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

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Zepler

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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 [4.3K] · Answer 1 · 2026-03-03T14:13:26+03:00

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 [3.9K] · Answer 2 · 2026-03-03T14:16:08+03:00

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)