MATH 282 - DE
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4.2 Sinusoidal Forcing
Today we are going to study forced equations where the forcing function is sinusoidal
(either sine or cosine). Let’s apply the guessing technique to sinusoidally forced linear
equations:
Example. Let’s calculate the general solution to the equation:
π2 π¦ ππ¦
+
+ 2π¦ = cos 2π‘
ππ‘ 2 ππ‘
Solution:
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Here are the graphs of three solutions:
Here is the graph of the steady-state solution:
A little translation:
Consider the 2nd order linear forced equation
π2π¦
ππ¦
π 2 +π
+ π π¦ = π(π‘)
ππ‘
ππ‘
where π and π are positive and π ≥ 0.
Engineering Terminology:
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forced response – any solution to the forced equation.
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steady-state response – behavior of the forced response over the long term.
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natural (or free) response – any solution of the associated homogeneous equation.
Why are initial conditions essentially irrelevant?
When we guess a solution of the form π¦π (π‘) = πΌπ πππ‘ and compute the complex number πΌ, we
have essentially determined everything we need to know about the steady-state solution.
Euler’s formula gives us a nice way of determining the amplitude, and phase-angle of the
steady-state solution immediately from πΌ.
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Why we prefer to calculate the steady-state solution using complex numbers?
Recall that we calculated the steady-state solution
1
π¦π (π‘) = − (cos 2π‘ − sin 2π‘)
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for the equation
π 2 π¦ ππ¦
+
+ 2π¦ = cos 2π‘,
ππ‘ 2 ππ‘
and we did so using computations that involved complex numbers. In fact, we found π¦π (π‘) as
the real part of
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π¦π (π‘) = − (1 + π)π (2π)π‘ .
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The complex number
1
πΌ = − (1 + π)
4
Tells us everything we need to know about the steady-state solution.
In order to see why, we use polar coordinates in the complex plane ( see pp. 745-747 in
Appendix C of the text).
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