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Zepler
3 months ago
10

Consider the vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + u = 8 co

s ωt, u(0) = 5, u'(0) = 7 (a) Find the solution for ω ≠ 1.
Mathematics
2 answers:
Svet_ta [12.7K]3 months ago
5 0

Step-by-step explanation:

We are dealing with the differential equation

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

This can be approached using the method of undetermined coefficients. The overall solution is

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

where uh(t) represents the solution to the homogeneous part, and up(t) is a particular solution.

1. To start, let's find uh(t) using the characteristic polynomial:

u'' +u=0

r^{2}+1=0

The roots are imaginary:

r1=+i

r2=-i

Thus, uh(t) can be expressed as

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

where A and B are constants that need to be determined.

2. Next, we establish up(t). Using the method of undetermined coefficients, we can guess that up(t)=Ccos(wt)+Dsin(wt). Taking its derivatives gives us

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

Substituting this into the differential equation yields

-\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)

From here, we obtain

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

As it turns out, we set D=0 since the solution does not concern sin functions.

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

3. Ultimately, the expression for u(t) is

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

When we apply the initial conditions, we find

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

PIT_PIT [12.4K]3 months ago
4 0

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, which leads to k² = -1, resulting in k = ±i.

The roots are k = i or -i.

The general solution takes the form  u(x)=C₁cosx+C₂sinx.

Applying the method of undetermined coefficients, we have

Uc(t) = Pcos wt  + Qsin wt

Calculating the derivatives gives us Uc’(t) = -Pwsin wt  + Qwcos wt

And differentiating again yields Uc’’(t) = -Pw^2cos wt  - Qw^2sin wt

With the equation U’’ + u = 8cos wt, we substitute:

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

This simplifies to (-Pw^2 + P) cos wt   + (-Qw^2 + Q) sin wt = 8cos wt.

From -Pw^2 + P = 8, we find P= 8  /(1- w^2).

From -Qw^2 + Q = 8, we can conclude Q = 0.

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

Combining gives us U(t) = uh(t ) + Uc(t)

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

Initial conditions yield:

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

Which leads us to C1 + 8 /(1- w^2) = 5

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

Next, taking the derivative:

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

Evaluating at t = 0 gives us:

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

Thus, c2 = 7.

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

   

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