Example 5

Find a particular solution of the differential equation

y" + 16y = - sin(4t).

The characteristic polynomial has roots ± i4. Then cos (4t) and sin (4t) are already solutions of the homogeneous equation, so our guess must have an extra factor of t. The guess for a solution is then

y(t) = Kt cos (4t) + Lt sin (4t).

Compute the first two derivatives of y(t).

y'(t) = K[-4t sin (4t) + cos (4t)] + L[4t cos (4t) + sin (4t)],

y"(t) = K[-16t cos (4t) - 8 sin (4t)] + L[-16t sin (4t) + 8 cos (4t)].

Now substitute into the original differential equation.

K[(-16t + 16t) cos (4t) - 8 sin (4t)] + L[(- 16t + 16t) sin (4t) + 8 cos (4t)] = -sin (4t).

-8K sin (4t) + 8L cos (4t) = - sin (4t).

From the sine terms we get -8K = - 1, so K = 1/8. From the cosine terms we get 8L = 0, so L = 0. The particular solution is therefore

y(t) = 0.125t cos(4t).

In Example 5, the particular solution oscillates more and more wildly as t approaches infinity, as shown in Figure 14.7.3. This happens because the forcing term cos (4t) has the same frequency as the solutions of the homogeneous equation, A cos (4t) + B sin (4t)- In this case the forcing term causes the oscillation to build up. This phenomenon is called resonance.

If, instead, the forcing term in Example 5 had a different frequency, -sin (ωt) where ω is not equal to 4, then the particular solution of the differential equation would be a simple oscillation of the form K cos (ωt) + L sin (ωt), whose amplitude does not change with time.

Figure 14.7.3 Exercise 5