10
1 First-Order Linear Differential Equations
)
A(t) = De−kt + De−k(t−T ) + · · · + De−k(t−(n−1)T
−kt
kT
k(n−1)T
1 + e + ··· + e
= De
knT
e
−1
= De−kt kT
, (n − 1)T ≤ t < nT.
e −1
(1.18)
From (1.18), it follows that A(t) is discontinuous at T, 2T, · · · . Moreover,
A(nT − 0) = D
1 − e−knT
ekT − 1
and A(nT + 0) = D
ekT − e−knT
.
ekT − 1
Hence, the medicine never exceeds the amount D/(1 − e−kT ). The minimum in an
interval occurs at the end of each interval. This also increases, but always lies below
D/(ekT − 1). This gives us at least some information as to what happens to the drugs
which we take on long term basis for the illnesses such as hypertension and diabetes.
For the short term medicinal courses (say, antibiotics), it indicates the length of the
period of residue remaining in the body.
Example 1.14 Newton’s second law states that the rate of change of momentum of a
body equals the force applied. This law can be applied to bodies with variable mass,
e.g., rockets. We shall demonstrate this for motion along a straight line. Let Mv be
the momentum of the body at time t, and let M(v + u) be the momentum which
is added to the body at time t because the body is joined by another body of mass
M which before joining had the velocity v + u (u is the velocity of M relative
to M). Further, assume that at the instant of joining, the velocity of M changes from
v to v + v and M takes on the velocity of M. Thus, the momentum at time t is
Mv + M(u + v), whereas at time t + t it is (M + M)(v + v). Hence, the
change of momentum (Mv) during the time t is
(Mv) = (M + M)(v + v) − Mv − M(u + v) = Mv − uM + Mv.
Therefore, from Newton’s second law, we have
Mv − uM + Mv =
t+t
F(s)ds.
t
Dividing this relation by t and then letting t → 0, and assuming that M, v
t+t
and t
F(s)ds tend to zero in such a way that the resulting limits exist, we obtain
M
dM
dv
−u
= F(t),
dt
dt
which is the same as
d(Mv)
dM
= (v + u)
+ F(t).
dt
dt
(1.19)
1 First-Order Linear Differential Equations
11
Now consider a rocket traveling vertically upward in such a way that its rate
of change of mass d M/dt is constant (−r ). The lost mass consists of burning the
fuel, and maintains the constant exhaust speed c of the rocket. The rocket is acted
upon by a gravitational force Mg, where g is gravitational constant, and starts with
initial velocity v0 and initial mass M0 . We need to find the velocity v and distance
traveled x as functions of time t. For this, substituting d M/dt = −r, M = M0 − r t,
u = −c, F = −Mg in (1.19), we find the differential equation
dv
cr
=
− g.
dt
M0 − r t
(1.20)
Since v(0) = v0 , an integration of (1.20) gives
rt
M0
− gt.
(1.21)
rt
1
1 − ln 1 −
− gt 2 .
M0
2
(1.22)
v(t) = v0 − c ln 1 −
Now since d x/dt = v and x(0) = 0, it follows that
rt
cM0
1− 1−
x(t) = v0 t +
r
M0
We shall now find the burnout velocity v1 of the rocket, i.e., the velocity with
which the rocket is traveling at time t1 , when the entire fuel supply is exhausted
and the remaining mass M1 of the rocket is that of its structure and payload. Since,
t1 = (M0 − M1 )/r , from (1.21), we have
v1 = v0 + c ln
M0
(M0 − M1 )g
.
−
M1
r
(1.23)
If M f , M p , and Ms , respectively, denote the mass of the fuel, mass of the payload,
and the mass of the rocket structure, then clearly M0 = M f + M p + Ms and M1 =
M p + Ms , and the relation (1.23) can be written as
v1 = v0 + c ln 1 +
Mf
Ms + M p
−
Mfg
.
r
(1.24)
Thus, for given fuel and payload, the higher the exhaust velocity c of the fuel and the
smaller the structural mass Ms , gives the higher burnout velocity v1 of the rocket.
Problems
1.1 A cable of a suspension bridge supporting a uniform load of W pounds per
horizontal foot and with horizontal tension H in the cable at the origin satisfies the
differential equation (see Fig. 1.3)
0
