Physics for Scientists and Engineers last updated: 9/22/1
PHYS-013 Lecture notes and addenda T. H¨ubsch
1. The differential equation (a body falling under the influence of gravity and resisted by air
drag)
mdv(t)
dt=mg −bv(t),or dv(t)
dt=g−
b
mv(t),(1)
may easily be solved by the ‘trial-and-adjust’ method. First, we note that the equation is
linear in v(t). This allows us to solve the equation by means of a superposition (addition) of
the solutions of the two simpler equations:
dv1(t)
dt=g , and dv2(t)
dt=−
b
mv(t).(2 a, b)
The first of these, Eq. (2a), is solved by v1(t) = v10 +gt (as seen before), while Eq. (2b) is
solved by remembering that a derivative of the exponential function is proportional to the
exponential function itself, so that v2(t) = v20e−bt/m . This indeed satisfies Eq. (2b), as is easy
to verify by substituting.
Then, we look for the general solution to the original equation (1) in the form
v(t) = A+Bt +C eDt .(3)
From this, we calculate the derivative to be
dv(t)
dt=B+CDeDt .(4)
Substituting (4) in the left-hand-side and (3) in the right-hand-side of Eq. (1), we obtain:
B+CDeDt =g−
b
mA+Bt +C eDt=g−
b
mA−
b
mBt −
b
mCeD t .
Equating the coefficients of t0= 1 = const.,t1=tand eDt, we have that
B=g−
b
mA , 0 = −
b
mB , and CD =−
b
mC . (5 a, b, c)
Of these, Eq. (5b) forces B= 0, after which Eq. (5a) sets A=mg
b, while Eq. (5c) sets
D=−b
m. This then produces
v(t) = mg
b+Ce−bt/m ,
which indeed has one integration constant: C, as should be the case for a first order differential
equation. To determine this last constant, we can use the initial condition, noticing that
v(0) = mg
b+Cdef
=v0,
is the initial (at t= 0) speed and so can write C=v0−mg
b, whereupon
v(t) = mg
b+v0−
mg
be−bt/m =mg
b1−e−bt/m+v0e−bt/m .
Finally, we know that the body has reached the terminal speed, vT, if it no longer
accelerates, i.e., when the gravitational force is cancelled by the air drag, mg =bvT, so that
vT=mg
b. Indeed, when t→+∞, then e−bt/m →0, and so v(+∞) = mg
b=vT. This
provides an auxiliary check for our general solution. Notice that the value of the terminal
speed, vT=mg
b, is independent of the initial speed, v0. Also, note that v(t) does not in fact
reach vTat any finite time, but merely approaches it asymptotically.
