4—Differential Equations 71as a constant and an exponential, it’s easy to verify that you add the results that you get for the two
cases separately. If the forcing function is too complicated for you to guess a solution then there’s a
general method using Green’s functions that I’ll get to in section4.6.
Choose a specific example
Fext(t) =F 0
[
1 −e−βt
]
(4.13)
This starts at zero and builds up to a final value ofF 0. It does it slowly or quickly depending onβ.
F 0
t
Start with the first term,F 0 , for external force in Eq. (4.12). Tryx(t) =Cand plug into that
equation to find
kC=F 0
This is simple and determinesC.
Next, use the second term as the forcing function,−F 0 e−βt. Guess a solutionx(t) =C′e−βt
and plug in. The exponential cancels, leaving
mC′β^2 −bC′β+kC′=−F 0 or C′=
−F 0
mβ^2 −bβ+k
The total solution for the inhomogeneous part of the equation is then the sum of these two expressions.
xinh(t) =F 0
(
1
k
−
1
mβ^2 −bβ+k
e−βt
)
The homogeneous part of Eq. (4.12) has the solution found in Eq. (4.6) and the total isx(t) =xhom(t) +xinh(t) =x(t) =Aeα^1 t+Beα^2 t+F 0
(
1
k
−
1
mβ^2 −bβ+k
e−βt
)
(4.14)
There are two arbitrary constants here, and this is what you need because you have to be able to
specify the initial position and the initial velocity independently; this is a second order differential
equation after all. Take for example the conditions that the initial position is zero and the initial
velocity is zero. Everything is at rest until you start applying the external force. This provides two
equations for the two unknowns.
x(0) = 0 =A+B+F 0
mβ^2 −bβ
k(mβ^2 −bβ+k)
x ̇(0) = 0 =Aα 1 +Bα 2 +F 0
β
mβ^2 −bβ+k
Now all you have to do is solve the two equations in the two unknownsAandB. Take the first,
multiply it byα 2 and subtract the second. This givesA. Do the same withα 1 instead ofα 2 to getB.
The results are