MA108 ODE: Basic Concepts and Examples
Preeti Raman
IIT Bombay
Preeti Raman, IITB MA108
Welcome
Suggested Books
1 Advanced Engineering Mathematics by E. Kreyszig, 9th
edition, Wiley Publishers.
2 Elementary Differential equations and Boundary Value
Problems by W. E. Boyce and R. C. DiPrima, 9th edition,
Wiley Publishers.
Preeti Raman, IITB MA108
Basic Concepts
Let y (x) denote a function in the variable x. An ordinary
differential equation is an equation containing one or more
derivatives of an unknown function y . Note that the equation may
contain y itself, and known functions of x (including constants).
In other words, an ODE is a relation between the derivatives
y , y 1 , . . . , y (n) and functions of x:
F (x, y , y 1 , . . . , y (n) ) = 0.
DE’s occur naturally in physics, engineering and so on. Can you
give some obvious examples? Velocity and acceleration being
derivatives, often give rise to DE’s.
Preeti Raman, IITB MA108
Examples
Example 1: A falling object.
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v 2 where v is the velocity and c is a
constant. Then
dv
m = mg − c · v 2 .
dt
An ODE of first order.
Preeti Raman, IITB MA108
Examples
Example 2: Radioactive decay.
A radioactive substance decomposes at a rate proportional to the
amount present. Let y (t) be the amount present at time t. Then
dy
=k ·y
dt
where k is a physical constant whose value is found by experiments
(−k is called the decay constant). ODE of first order.
Preeti Raman, IITB MA108
Examples
Example 3: Electrical circuits.
Consider a basic RLC circuit:
resistance of the resistor - R ohms,
inductance of the inductor - L henrys,
capacitance of the capacitor - C farads.
These are wired and connected to an electromotive force V (t)
volts.
Let Q(t) (coulombs) be the total charge in the
capacitor at time t.
dQ By
I (t) = = current.
dt
dI Q(t)
Kirchhoff’s voltage law, L dt + RI + C = V (t), i.e.,
d 2Q dQ 1
L 2
+R + · Q = V (t).
dt dt C
ODE of second order.
Preeti Raman, IITB MA108
Linear equations
Definition (Linear ODE)
The ODE F (x, y , y 0 , . . . , y (n) ) = 0 is called linear if F is a linear
function of the variables y , y 0 , . . . , y (n) . A linear ODE of order n is
of the form
a0 (x)y (n) + a1 (x)y (n−1) + . . . + an (x)y = b(x)
where a0 , a1 , . . . , an , b are functions of x and a0 (x) 6= 0.
Check list : If the dependent variable is y , derivatives occur upto
first degree only, no products of y and/or its derivatives are there.
Preeti Raman, IITB MA108
Examples Revisited
Example 1: A falling object.
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v 2 where v is the velocity and c is a
constant. Then
dv
m = mg − c · v 2 .
dt
An ODE of first order. Linear or non-linear? (NL)
Preeti Raman, IITB MA108
Examples Revisited
Example 2: Radioactive decay.
A radioactive substance decomposes at a rate proportional to the
amount present. Let y (t) be the amount present at time t. Then
dy
=k ·y
dt
where k is a physical constant whose value is found by experiments
(−k is called the decay constant). ODE of first order. Linear.
Preeti Raman, IITB MA108
Examples Revisited
Example 3: Electrical circuits.
Consider a basic RLC circuit:
resistance of the resistor - R ohms,
inductance of the inductor - L henrys,
capacitance of the capacitor - C farads.
These are wired and connected to an electromotive force V (t)
volts.
Let Q(t) (coulombs) be the total charge in the
capacitor at time t.
dQ By
I (t) = = current.
dt
dI Q(t)
Kirchhoff’s voltage law, L dt + RI + C = V (t), i.e.,
d 2Q dQ 1
L 2
+R + · Q = V (t).
dt dt C
ODE of second order. Linear or non-linear? (L)
Preeti Raman, IITB MA108
Examples
Example 4: The motion of an oscillating pendulum.
Consider an oscillating pendulum of length L. Let θ be the angle it
makes with the vertical direction.
Then:
d 2θ g
+ sin θ = 0.
dt 2 L
(Recall: torque = m.i × a.a = - force × length.)
ODE of second order. Non-linear.
Preeti Raman, IITB MA108
Examples
Example 5: Given an amount of a radioactive substance, say 1 gm,
find the amount present at any later time.
The relevant ODE is
dy
= k · y.
dt
Initial amount given is 1 gm at time t = 0. i.e.,
y (0) = 1.
By inspection, y = ce kt , for an arbitrary constant c, is a solution
of the above ODE. The initial condition determines c = 1. Hence
y = e kt
is a particular solution to the above ODE with the given initial
condition.
Preeti Raman, IITB MA108
Examples
Example 6: Find the curve through the point (1, 1) in the xy -plane
having at each of its points, the slope − yx .
The relevant ODE is
y
y1 = − .
x
By inspection,
c
y=
x
is its general solution for an arbitrary constant c; i.e., a family of
hyperbolas.
The initial condition given is
y (1) = 1,
which implies c = 1. Hence the particular solution for the above
problem is
1
y= .
x
Preeti Raman, IITB MA108
Geometric Meaning of Solutions of y 1 = f (x, y )
Consider the first order ODE
dy
= f (x, y ).
dx
Suppose that f (x, y ) is defined in a region D ⊆ R2 . If y = φ(x) is
a solution curve and (x0 , y0 ) is a point on it, then the slope at
(x0 , y0 ) is f (x0 , y0 ).
Along the curves f (x, y ) = c, where c is a constant, the slopes are
constant. These curves are called isoclines.
At each point (a, b) ∈ D, assign a unit vector with slope f (a, b).
The vector field H : D → R2 given by
H(a, b) = (1, f (a, b))
is called the direction field. A drawing of the vector field H at a
large number of points of D gives us approximate solution curves.
Preeti Raman, IITB MA108
Geometric Meaning of Solutions of y 1 = f (x, y )
Example 7: Consider the first order linear ODE
y 1 = xy .
The isoclines are the hyperbolas xy = k, k 6= 0, and the
coordinate axes.
The direction fields are given by
H(x, y ) = (1, xy ).
Check that the solutions for the above ODE are
2 /2
y = c · ex ,
where c is a constant.
Preeti Raman, IITB MA108
Isoclines & Direction Fields
Isoclines are in blue, direction fields are in black, and solution
curves are in red. Note that solution curves are approximated by
direction fields.
Preeti Raman, IITB MA108
