MA1002: (Ordinary Differential Equations)

Part-I: First oder ODEs

Department of Mathematics
National Institute of Technology Rourkela
Syllabus
I ODEs of first order: Separable equations, Exact equations, Integrating factors,
Linear equations, Bernoulli equations, orthogonal trajectories, Existence and
uniqueness of IVP and Applications.
I ODEs of higher order: Fundamental/ general solutions of order two,
homogeneous equations , Wronskian, reduction of order, Solution of
non-homogeneous equations by method of undetermined coefficients and
variation of parameters. Extension to higher order differential equations,
Euler-Cauchy equation, Power series method and Applications
I Laplace Transform: Laplace/inverse Laplace transforms, existence, first shifting
theorem, transforms of derivative/integral, second shifting theorem,
differentiation/ integration, Convolution theorem, Solution of ODEs
I Matrix Theory: Gauss elimination/ Gauss-Jordon methods, Vector space,
subspace, linear span, linear dependence/ independence, Basis/dimension, Row
and column spaces, Rank and Nullity Theorem, Inner product spaces, Matrix
representation of Linear Transformations, Solvability of systems of linear
equations, Eigen values, Eigen vectors, Diagonalization of matrices, Reduction
of a quadratic form to canonical form.
I One can find the detailed syllabus: NITRIS → CURRICULUM → VIEW
CURRICULLA AND SYLLABI.

Some topics may (not) be discussed

 Recommended Books

I Erwin Kreyszig, Advanced Engineering Mathematics, Wiley;

10th edition. Chapters 1,2 3, 5 6, 7, 8 (Respective topics)
Reference Books
I Shepley L. Ross, Differential Equations, Wiley; 3rd edition.
I George Simmons, Differential Equations with Applications and
Historical Notes,McGraw Hill Education; 2nd edition.
I Earl A. Coddington, An Introduction to Ordinary Differential
Equations, Dover Publications.
I TA: 20 marks, Mid-Sem: 30 marks, End-Sem: 50 marks
I Attendance policy: Attendance of all the students will be
taken during class hours only.
Introduction

Many a phenomenon of the natural world is governed by

principles which are relations involving rates at which things
happen.
In mathematical terms, the relations are equations and the
rates are derivatives.
Such equations involving derivatives are called differential
equations.

Definition
An equation involving derivatives of one or more dependent
variables with respect to one or more independent variables is
called a differential equation (DE). If we have only one
independent variable then we call the DE an ordinary differential
equation (ODE). If there are more than one independent
variables, we call it a partial differential equation (PDE).
Examples of DE in Natural Phenomena

The equation of motion for a falling body is

d2 x
= −g, (1)
dt2

where g is the gravitational constant

The decay of a radioactive compound is governed by

dm
= −λm, (2)
dt
where m denotes mass of the compound, and λ is a constant
depending on the compound.

The equation of motion of a simple pendulum is governed by

d2 θ g
+ sin θ = 0, (3)
dt2 L
where θ denotes the angle with the vertical of the pendulum of
length L.
Classification of ODE

Definition
The order of an ODE is the order of the highest derivative
appearing in the equation.

For example, the order of the following ODE is 3:

 dy 2 d2 y  d3 y 
+ + + sin = x17 + ex .
dx dx2 dx3

Definition
A crucial classification of ODE is whether it is linear or not. An
ODE
F (x, y, y (1) , y (2) , . . . , y (n) ) = 0
is said to be linear if F is a linear function of the variables
di y
y, y (1) , y (2) , . . . , y (n) , where y (i) = dx i.
Classification of ODE

A linear ODE is of the form

an (x)y (n) + an−1 (x)y (n−1) + . . . + a0 (x)y = g(x). (4)

If g(x) is zero for all x in the domain, we say that (4) is a linear
homogenous ODE. Else, we call it non-homogenous.

For example, consider the following DE for x ∈ (1, 2):

1
(sin x)y (2) + ex y (1) + y=0 (5)
x
1
(sin x)y (2) + ex y (1) + y = x (6)
(2) x (1) 1 x 2
x sin y + e y + y = x + 1 (7)
x
Then (5) is a linear homogenous ODE, (6) is linear
non-homogenous ODE and (7) is a non-linear ODE. Linear ODE’s
are ‘far simpler’ than non-linear ones.
Solutions of ODE

A solution of the ODE

F (x, y, y (1) , y (2) , . . . , y (n) ) = 0

in the interval (α, β) is a function y = φ(x) such that

φ(1) , φ(2) , . . . , φ(n) exist and satisfy

F (x, φ, φ(1) , φ(2) , . . . , φ(n) ) = 0

for every x ∈ (α, β). In this course we will be concerned with

real valued solutions only.

For example, y1 (x) = cos x is a solution of y (2) + y = 0, and

so is y2 (x) = sin x. It is easy to see that any linear
combination c1 y1 (x) + c2 y2 (x) of y1 and y2 is again a solution
of y (2) + y = 0.
Solutions of ODE

The solution involving the arbitrary constants c1 and c2 is called a

general solution of the equation. A solution obtained by assigning
specific values to the arbitrary constants in the general solution is
called a particular solution. For example, y(x) = 100 cos x − 1.2 sin x
is a particular solution of y (2) + y = 0.

For a linear ODE with continuous coefficients, every solution of the

equation can be obtained by assigning specific values to the
arbitrary constants in the general solution.

But for non-linear equations, this need not be the case. For
example, consider Clairut’s equation:

(y (1) )2 − xy (1) + y = 0. (8)

We get a family of solutions y = cx − c2 for an arbitrary constant c

2
as can be verified readily. But y = x4 is also a solution, and we can
not get it by assigning any specific value to the constant c. Such a
solution is called a singular solution.
IVP: Initial Value Problem

An IVP consists of an ODE

F (x, y, y 1 , y (2) , · · · , y (n) ) = 0, x ∈ I

together with a set of supplementary conditions

y(x0 ) = a0 , y (1) (x0 ) = a1 , · · · , y (n−1) (x0 ) = an−1

for some fixed x0 in the interval I and fixed values

a0 , · · · , an−1 .

For example,
y (2) + y = 0, y(0) = 1, y (1) (0) = 0, I = [−π, π].

As we know the general solution y(x) = c1 cos x + c2 sin x, we

can solve for c1 and c2 using the initial conditions and obtain
c1 = 1 and c2 = 0, so that that the IVP has a solution cos x.
BVP: Boundary Value Problem

A BVP consists of an ODE

F (x, y, y 1 , y (2) , . . . , y (n) ) = 0, x ∈ I

together with a set of supplementary conditions about the

values of y, y (1) , y (2) , . . . , y (n) at two different values of x in
the interval I under consideration.

For example,

y (2) + y = 0, y(0) = 1, y(π) = 2, I = [−π, π]

is a BVP. This particular BVP’s has no solution at all, which

can be verified by considering a general solution.

In general, BVPs are more difficult to handle than IVP’s.

Existence and Uniqueness of Solutions

There is no guarantee in general that a given IVP has a solution or

that the solution is unique.

For example, the IVP

xy (1) = 4y, y(0) = 1

has no solution, whereas the IVP

p
y (1) = |y|, y(0) = 0

has two solutions: y ≡ 0 and

2 2
y = x4 for x ≥ 0, y = − x4 for x < 0.

But it will be proved later that if the equation is ‘sufficiently

well-behaved’, a solution exists uniquely.

We need to consider the question of existence and uniqueness of

first order ODE only, as all other cases ‘can be reduced’ to this basic
case.
An Application: Finding Orthogonal Trajectories

Let F (x, y, c) be a one parameter family of curves(c is a parameter

which can take arbitrary values). A curve which intersects each
member of the given family at right angles is called an orthogonal
trajectory of the given family.

The angle of intersection of two curves is the angle between their

tangents at the point of intersection, and dx
dy
represents the slope of
the tangent, so we can use ODE’s to find orthogonal trajectories.

Example: Consider the family of hyperbolas given by xy = c. The

slope of the tangent to the curve at any point (x, y) is
dy
dx
= − xc2 = − xy
x2
= − xy .
Then the ODE for the orthogonal family of trajectories is
dy x
= =⇒ xdx − ydy = 0.
dx y

Hence, the orthogonal family is x2 − y 2 = a (a an arbitrary

contstant).
Methods for Solving First Order ODE

An equation of the form

dy M (x)
M (x)dx + N (y)dy = 0, i.e., =− (9)
dx N (y)

is called a separable equation. One can integrate directly to obtain

(sometimes implicit) solution.
Examples:
1 (y + ey )dy − (x − e−x )dx = 0
R R
⇒ (y + ey )dy − (x − e−x )dx = 0
y2 x2
⇒ 2
+ ey − 2
− e−x = c, (c is an arbitrary constant).

2 (x2 y 3 + 2y)dx + (2x − 2x3 y 2 )dy = 0

⇒ 2(ydx + xdy) + x2 y 3 dx − 2x3 y 2 dy = 0
d(xy)
⇒ 2 (xy)3 +
dx
x
− 2 dy
y
= 0,

⇒ −(xy)−2 + ln | x | −2 ln | y |= c.
Homogenous ODE

A homogenous ODE is of the form

dy F (x, y)
= ,
dx G(x, y)

where F and G are homogenous function of x, y of the same

degree n, i.e.,

F (λx, λy) = λn F (x, y), G(λx, λy) = λn G(x, y) ∀λ.

For example,
dy −2x2 + y 2
= .
dx x2

We reduce homogenous ODEs reducible to separable ones

by substituting y = ux. We solve the resulting separable
equation for u, and then substitute back u = xy .
Homogenous ODE: an example

dy −2x2 +y 2
Example: Find general solution of dx = x2 .

Solution:
dy
y = ux ⇒ dx = u + x du
dx = −2 + u
2

du dx
=⇒ u2 −u−2 = x (separablef orm)
1 u−2
=⇒ 3 ln | u+1 |= ln |x| + c
y−2x
=⇒ | y+x |= ax3

for some arbitrary constant a.

ODE’s Reducible to Homogenous Form

Equations of the type

(a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0 (10)

can be also reduced to separable form.

Case 1: When a1 b2 6= a2 b1 , we substitute

x = X + h, y = Y + k; where h, k satisfy

a1 h + b1 k + c1 = 0, a2 h + b2 k + c2 = 0,
to express (10) as a homogenous ODE

(a1 X + b1 Y )dX + (a2 X + b2 Y )dY = 0,

and solve as before.

Solutions of ODE

Example: Consider (x − 2y + 1)dx + (4x − 3y − 6)dy = 0.

Let x = X + h, y = Y + k,
where h − 2k + 1 = 0, and 4h − 3k − 6 = 0.
Thus, h = 3, and k = 2.

Substituting, we get

(X − 2Y )dX + (4X − 3Y )dY = 0 (homogenousf orm)

dY 2Y − X
⇒ = .
dX 4X − 3Y
Now using Y = uX, we can obtain a separable form and then
solve for u, hence for y.
Case 2: a1 b2 = a2 b1

Case 2: Consider the ODE

(a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0, a1 b2 = a2 b1 .

We can substitute z = a1 x + b1 y to obtain a separable equation.

Example: Find general solution of

(x + 2y + 3)dx + (2x + 4y − 1)dy = 0.
dy
Substituting x + 2y = z ( =⇒ dz
dx = 1 + 2 dx ), we obtain
 dz 
−1
(z + 3) + (2z − 1) dx2 =0

=⇒ 7dx + (2z − 1)dz = 0 (separablef orm)

=⇒ 7x + z 2 − z = c
=⇒ 7x + (x + 2y)2 − (x + 2y) = c,
where c is an arbitrary constant.
Exact ODE

Let φ(x, y) be a function of x, y such that it has first order

partial derivatives φx and φy is some domain D. The total
differential of φ is defined as

dφ = φx dx + φy dy.

M (x, y)dx + N (x, y)dy is called an exact differential in a

domain D if there exists a function φ of x and y such that

dφ=M (x,y)dx+N (x,y)dy ∀x,y∈D, i.e.,φx =M (x,y), φy =N (x,y).

In such a case, we say M (x, y)dx + N (x, y)dy = 0 is an exact

ODE.
For example, 2xydx + x2 dy = 0 is an exact ODE: we can take
φ(x, y) = x2 y. In this case, the solution is x2 y = c, where c is
an arbitrary constant.
Condition for Exactness

Let M , N , My and Nx be continuous in a rectangular domain

D : 0 <| x − x0 |< α, 0 <| y − y0 |< β. Then
M (x, y)dx + N (x, y)dy = 0 is an exact ODE if and only if
My (x, y) = Nx (x, y) ∀x, y ∈ D.

It is easy to see the necessary part: if the equation is exact,

we can find a function φ(x, y) on D such that φx = M and
φy = N . Then φxy = My and φyx = Nx . But φxy (= My ) and
φyx = Nx are continuous, so φxy = φyx , and hence My = Nx .
Sufficient Condition for Exactness

We will give a constructive proof that My (x, y) = Nx (x, y)

∀x, y ∈ D = {(x, y) | 0 <| x − x0 |< α, 0 <| y − y0 |< β} (a
rectangular domain) implies that M dx + N dy = df for some
f (x, y).
We need f (x, y) to satisfy fx = M (x, y), and fy = N (x, y)
Integrating the former with respect to x treating y as
constant, we find that
Z x
f (x, y) = M (x, y) + h(y), (1)
x0

where the function h is an arbitrary function of y, playing the

role of an arbitrary constant. It is clear from (1) that
fx = M (x, y).
Now we must show that it is possible to choose h(y) so that
fy = N .
Sufficient Condition for Exactness

From (1),
Rx
fy = ∂
∂y
M (x, y)dx + h0 (y)
x0

Rx
fy = ∂
∂y
M (x, y)dx + h0 (y)
x0

Since we want fy = N , we find that h0 (y) should satisfy

Zx
0
h (y) = N (x, y) − My (x, y)dx. (2)
x0

To determine h(y) from (2), we must show that the right side is a
function of y alone. Differentiating the RHS of (2) with respect to x,
we obtain:
∂
Rx
Nx (x, y) − ∂x My (x, y)dx = Nx (x, y) − My (x, y)= 0.
x0
Sufficient Condition for Exactness

We have shown that the RHS of

Zx
0
h (y) = N (x, y) − My (x, y)dx
x0

does not involve x, so we can integrate both sides with respect to

y, and solve for h(y). The resulting function
y
h i
x R x R x R R
f (x,y)= M (x,y)+h(y) = M (x,y)+ N (x,y)− My (x,y)dx dy
x0 x0 y0 x0

has the desired property that fx = M and fy = N . 

We require the domain D, more generally, to be simply

connected, i.e., without any holes so that integration is
‘well-defined’. For example, an annular disc will not qualify as
simply connected domain.
Solution of Exact ODE

Example: Find general solution of the ODE

(y cos x + 2xey )dx + (sin x + x2 ey − 1)dy = 0.

Solution:
Here, M = y cos x + 2xey and N = sin x + x2 ey − 1. We have
My = cos x + 2xey = Nx , hence the given equation is exact.

Let φ(x, y) be the function such that φx = M and φy = N .

Then, integrating φx = y cos x + 2xey with respect to x
treating y as constant, we get

φ(x, y) = y sin x + x2 ey + h(y),

where the constant of integration depends on y, which we

write as h(y). Once we find h(y), φ(x, y) = c will give the
family of solutions.
Soution of Exact ODE

To find h(y), we use φy = N . When the condition for

exactness is satisfied, we always get h0 (y) as a function of y
alone after substituting in φy = N , and that allows us to
determine h(y), and hence φ(x, y), completely. Thus,

sin x + x2 ey + h0 (y) = sin x + x2 ey − 1 =⇒ h(y) = −y.

Hence the solution is given by φ(x, y) = c, i.e.,

y sin x + x2 ey − y = c.
Integrating Factors

Sometimes an equation M (x, y)dx + N (x, y)dy = 0 need not

be exact to start with, but may become so after multiplication
by a suitable function µ = µ(x, y). Such a function µ for which
µM dx + µN dy = 0 is exact is called an integrating factor (IF).

To find the IF , we again use the condition for exactness:

(µM )y = (µN )x . We deduce that

µy M + µMy = µx N + µNx ⇔ µ(My − Nx ) = µx N − µy M.

The RHS is easy to solve when µ is a function of x alone (or

y alone). Then, we have
µx N dµ  My − Nx 
µ= =⇒ = µ,
M y − Nx dx N
M −N
and we can solve for µ easily when yN x is a function of x
M −N
alone (or, − yM x is a function of y alone).
Integrating Factors

Example: Find general solution of (3xy + y 2 )dx + (x2 + xy)dy = 0.

1 Here M = 3xy + y 2 , N = x2 + xy, My 6= Nx . So the given equation
is not exact.
My −Nx
2 But N
= x+y
x2 +xy
= x1 is a function of x alone. Hence there is an
My −Nx
IF µ given by dµ
dx
= N
µ = x1 µ, i.e., µ = x.

3 Multiplying the ODE by the IF, we get

(3x2 y + xy 2 )dx + (x3 + x2 y)dy = 0, µM = 3x2 y + xy 2 ,
µN = x3 + x2 y, (µM )y = 3x2 + 2xy = (µN )x . Hence the resulting
ODE is exact.

4 Integrating φx = µM with respect to x treating y as constant, we

2 2
obtain φ(x, y) = x3 y + x 2y + h(y).
5 Now we need only to determine h(y), for which we use φy = µN to
get x3 + x2 y + h0 (y) = x3 + x2 y =⇒ h0 (y) = 0 ⇒ h(y) = 0.
x2 y 2
6 Hence the solution is given by x3 y + 2
= c, where c is an
arbitrary constant.
First Order Linear Non-Homogenous ODE

The general form of a first-order linear homogenous ODE is

dy
+ p(x)y = q(x).
dx
R
Multiplying by e p(x)dx
, we obtain
d
R
p(x)dx
R
p(x)dx
R
p(x)dx
R R
p(x)dx
dx (e y)=q(x)e =⇒ e y= q(x)e dx+c,

which gives the solution as

R R R
y=e− p(x)dx
y( q(x)e p(x)dx
dx+c).

Example: Consider
R
dy 2x+1 −2x p(x)dx
dx + x y=e , e =e2x+ln x =xe2x .

Multiplying the given ODE by xe2x , we obtain

2 2
d
dx (xe
2x
y)=x =⇒ xe2x y= x2 +c =⇒ y=x−1 e−2x ( x2 +c). 
First Order Non-linear ODE

Certain non-linear ODEs can be reduced to linear ones after

a suitable substitution for the dependent variable.

Bernoulli equations and Ricatti equations belong to that

category.

Definition
Bernoulli equation is a first order ODE of the form

dy
+ p(x)y = q(x)y n , n 6= 0, 1
dx
We can reduce the above non-linear ODE to a linear equation by
substituting z = y 1−n .
Example: Bernoulli Equation

dy
Example: Find general solution of dx + xy = xy −1 .

Solution: In the given ODE, substitute

dy
dz
z = y 1−(−1) = y 2 ( dx = 2y dx ) to obatin
dy dz
2y dx + 2xy 2 = 2x =⇒ dx + 2xz = 2x.
R 2
Multiplying by e 2xdx
= ex , we obtain
2 2 2
dz
ex dx + 2xex z = 2xex
d x2 2
= 2xex
dx (e z)
2 R 2 2
=⇒ ex z = 2xex dx + c = ex + c
2
z = 1 + ce−x
2
y 2 = 1 + ce−x .