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Ordinary Differential Equations
Swaroop Nandan Bora
swaroop@iitg.ernet.in
Department of Mathematics
Indian Institute of Technology Guwahati
Guwahati-781039
Teachers’ Training Camp, IITG, July 4, 2016
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
A first-order differential equation is an equation
dy
dx= f(x, y)
in which
f(x, y) is a function of two variables defined on a region in the xy-plane.
A solution of the given equation is a differentiable function defined
on an interval I of x-values (perhaps infinite) such that
d
dxy(x) = f(x, y(x))
on that interval.
The general solution to a first-order differential equation is a solution that contains all
possible solutions.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
The general solution always contains an arbitrary constant
but having this property doesnt mean a solution is the general solution.
Establishing that a solution is the general solution may require deeper results
from the theory of differential equations.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
We often need a particular rather than the general solution to a first-order
differential equation
dy
dx= f(x, y).
The particular solution satisfying the initial condition y(x0) = y0 is the solution
whose value is y0 when x = x0.
Thus the graph of the particular solution passes through the point (x0, y0) in the
xy-plane.
A first-order initial-value problem is a differential equation
dy
dx= f(x, y) whose solution must satisfy an initial condition y(x0) = y0.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Every differential equation need not have a solution.
Even if solution exists, that may not be unique.
Picard’s theorem:
Suppose f(x, y) and∂f
∂yare continuous on the interior of a rectangle R, and that
(x0, y0) is an interior point of R, then the IVP
dy
dx= f(x, y), y(x0) = y0
has unique solution y = y(x) for x in some open interval containing x0.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Applications
Orthogonal trajectories
Let F (x, y, c) = 0 be a given one-parameter family of curves in the xy-plane.
A curve that intersects the curves of the above family at right angles
is called an orthogonal trajectory of the given family.
Main point:
Suppose that the differential equation of above family can be obtained as
dy
dx= f(x, y).
Then the orthogonal trajectory must have a slope
−1/(f(x, y)).
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Examples
Example 1
Given a family of circles: x2 + y2 = a2.
Its differential equation is obtained as
dy
dx= −
x
y
Differential equation of the orthogonal trajectory
dy
dx=
y
x
Giving the orthogonal trajectory as
y = kx,
all are straight lines passing through the origin.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Example 2
Given a family of parabolas:
y = cx2.
The differential equation is
dy
dx=
2y
x.
The differential equation of the orthogonal trajectory
dy
dx= −
x
2y.
which gives us
x2 + 2y2 = k2.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Problems in Mechanics
Newton’s 2nd law of motion
mdv
dt= F.
Example
A body of weight 8 pounds falls from rest toward the earth from a great height. As it
falls, air resistance acts upon it which is numerically equal to 2v, where v is the
velocity in feet per second. To find the velocity and distance covered in time t.
Differential equation is
mdv
dt= F1 + F2.
g = 32, w = 8, m = 8/32 = 1/4
1
4
dv
dt= 8− 2v.
Since the body falls from rest,
initial condition is v(0) = 0.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Example 2
Solving it and using the initial condition
v = 4(1 − e−8t)
For finding the distance, the differential equation is
dx
dt= 4(1 − e−8t)
Solving and using the initial condition
x = 4(t+1
8e−8t
−
1
8)
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Interpretation of results
The solution for v tells us that
as t → ∞, then v reaches the limiting velocity 4 (ft/sec).
Again the solution for x tells us that
as t → ∞, then x also → ∞.
Does this imply that the body will pierce through the earth and continue
forever??
Of course not.
Because
when the body reaches the earth’s surface, the above differential equations no longer
apply.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Rate of growth and decay
Example
The rate at which a radioactive nuclei decay is proportional to the number of such
nuclei that are present in a given sample. Half of the original number of radioactive
nuclei have undergone disintegration in a period of 1500 years.
What percentage of the original radioactive nuclei will remain after 4500 years?
Formulation:
Let x be the amount of radioactive nuclei present after t years.
The differential equation is
dx
dt= kx.
The initial condition is, given x0 is the amount present initially,
x(0) = x0.
But according to the given condition
x(1500) =1
2x0.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Rate of growth and decay
Solution:
The solution is:
x = ce−kt.
By using the condition of half-life
e−k = (1
2)1/1500.
Therefore the solution is
x = x0[(1
2)1/1500]t.
For (i)
x = x0(1
2)3 =
1
8x0.
that is, 12.5 percent of original number remain after 4500 years.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Population growth
Growth of population,
such as human, animal species, bacteria colony.
Assumptions:
• The population increase is approximately continuous.
• In fact this increase is also a differentiable function of time.
Given a population, we let x be the number of individuals at time t.
Assuming that the rate of change of population is proportional to the number of
individuals,
we get the differential equation
dx
dt= kx.
If the number of individuals was x0 at time t0,
we have the initial condition: x(t0) = x0.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Population growth
The solution is given by
x(t) = x0ek(t− t0).
The population governed by this model (increasing exponentially with time) is
known as
Malthusian law.
But more realistically in many cases
the number of individuals x at time t is described by a differential equation
dx
dt= kx− λx2.
The additional term −λx2
is the result of some cause that tends to limit the ultimate growth of the population.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Mixture problem
A substance S is allowed to flow into a certain mixture in a container at a certain
rate. The mixture is kept uniform by stirring.
Further, in one such situation, this uniform mixture simultaneously flows out of the
container at another (generally different rate.
Letting x denote the amount of S present at time t, the derivative dx/dt denotes the
rate of change of x with respect to t.
If IN denotes the rate at which S enters the mixture and OUT denotes the rate at
which it leaves,
Then we have the differential equation
dx
dt= IN −OUT.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Example
A tank initially contains 50 gallons of pure water. Starting at time t = 0, a brine
containing 2 lb of dissolved salt per gallon flows into the tank at the rate of 3 gal/min.
The mixture is kept uniform by stirring and the well-mixed mixture simultaneously
flows out of the tank at the same rate. To find the amount of salt at time t.
Formulation:
dx
dt= IN −OUT.
IN= (2 lb/gal)(3 gal/min)= 6 lb/min
OUT =( x
50lb/gal
)
(3gal/min) =3x
50lb/min.
The initial value problem is
dx
dt= 6−
3x
50, x(0) = 0.
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
Mixture problem
Solution
x = 100 + ce−3t/50.
Using the initial condition
x = 100(1 − e−3t/50).
How much salt is present at the end of 25 minutes?
x(25) = 100(1 − e−1.5) ≡ 78lb.
What happens when t → ∞?
Swaroop Nandan Bora swaroop@iitg.ernet.in (Department of MathematicsIndian Institute of Technology GuwahatiGuwahati-781039 )IITG, Differential EquationsTeachers’ Training Camp, IITG, July 4, 2016
/ 19
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