Question

Consider the vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + u = 8 cos ωt, u(0) = 5, u'(0) = 7 (a) Find the solution for ω ≠ 1.

Answer 1

Answer:

  u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

Step-by-step explanation:

The characteristic equation is     k²+1 = 0,⇒k²=−1,⇒ k=±i.

the roots are k = i or -i

the solution has the form  u(x)=C₁cosx+C₂sinx.

Using undetermined coefficient method

Uc(t) = Pcos wt  + Qsin wt

Uc’(t) = -Pwsin wt  + Qwcos wt

Uc’’(t) = -Pw^2cos wt  - Qw^2sin wt

U’’ + u = 8cos wt

-Pw^2cos wt  - Qw^2sin wt + Pcos wt  + Qsin wt = 8cos wt

(-Pw^2 + P) cos wt   + (- Qw^2 + Q ) sin wt = 8cos wt

-Pw^2 + P = 8  which implies P= 8  /(1- w^2)

- Qw^2 + Q = 8 which implies Q = 0

Uc(t) = Pcos wt  + Qsin wt = 8 cos wt /(1- w^2)

U(t) = uh(t ) + Uc(t)

     = C1cos t + c2 sin t + 8 cos wt /(1- w^2)

Initial value problem

U(0) = C1cos(0) + c2 sin (0) + 8 cos (0) /(1- w^2)

C1 + 8 /(1- w^2) = 5

C1 = 5 -8 /(1- w^2) = -(3 + w^2 ) /(1- w^2)

U’(t) = -C1 sin t + c2 cos t - 8 w sin wt /(1- w^2)

U’(0) = -C1 sin (0) + c2 cos (0) - 8 w sin (0) /(1- w^2) = 7

 c2 = 7

  u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

   

Answer 2

Step-by-step explanation:

We have the differential equation

$u'' +u=8cos(\omega t)$

This differential equation can be solved by the undetermined coefficient method. The general solution is

u(t) = uh(t) + up(t)

where uh(t) is the solution to the homogeneous equation and up(t) is a particular solution

1. First of all, we calculate uh(t) by using the characteristic polynomial:

$u'' +u=0$

$r^{2}+1=0$

The solution are imaginary:

r1=+i

r2=-i

hence, uh(t) is

$u_{h}(t)=Acos(t)+Bsin(t)$

where A and B are constant that we have to calculate.

2. Now we calculate up(t). In the undetermined coefficient method we can assume that the solution is up(t)=Ccos(wt)+Dsin(wt). By taking derivatives of up(t) we have

$u_{p}'(t)=-\omega Csin(\omega t)+\omega Dcos(\omega t)\\u_{p}''(t)=-\omega ^{2}Ccos(\omega t)-\omega ^{2}Dsin(\omega t)$

and by replacing in the differential equation

$-\omega ^{2}Ccos(\omega t)-\omega ^{2}Dsin(\omega t)+Ccos(\omega t)+Dsin(\omega t)=8cos(\omega t)\\(-\omega ^{2}C +C)cos(\omega t) + (-\omega ^{2}D+D)sin(\omega t)=8cos(\omega t)$

Hence we have

$-\omega^{2}C+C=8\\-\omega^{2}D+D=0\\C=\frac{8}{1-\omega^{2}}\\D=0$

where we have taken D=0 because it is clear that the solution does not depended of sin functions.

$u_{p}(t)=\frac{8}{1-\omega^{2}}cos(\omega t)$

3. Finally the u(t) is

$u(t)=Acos(t)+Bsin(t) +\frac{8}{1-\omega^{2}}cos(\omega t)$

and by applying the initial conditions we have

$u(0)=A+\frac{8}{1-\omega^{2}} = 5\\u'(0)=B = 7\\B = 7\\A = 5-\frac{8}{1-\omega^{2}}$