first order differential equation (applications)
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iit guwahati lecturesTRANSCRIPT
Applications of Linear Equations: Growth and Decay
Problems
The initial value problem
dx
dt= kx, x(t0) = x0 (1)
where k is a constant, occurs in many physical theories involving either
growth or decay. For instance, in biology it models the growth of bacteria or a
small population of animals that increases at a rate proportional to the amount
present at any time.
In physics the IVP (1) models the process of estimating the amount of
radioactive substance remaining at any time t or the temperature of a cooling
body. Similarly in chemistry, it may be used to estimate the amount os
chemicals that remain at any time during certain chemical processes.
Applications of First Order Ordinary Differential Equations – p. 1/18
Bacterial Growth
Problem: A culture has N0 number of bacteria. At t = 1 hour the number of
bacteria is measured to be (3/2)N0. If the rate of growth is proportional to the
number of bacteria present, determine the time necessary for the number of
bacteria to triple.
Answer: The associated initial value problem is
dN
dt= kN, N(0) = N0 (2)
which has the solution N(t) = N0ekt. Since N(1) = 3
2N0, we have
3
2N0 = N0e
k⇒ k = ln( 3
2) = 0.4055. Thus N(t) = N0e
0.4055t.
Suppose that the bacteria population triples in time t = t1. Then
3N0 = N0e0.4055t1
⇒ t1 =ln3
0.4055≈ 2.71 hours .
Applications of First Order Ordinary Differential Equations – p. 2/18
Carbon Dating
Problem: A fossilized bone is found to contain 1/1000 the original amount of
radioactive carbon C-14. Given the the half-life of C-14 in approximately
5600 years, determine the age of the fossil.
Answer: If A0 be the original amount of C-14, then the associated IVP is once
again of the same type as in the previous problem and it has the solution
A(t) = A0ekt. Since A(t) = A0/2 when t = 5600 years, we have
A0
2= A0e
5600t⇒ k = −
ln2
5600= −0.00012378.
Therefore the process satisfies the equation, A(t) = A0e−0.00012378t.
Thus if the age of the fossil be t1 years, then
A0
1000= A0e
−0.00012378t1⇒ t1 =
ln1000
0.00012378≈ 55, 800 years.
Applications of First Order Ordinary Differential Equations – p. 3/18
Electrical Circuits
Problem: A 12-volt battery is connected to a simple series circuit in which the
inductance is 1/2 henry and the resistance is 10 ohms. Determine the current i
if the initial current is zero.
In a series circuit containing only a resistor and an inductor, Kirchoff’s second
law states that the sum of the voltage drop across the inductor (L didt ) and the
voltage drop across the resistor (iR) is the same as the impressed voltage
(E(t)) on the circuit. Thus the current flow i(t) satisfies the linear equation
Ldi
dt+ Ri = E(t) (3)
Thus the given problem gives rise to the IVP
1
2
di
dt+ 10i = 12, i(0) = 0 (4)
Upon solving it we get the current flow as i(t) = 6
5−
6
5e−20t.
Applications of First Order Ordinary Differential Equations – p. 4/18
Fluid Mixtures
Problem: Initially 50 pounds of salt is dissolved in a large tank holding 300
gallons of water. A brine solution is pumped into the tank at a rate of 3 gallons
per minute and a well-stirred solution is then pumped out at the same rate. If
the concentration of the solution entering is 2 pounds per gallon, determine
the amount of salt in the tank at any time. How much salt is present after 50
minutes? After a long time?
Answer: If A(t)be the amount of salt in the tank at any time t, then
dA
dt= (Rate of entry) − (Rate of exit) = R1 − R2.
For the given problem the rate at which salt enters the tank is
R1 = (3gal/min) × (2lb/gal) = 6lb/min
and the rate at which salt leaves the tank is
R2 = (3gal/min) × (A
300lb/gal) =
A
100lb/min.
Applications of First Order Ordinary Differential Equations – p. 5/18
Fluid Mixtures
Thus the associated IVP is
dA
dt= 6 −
A
100, A(0) = 50 (5)
It has the solution A(t) = 600 − 550e−t/100. Thus at t = 50,
A(50) = 266.41 lbs. Also as t → ∞, we have A → 600. Thus after a long
period of time the amount of salt in the solution is 600 lbs.
If in the preceding example, the well stirred solution is pumped out at a slower
rate of 2 gallons per minute, then the solution is accumulating at a rate of
(3− 2)gal/min = 1 gal/min. After t minutes there are 300 + t gallons of brine
in the tank so that R2 = (2gal/min) ×(
A300+t lb/gal
)
= 2A300+t lb/min and
the IVP (5) takes the formdA
dt+
2A
300 + t= 6, A(0) = 50, (6)
This has the solution A(t) = 2(300 + t) − (4.95 × 107)(300 + t)−2.
Applications of First Order Ordinary Differential Equations – p. 6/18
Applications of nonlinear equations: Population
Growth
If the rate of growth in a population P is described by the equationdP
dt= kP, k > 0 then the population exhibits unbounded exponential
growth. This is an unrealistic model of growth in many cases, especially in
those where the initial population is large. In such cases overcrowded
conditions with the resulting detrimental effects of such as pollution and
excessive and competitive demand for resources inhibit growth. In 1840 a
Belgian mathematician-biologist P. F. Verhulst proposed a different equation
of the formdP
dt= P (a − bP )
where a and b are positive constants determined by the circumstances. This
equation is referred to as the logistic equation and its solution as the logistic
function.
Applications of First Order Ordinary Differential Equations – p. 7/18
Population Growth
In particular if a(> 0) is the average birthrate and the average death rate is
proportional to the population P (t) at any time t, then the rate of growth per
individual (1
P
dP
dt) satisfies
1
P
dP
dt= (average birth rate − average death rate) = a − bP.
Thus if P0 be the initial population, this gives rise to the IVP
dP
dt= aP − bP 2, P (0) = P0 (7)
This has the solution
P (t) =aP0
bP0 + (a − bP0)e−at.
Logistic curves are quite accurate predictors of growth patterns in a limited
space of certain types of bacteria, protozoa, water fleas and fruit flies.Applications of First Order Ordinary Differential Equations – p. 8/18
Spread of contagious diseases.
In the spread of contagious diseases, it is reasonable to expect that the rate
dx/dt at which the disease spreads is proportional not only to the number of
people x(t) that have already contracted disease, but also to those y(t) which
have not yet been affected. Thus
dx
dt= kxy
where k is the constant of proportionality.
Thus if one infected person is introduced into a population of n people, then
we have x + y = n + 1 so that the rate of spread of the disease is given bydx
dt= kx(n + 1 − x). This gives rise to the obvious initial value problem
dx
dt= kx(n + 1 − x), x(0) = 1. (8)
Applications of First Order Ordinary Differential Equations – p. 9/18
Spread of contagious diseases.
Problem: Suppose a student carrying a flu virus returns to an isolated college
campus of 1000 students. Determine the number of infected students after 6
days if it is observed that 50 students are affected after 4 days.
Answer: Assuming that nobody leaves the campus throughout the duration of
the disease, we seek the solution of the following IVP.
dx
dt= kx(1000 − x), x(0) = 1.
This has the solution x(t) =1000
1 + 999e−1000kt. Using the fact that x(4) = 50
we have k =−1
4000ln
19
999= 0.0009906 so that x(t) =
1000
1 + 999e−0.0009906t.
Thus the number of students affected after 6 days is
x(6) =1000
1 + 999e−5.9436= 276 students.
Applications of First Order Ordinary Differential Equations – p. 10/18
Problems from mechanics: Falling bodies
It is well known that free-falling bodies close to the surface of the earth
accelerate at a constant rate g. Since acceleration is the first derivative of
velocity which in turn is the first derivative of the distance s(t) covered in
time t, the vertical distance covered by the body is described by the equation
d2s
dt2= g.
If v(t) be the velocity at any time t thend2s
dt2=
dv
dt=
dv
ds
ds
dt= v
dv
ds. Thus if
the body is falls from rest from a height h close to the earth’s surface, then
this gives rise to the following first order IVP in v.
vdv
ds= g, v(h) = 0 (9)
Applications of First Order Ordinary Differential Equations – p. 11/18
Falling bodies
If a falling body of mass m encounters air resistance which is proportional to
its velocity v(t) at any time time then the force acting on the body is mg − kv
where k is a constant of proportionality and the negative sign is due to the fact
that resistance opposes the motion.
Then Newton’s second law of motion implies that
mdv
dt= mg − kv.
Thus if the body falls from a height h, then this gives rise to the initial value
problem
vdv
ds+ v
k
m= g, v(h) = 0 (10)
Applications of First Order Ordinary Differential Equations – p. 12/18
Projectiles
A rocket of mass m is propelled from the surface of the earth. its motion is
resisted by gravitational force which is inversely proportional to the square of
the distance covered at any time t. Thus its motion is described by the
equation
md2s
dt2= −
k
s2
where k is the constant of proportionality.
Using the fact that the acceleration due to gravity on the surface of the earth is
g if R be the radius of the earth, this yields the equation mg = k/R2. Thus if
v0 be the initial velocity of the rocket, then the velocity v(s) at any height s is
the solution of the IVP
vdv
ds= −
gR2
s2, v(R) = v0 (11)
Applications of First Order Ordinary Differential Equations – p. 13/18
Motion of a pendulum
A pendulum of mass m is fixed to the end of a rod of length a which has
negligible mass. If the bob is pulled to one side through an angle α and
released, then by the principle of conservation of energy.
1
2mv(t)2 = mg(a cosα − a cos θ) where v(t)(=
ds
dt) is the
velocity at time t and θ is the angular displacement of the pendulum from the
vertical position at any time t. Since s = aθ, this gives rise to the IVP1
2a
(
dθdt
)2= g(cos θ − cosα), θ(0) = α (12)
Applications of First Order Ordinary Differential Equations – p. 14/18
The shape of a hanging wire
Suppose that a suspended wire hangs under its own weight. Then the curve
described by the shape of the wire is the solution of an IVP of second order
which may be reduced to two IVPs of first order.
We examine a portion of the wire between the lowest point P1 and any other
point P2. This portion of the wire is at an equilibrium under the action of three
forces, namely, the weight of the segment P1P2, the tensions T1 and 2 in the
wire at P1 and P2 respectively. If w be the linear density of the wire and s the
length of P1P2, then the weight of P1P2 is ws.
Applications of First Order Ordinary Differential Equations – p. 15/18
The shape of a hanging wire
Resolving the tension T2 into its horizontal and vertical components T2 cos θ
and T2 sin θ respectively (see figure), the following equation arise from the
equilibrium condition.
T1 = T2 cos θ (13)
ws = T2 sin θ (14)
Since dx = ds cos θ, the length s of the arc P1P2 is given by
s =
∫ x
0
√
1 +
(
dy
dx
)2
(15)
From the fundamental theorem of integral calculus we have
ds
dx=
√
1 +
(
dy
dx
)2
(16)
Applications of First Order Ordinary Differential Equations – p. 16/18
The shape of a hanging wire
But from (13) and (14) we have tan θ =ws
T1
so that
dy
dx=
ws
T1
(17)
Differentiating the above equation with respect to x we haved2y
dx2=
w
T1
ds
dx.
Now using (16) we have
d2y
dx2=
w
T1
√
1 +
(
dy
dx
)2
.
Thus the shape of the wire is the solution of the IVP
d2y
dx2=
w
T1
√
1 +
(
dy
dx
)2
, y(0) = 0,dy
dx= 0 when (x, y) = (0, 0) (18)
Applications of First Order Ordinary Differential Equations – p. 17/18
The shape of a hanging wire
Setting p =dy
dx, we get the IVP
dp
dx=
w
T1
√
1 + p2, p(0) = 0, (19).
The solution of this IVP is the first order ODEdy
dx= tan
wx
T1
. The latter gives
rise to the second IVP
dy
dx= tan
wx
T1
, y(0) = 0, (20).
Thus the shape of the wire obtained by solving this problem is
y =T1
wln
(
secwx
T1
)
.
Applications of First Order Ordinary Differential Equations – p. 18/18