4=\dfrac{1}{4}, c_5=-1

y(0)=5=>2c3+4c44c5=5y''(0)=5=>2c_3+4c_4-4c_5=5

=>c3=0=>c_3=0

y(0)=12=>c22c4+c5=12y'(0)=-12=>c_2-2c_4+c_5=-\dfrac{1}{2}

=>c2=1=>c_2=1

y(0)=0=>c1+c4=0y(0)=0=>c_1+c_4=0

c1=14c_1=-\dfrac{1}{4}

The following is the solution of the initial value problem

(c)


y=14+t+14e2tte2ty = -\dfrac{1}{4}+ t + \dfrac{1}{4}e ^{−2t} − te^{−2t}


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