Answer to Question #285963 in Differential Equations for hector12

Question

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

Answer 1

$$y^{(5)}+4y^{(4)}+4y^{(3)}=0$$

Characteristic (auxiliary) equation

$$r^5+4r^4+4r^3=0$$

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

$$r_1=r_2=r_3=0, r_4 = r_5=-2$$

The general solution of the homogeneous differential equation is

$$y=c_1+c_2t+c_3t^2+c_4e^{-2t}+c_5te^{-2t}$$

$$y'=c_2+2c_3t-2c_4e^{-2t}+c_5e^{-2t}-2c_5te^{-2t}$$

$$y''=2c_3+4c_4e^{-2t}-4c_5e^{-2t}+4c_5te^{-2t}$$

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

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

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

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

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

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

$$c_4=\dfrac{1}{4}, c_5=-1$$

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

$$=>c_3=0$$

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

$$=>c_2=1$$

$$y(0)=0=>c_1+c_4=0$$

$$c_1=-\dfrac{1}{4}$$

The following is the solution of the initial value problem

(c)

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