Further Applications of First Order
Equations
Mixture Problems
Mixture problems generally concern a tank, or reservoir, containing a
solution of some substance, being filled at a certain rate with another
solution of the same substance, instantaneously mixed with the
solution in the tank, and at the same time being drained at a certain
rate. Typically the incoming solution has a different concentration
than the solution already in the tank.
At a given time, t, typically measured in hours or minutes, the tank
contains a volume, V (t), of solution. Dissolved in this volume is a
quantity, Q(t), of some substance. Volume is ordinarily measured in
gallons or liters while the dissolved quantity is measured in pounds,
grams, kilograms, etc. The concentration, relative to the units used,
is then
C(t) = Q(t)/V (t).
In all mixture problems it is important to understand whether it is the
quantity or concentration of the dissolved substance that is to be
found.
The Differential Equations Suppose the volume of solution
in the tank is V (t) and the tank is being filled at a rate of Ri
volume units per unit time and drained at a rate of Ro units per
unit time. Then it
is clear that we have
dV
= Ri Ro.
dt
If the input concentration is Ci (usually constant) and the output concentration is taken to be the concentration C(t) in the tank, then the
rate of change of quantity of solute is
dQ
1
dt
= Ci Ri C(t) Ro = Ci Ri
Q(t)
Ro .
V
(t)
This is a linear differential equation in Q(t); the one to be used if the
quantity of solute in the tank is the variable of interest. If it is the
concentration in the tank, C(t), which is of interest, we have
dC
d
=
dt
dt
Q(t)
V (t)
dV
dQ
1 V (t) dt dt
V
Q(t)
(t)2
= V (Ci Ri C(t) Ro)
(t)
which reduces to
dC
(Ro + dV
=
) d
dt
V (t)t
dV
dt
V
(t)
C(t),
C(t) + Ci Ri
V
.(t)
Since dV
= Ri and the differential
= Ri Ro, we have Ro + dV
d
d
t
t
equation
(C(t) Ci).
dC
Ri
is
= V
dt
(t)
If Ri = Ro the volume, V (t) V , is constant and the concentration
equation simplifies to the constant coefficient equation
dC
dt
Ri
V
(C(t) Ci).
In a given problem, the quantity equation or the concentration
equation might be the easier one to work with, depending on the data
and what is asked for.
Example 1 At time t = 0 the tank contains 50 gallons of brine
with a concentration of 5 lbs. of salt per gallon. Pure water is
added at the rate of 10 gallons per minute while the well-mixed
solution is drained out, also at 10 gallons per minute. When does
the tank contain exactly 100 lbs. of salt?
Note that the problem is posed in terms of the quantity of
salt.
Therefore we use the differential equation for the quantity
dQ
Q(t)
R o.
= Ci Ri V
dt
(t)
Since the input and output rates are the same, V (t) 50 in this
problem. Taking T0 = 0, we have Q(0) = C(0) V = 5 50 = 250
lbs.. Further; RV0 = 10
5 = .2. Thus the initial value problem is
dQ
= 0 10 .2 Q(t), Q(0) = 250.
dt
The general solution is Q(t, c) = c e.2t. To match the initial condition
we clearly must have c = 250 so that
Q(t) = 250 e.2 t.
If we are to have Q(T ) = 100 we clearly need
250 e.2 T = 100, or
.2 T = log
2
5
and therefore
.9163
log 2 log 5
= .2 = 4.58 minutes.
.2
T =
It would also be possible to use the concentration equation. Then we
would have C(0) = 5 and we would use the differential equation for
C(t). To have 100 lbs. at time T we need C(T ) 50 = 100,
i.e.,
C(T ) = 2 (lbs. per gallon); worked this way we get the same result.
Example 2 The tank initially contains 100 liters of water with a
concentration of 1 gram of salt per liter. A 5 grams per liter solution is
added at the rate of 4 liters per minute while the well mixed solution
is drained out at the rate of 8 liters per minute. At what time is the
concentration of the salt solution equal to 3 grams per liter and how
many liters are in the tank at that point?
Again we let t = 0 be the initial time and t = T the time when the
concentration reaches 3 gm/liter. Here V (0) = 100 and
dV
= 4 8 = 4 gm/liter
dt
so we have V (t) = 100 4 t, V (T ) = 100 4 T . We have ro = 8 so
the differential equation satisfied by the concentration is
dC
4
=
dt
100
4t
and C(0) = 1.
5 4
100 4t
C(t) +
Here we need to use the general formula because p(t) =
4
100 4t
is
not constant. We have P (t) = log(100 4t) so eP (t) = 100
4t, eP (s) =
1
100
4s
. Thus the general solution can be written as
t
C(t, c) = c (100 4t) + (100
4t)
We need c =
1
10
0
C(t) =
20
100 4s 100 4s
ds.
to match the initial condition C(0) = 1. So
100
4t
100
+ 20 (100 4t)
1
(100
4s)2
Integrating and setting C(T ) = 3, we obtain
3 = C(T ) =
100 4T
2000
. .
.T
.
.
.
+
Thus we have the equation 100
4 (100 4s).0
ds.
.
..
100 4T
1
1
= 5
1 + 500
3 =
100
100
100 4T
4
2=4 (
100 4T
100 4T
)
=
100
.
25
100 4T
.
100
This gives
50 = 100 4T 4T = 50 T = 12.5 minutes.
At T = 12.5 the volume is V (12.5) = 100 4 12.5 = 50 liters.
A Chemical Reaction Problem Let us suppose that two
chemi- cals, P and Q, interracting in solution, produce a third
chemical X. If P and Q are present in initial concentrations p and q,
respectively, at time t = 0 and if x = x(t) denotes the time history of
the X concen- tration, one model for the way in which x(t) develops is
provided by the differential equation, wherein is a positive constant,
dx
= (p x)(q x).
dt
This is a separable equation which we may rewrite as
1
(x p)(x dx = dt,
q)
We have to treat two separate cases, depending on how p and q are
related.
Case i; p = q:
Application of the method of partial fractions
decom- position gives
1
(x p)(x
q)
a
x
p
b
x
q
(a + b)x (aq +
bp)
.
(x p)(x q)
Comparison of the first expression with the last gives a + b = 0, aq +
bp = 1. Solving for a and b we have a = 1 , b = 1 . Thus we
pq
qp
have
)
dx
=
dt.
1
1
1
(
p x
x
q
q
p
Integrating we have
1
pq
log
|x |x
q|
p|
= t + c
= e(pq)c e(pq)t c e(pq)t
so that
x
x
p
q
from which we obtain
p c e(pq)t q
x = x(t) =
(pq)t
If p < q then limt
e
q < p we can rewrite
1 c e(pq)t
= 0 and we have limt x(t) = p. If
p e(qp)t c q
e(qp)t c
and assuming c = 0, we have limt x(t) = q. (In other words, the
smaller quantity, p or q, gets used up first as X is produced.)
x(t) =
Case
ii; p
Here we have
q:
1
(x p)2 dx = dt
and we can integrate immediately to obtain
1
c.
=t
xp
+
Letting c c we obtain
x = x(t) = p +
1
c t
from which it is easy to see that limt x(t) = p.
Problems Involving Acceleration Newtons law of motion
states that m a = f , the mass of a body multiplied by its acceleration
is equal to the applied force. In this section we will consider
accelerated motion in a one-dimensional setting; higher dimensional
settings require the study of systems of differential equations
to be studied later.
Let us use t, x to represent time and the (one-dimensional)
position
of the body under discussion. The velocity of the body is then
dx
d
t
and
the acceleration isdtd22x . If we assume the applied force can depend
on any, or all, of t, x andd dx , Newtons equation becomes
t
dm2
xdt2
dx
= f t,
x,
dt
Since this differential equation is a second order equation we have to
specify
dx
x (t0) =
(t0) = v0
x 0,
dt
in order to define an initial value problem. We can always take t0 0;
if it is not originally 0 we can set = t t0 and take = 0 as
the initial time. Then t = t0 corresponds to = 0. The differential
equation becomes
2
.
d
m
xd 2 = f + t0,
x,
dx
d
f ,
x,
dx
d
We can then rename as t
again,
f and f again and proceed with
the
earlier differential equation with initial data at t = 0.
So far we have learned only how to solve first order differential
equa- tions of certain well-defined types. Our emphasis in this section
will lie with solving the given second order differential equation for
x(t) using only first order techniques. This will require that we
consider specific
forms for f .t, x, dx
..
d
t
Case i);
v(t)
dx
f .t, x,
dt
dx
= f .t,
dx
In this case we can define
dt
. The initial value problem then consists of the differential equation
dt
and initial value
dv
= f (t, v), v(0) = v0,
dt
a first order differential equation in v. Let us assume we can find a
solution v(t). If, in fact, it is x(t), or x(T ) for some fixed T > 0, which
is required, we can then use
t
x(t) = x0
+
v(s) ds
to determine x(t).
Example 3; A Rocket Sled Problem
A high speed
rocket sled is accelerated to a speed v0 and then, at time t = 0
and position x0, begins to decelerate due to track friction and air
resistance. The differential equation for the slowing motion may be
taken to be
dv
m 3 = k v
A
v . dt
It is required to give v = v(t) as a function of t and then integrate to
obtain x(t).
Setting p =
, q =
A
m
we have
1
dv = dt.
q v3 + p v
Using the method of partial fractions decomposition we proceed as
follows:
1
1
1
1 a
bv + c
=
=
+
q v3 + p v q v (v2 + qp )
v + q
q vp p 2
2
= 1 (a + b)v + c v + a .
q
q
v (v2 + qp )
The first and last expressions agree just in case a = p , c = 0, b = q .
Thus we have
1p
1
q v
p
dv
Integrating we have
= dt.
q
v2 +
p
.
log |v|
2
log v + p
q
2p
= t + c.
Applying the exponential function we obtain
v
epc
v2 + pq
e
t
ce
pt
v2
= c2
v2 +
e2pt
p
q
so that, taking our definition of p and q into account,
p
2 2pt
.k
c2
e2pt
v = v(t) =
,
x(t) = x0 +
ps
ds =
1 c2
.
1.
1
,
k
A
ept
e2ps
so that, with c as above,
1. k
x(t) = x0 +
A
1 c2 e2pt
ps
ce
ds.
1 c2
e2ps
we have du = p eps ds and the integral becomes
c
A
give x(0)0= x0 we have
2
.k/A + v . If in addition we
.k
.k
c
e
The condition v(0) = v0 requires c =
v0
Setting u = eps
pt
c
du
1 c 2 u2
A
.
1
1
.
sin (c) sin
e
pt
.
1.
k,
sin
1
(c
.1
u)..
.
.
e pt
..
k
p =
Case ii); f .t, x,
the force f
dx
= f .x,
dt
dx
This is the case where
dt
has no direct dependence on the time t. In this case, to get a first order
differential equation we replace t by x as the independent variable:
0 =
m
= m
d2
f
x
dt2 x,
dx
dv
dt
f (x, v)
= m
dt
dv dx
dv
dx dt
f (x, v) = m v
dx
f (x, v)
and thus we obtain the first order differential equation
dv
m v . dx= f (x, v)
In this case an initial value problem is posed by choosing an initial
position x0 and specifying the velocity at that initial position: v (x0) =
v0. On solution of the differential equation we then obtain v = v(x),
giving velocity as a function of position. If we wish to obtain x as a
function of t, and hence v(t) = v(x(t)), we can do this by solving
dx
= v(x), x (t0) = x0,
dt
where t0 is the initial time.
Example
4It is known that the gravitational force exerted by
the earth on an object above the surface of the earth is given by the
formula
g m R2
f = f (r) =
,
r2
where m is the mass of the object, r denotes distance from the earths
center, g is the acceleration of gravity at the earths surface and R
denotes the radius of the earth, i.e., the distance of the earths surface
from its center. In this case, therefore, we have
dv
2
v
= gR
r. 2
dr
Let us suppose an object of mass m leaves the earths surface, r = R,
with a velocity v(R) = vR > 0. The differential equation, as shown,
is separable; we have
v dv = g R2
v2
dr
r2
Using the initial condition we have
c =
vR2
g R2
2r
+ c.
gR
and therefore
.
.2
,
g
2
v(r) = R2
+ vR 2 g R.
r
t t
The + sign applies when the object is rising and the tt sign applies
when it is falling. Let us enquire as to whether or not there is an
altitude rmax > 0 for which v(rmax) = 0. This requires
2 g R2 + v2R
2 g R = 0 rmax =
rmax
2 g R2
2 g R R
v2
If v2 < 2 g R this formula gives a positive value of rmax such that
R
2
v(rmax) = 0. However,
if v R 2 g R no such value is obtained; the
initial velocity vR = 2 g R is called the escape velocity; it
corresponds
object always has a positive velocity
to rmax = ; the
with that
we
2
have
velocity v(r) = 2 g R r tending .to 0 as r . If vR > 2 g R
lim v(r) = v2 2 g R.
r
R
Suppose vR2 < 2 g R so that the initial velocity is less than the escape
velocity. We can then ask when the object will return to the earths
surface, assuming it left the earths surface at t = 0. Clearly this time,
T , is twice the time, , required for the object to rise to the altitude
where its velocity is 0. When rising
dr
dt
R2
= v(r) = .
,
2g
+ v2R 2 g R
.
2g
R2
dr 2 g
R
+ v2R
= dt.
Thus T = 2 where, setting r = 2 , k 2 = (2gR v2 )/2gR2 ,
rmax
R
dr
. 2 g R2
2
= 2gR
rmax
2
d
1 k2 2 =
+ vR 2 g
(fromMATLAB) = 1 3
2gRk
tan
1.
k
1
k2 2
rmax
.. .
2 2 .
k .
.
R
