1. MA 108 - Ordinary Differential Equations
Neela Nataraj
Department of Mathematics,
Indian Institute of Technology Bombay,
Powai, Mumbai 76
neela@math.iitb.ac.in
February 29, 2016
Neela Nataraj Lecture 1: D3

2. Outline of the lecture
Basic Concepts
Classification based on
1 Type
2 Order
3 Linearity
Solution(s)
1 Explicit
2 Implicit
3 Formal
First Order ODEs, IVP
Separable ODEs
Neela Nataraj Lecture 1: D3

3. 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).
1
Dennis G. Zill, Differential Equations
Neela Nataraj Lecture 1: D3

4. 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).
DE’s occur naturally in physics, engineering, biology, economics and so
on.
1
Dennis G. Zill, Differential Equations
Neela Nataraj Lecture 1: D3

5. 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).
DE’s occur naturally in physics, engineering, biology, economics and so
on.
A few examples of physical systems which lead to DE 1
:
Radio active decay
Population dynamics
Newton’s law of cooling
Spread of a contagious disease
Chemical reactions
Falling bodies
Suspended cables
1
Dennis G. Zill, Differential Equations
Neela Nataraj Lecture 1: D3

6. Classification based on TYPE: ODE/PDE
Definition
Let y(x) denote a function in the variable x. An ordinary
differential equation (ODE) is an equation containing one or more
derivatives of an unknown function y.
Neela Nataraj Lecture 1: D3

7. Classification based on TYPE: ODE/PDE
Definition
Let y(x) denote a function in the variable x. An ordinary
differential equation (ODE) is an equation containing one or more
derivatives of an unknown function y.
In general, a differential equation involving one or more dependent
variables with respect to a single independent variable is called an
ODE.
Neela Nataraj Lecture 1: D3

8. Classification based on TYPE: ODE/PDE
Definition
Let y(x) denote a function in the variable x. An ordinary
differential equation (ODE) is an equation containing one or more
derivatives of an unknown function y.
In general, a differential equation involving one or more dependent
variables with respect to a single independent variable is called an
ODE.
Definition
A differential equation involving partial derivatives of one or more
dependent variables with respect to more than one independent
variable is called a partial differential equation (PDE).
Neela Nataraj Lecture 1: D3

9. Examples
dy
dx
+ 5y = ex
,
d2y
dx2
−
dy
dx
+ 6y = 0,
ODEs with one dependent variable
Neela Nataraj Lecture 1: D3

10. Examples
dy
dx
+ 5y = ex
,
d2y
dx2
−
dy
dx
+ 6y = 0,
ODEs with one dependent variable
dx
dt
+
dy
dt
= 2x + y.
an ODE with two dependent variables
Neela Nataraj Lecture 1: D3

11. Examples
dy
dx
+ 5y = ex
,
d2y
dx2
−
dy
dx
+ 6y = 0,
ODEs with one dependent variable
dx
dt
+
dy
dt
= 2x + y.
an ODE with two dependent variables
∂2u
∂x2
+
∂2u
∂y2
= 0.
PDE with two independent variables- Laplace equation
Neela Nataraj Lecture 1: D3

12. Examples
dy
dx
+ 5y = ex
,
d2y
dx2
−
dy
dx
+ 6y = 0,
ODEs with one dependent variable
dx
dt
+
dy
dt
= 2x + y.
an ODE with two dependent variables
∂2u
∂x2
+
∂2u
∂y2
= 0.
PDE with two independent variables- Laplace equation
∂2u
∂t2
− (
∂2u
∂x2
+
∂2u
∂y2
) = 0.
PDE with three independent variables- Wave equation
Neela Nataraj Lecture 1: D3

13. ODE- general form
Along with derivatives of y (the dependent variable) with respect
to x (the independent variable), note that, an an ODE may
contain
y itself (the 0th
derivative), and
known functions of x (including constants).
Neela Nataraj Lecture 1: D3

14. ODE- general form
Along with derivatives of y (the dependent variable) with respect
to x (the independent variable), note that, an an ODE may
contain
y itself (the 0th
derivative), and
known functions of x (including constants).
In other words, an ODE is a relation between the derivatives y, y
or
dy
dx
, . . . , y(n)
or
dny
dxn
Neela Nataraj Lecture 1: D3

15. ODE- general form
Along with derivatives of y (the dependent variable) with respect
to x (the independent variable), note that, an an ODE may
contain
y itself (the 0th
derivative), and
known functions of x (including constants).
In other words, an ODE is a relation between the derivatives y, y
or
dy
dx
, . . . , y(n)
or
dny
dxn
and functions of x:
F(x, y, y , . . . , y(n)
) = 0.
Neela Nataraj Lecture 1: D3

16. ODE- general form
Along with derivatives of y (the dependent variable) with respect
to x (the independent variable), note that, an an ODE may
contain
y itself (the 0th
derivative), and
known functions of x (including constants).
In other words, an ODE is a relation between the derivatives y, y
or
dy
dx
, . . . , y(n)
or
dny
dxn
and functions of x:
F(x, y, y , . . . , y(n)
) = 0.
Neela Nataraj Lecture 1: D3

17. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0
Neela Nataraj Lecture 1: D3

18. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE,
Neela Nataraj Lecture 1: D3

19. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
Neela Nataraj Lecture 1: D3

20. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t
Neela Nataraj Lecture 1: D3

21. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE,
Neela Nataraj Lecture 1: D3

22. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE, 4th order)
Neela Nataraj Lecture 1: D3

23. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE, 4th order)
3
∂v
∂t
+
∂v
∂s
= v
Neela Nataraj Lecture 1: D3

24. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE, 4th order)
3
∂v
∂t
+
∂v
∂s
= v (PDE,
Neela Nataraj Lecture 1: D3

25. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE, 4th order)
3
∂v
∂t
+
∂v
∂s
= v (PDE, 1st order)
Neela Nataraj Lecture 1: D3

26. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE, 4th order)
3
∂v
∂t
+
∂v
∂s
= v (PDE, 1st order)
4
∂2u
∂x2
+
∂2u
∂y2
+
∂2u
∂z2
= 0
Neela Nataraj Lecture 1: D3

27. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE, 4th order)
3
∂v
∂t
+
∂v
∂s
= v (PDE, 1st order)
4
∂2u
∂x2
+
∂2u
∂y2
+
∂2u
∂z2
= 0 (PDE,
Neela Nataraj Lecture 1: D3

28. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE, 4th order)
3
∂v
∂t
+
∂v
∂s
= v (PDE, 1st order)
4
∂2u
∂x2
+
∂2u
∂y2
+
∂2u
∂z2
= 0 (PDE, 2nd order)
Neela Nataraj Lecture 1: D3

29. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE, 4th order)
3
∂v
∂t
+
∂v
∂s
= v (PDE, 1st order)
4
∂2u
∂x2
+
∂2u
∂y2
+
∂2u
∂z2
= 0 (PDE, 2nd order)
5
dx
dt
= f (x, y),
dy
dt
= g(x, y), x = x(t), y = y(t).
Neela Nataraj Lecture 1: D3

30. Classification based on ORDER
Further classification according to the appearance of the highest
derivative appearing in the equation is done now.
Definition
The order of a differential equation is the order of the highest
derivative in the equation.
Examples :
1
d2y
dx2
+ xy
dy
dx
2
= 0 (ODE, 2nd order)
2
d4x
dt4
+ 5
d2x
dt2
+ 3x = sin t (ODE, 4th order)
3
∂v
∂t
+
∂v
∂s
= v (PDE, 1st order)
4
∂2u
∂x2
+
∂2u
∂y2
+
∂2u
∂z2
= 0 (PDE, 2nd order)
5
dx
dt
= f (x, y),
dy
dt
= g(x, y), x = x(t), y = y(t). (System of
ODEs, 1st order)
Neela Nataraj Lecture 1: D3

31. Linear equations
Linear equations
Neela Nataraj Lecture 1: D3

32. Linear equations
Linear equations - F(x, y, y , . . . , y(n)
) = 0 is linear if F is a linear
function of the variables y, y , . . . , y(n)
.
Neela Nataraj Lecture 1: D3

33. Linear equations
Linear equations - F(x, y, y , . . . , y(n)
) = 0 is linear if F is a linear
function of the variables y, y , . . . , y(n)
.
Thus, a linear ODE of order n is of the form
Neela Nataraj Lecture 1: D3

34. Linear equations
Linear equations - F(x, y, y , . . . , y(n)
) = 0 is linear if F is a linear
function of the variables y, y , . . . , y(n)
.
Thus, a linear ODE of order n is of the form
a0(x)y(n)
+ a1(x)y(n−1)
+ . . . + an(x)y = b(x)
Neela Nataraj Lecture 1: D3

35. Linear equations
Linear equations - F(x, y, y , . . . , y(n)
) = 0 is linear if F is a linear
function of the variables y, y , . . . , y(n)
.
Thus, 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) = 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.
Neela Nataraj Lecture 1: D3

36. Example : Radioactive decay
A radioactive substance decomposes at a rate proportional to the
amount present.
Neela Nataraj Lecture 1: D3

37. Example : Radioactive decay
A radioactive substance decomposes at a rate proportional to the
amount present. Let y(t) be the amount present at time t.
Neela Nataraj Lecture 1: D3

38. Example : 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
dt
= −k · y
Neela Nataraj Lecture 1: D3

39. Example : 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
dt
= −k · y
where k is a physical constant whose value is found by experiments
Neela Nataraj Lecture 1: D3

40. Example : 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
dt
= −k · y
where k is a physical constant whose value is found by experiments
(−k is called the decay constant).
Neela Nataraj Lecture 1: D3

41. Example : 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
dt
= −k · y
where k is a physical constant whose value is found by experiments
(−k is called the decay constant).
Linear ODE of first order.
Neela Nataraj Lecture 1: D3

42. Examples - The motion of an oscillating pendulum
Consider an oscillating pendulum of length L.
Neela Nataraj Lecture 1: D3

43. Examples - The motion of an oscillating pendulum
Consider an oscillating pendulum of length L. Let θ be the angle it
makes with the vertical direction.
Neela Nataraj Lecture 1: D3

44. Examples - The motion of an oscillating pendulum
Consider an oscillating pendulum of length L. Let θ be the angle it
makes with the vertical direction.
Neela Nataraj Lecture 1: D3

45. Examples - The motion of an oscillating pendulum
Consider an oscillating pendulum of length L. Let θ be the angle it
makes with the vertical direction.
d2θ
dt2
+
g
L
sin θ = 0.
Neela Nataraj Lecture 1: D3

46. Examples - The motion of an oscillating pendulum
Consider an oscillating pendulum of length L. Let θ be the angle it
makes with the vertical direction.
d2θ
dt2
+
g
L
sin θ = 0.
Neela Nataraj Lecture 1: D3

47. Examples - The motion of an oscillating pendulum
Consider an oscillating pendulum of length L. Let θ be the angle it
makes with the vertical direction.
d2θ
dt2
+
g
L
sin θ = 0.
ODE of
Neela Nataraj Lecture 1: D3

48. Examples - The motion of an oscillating pendulum
Consider an oscillating pendulum of length L. Let θ be the angle it
makes with the vertical direction.
d2θ
dt2
+
g
L
sin θ = 0.
ODE of second order.
Neela Nataraj Lecture 1: D3

49. Examples - The motion of an oscillating pendulum
Consider an oscillating pendulum of length L. Let θ be the angle it
makes with the vertical direction.
d2θ
dt2
+
g
L
sin θ = 0.
ODE of second order. not linear - Non-linear DE.
Neela Nataraj Lecture 1: D3

50. Example : A falling object
A body of mass m falls under the force of gravity.
Neela Nataraj Lecture 1: D3

51. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is
Neela Nataraj Lecture 1: D3

52. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
Neela Nataraj Lecture 1: D3

53. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant
Neela Nataraj Lecture 1: D3

54. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
Neela Nataraj Lecture 1: D3

55. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
An ODE of first order.
Neela Nataraj Lecture 1: D3

56. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
An ODE of first order. Linear or non-linear?
Neela Nataraj Lecture 1: D3

57. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
An ODE of first order. Linear or non-linear? (NL)
Neela Nataraj Lecture 1: D3

58. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
An ODE of first order. Linear or non-linear? (NL)
Examples :
1 y + 5y + 6y = 0
Neela Nataraj Lecture 1: D3

59. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
An ODE of first order. Linear or non-linear? (NL)
Examples :
1 y + 5y + 6y = 0 - 2nd order, linear
Neela Nataraj Lecture 1: D3

60. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
An ODE of first order. Linear or non-linear? (NL)
Examples :
1 y + 5y + 6y = 0 - 2nd order, linear
2 y(4)
+ x2
y(3)
+ x3
y = xex
Neela Nataraj Lecture 1: D3

61. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
An ODE of first order. Linear or non-linear? (NL)
Examples :
1 y + 5y + 6y = 0 - 2nd order, linear
2 y(4)
+ x2
y(3)
+ x3
y = xex
- 4th order, linear
Neela Nataraj Lecture 1: D3

62. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
An ODE of first order. Linear or non-linear? (NL)
Examples :
1 y + 5y + 6y = 0 - 2nd order, linear
2 y(4)
+ x2
y(3)
+ x3
y = xex
- 4th order, linear
3 y + 5(y )3
+ 6y = 0
Neela Nataraj Lecture 1: D3

63. Example : A falling object
A body of mass m falls under the force of gravity. The drag force
due to air resistance is c · v2
where v is the velocity and c is a
constant Then,
m
dv
dt
= mg − c · v2
.
An ODE of first order. Linear or non-linear? (NL)
Examples :
1 y + 5y + 6y = 0 - 2nd order, linear
2 y(4)
+ x2
y(3)
+ x3
y = xex
- 4th order, linear
3 y + 5(y )3
+ 6y = 0 - 2nd order, non-linear.
Neela Nataraj Lecture 1: D3

64. Can we solve it?
Given an equation, you would like to
Neela Nataraj Lecture 1: D3

65. Can we solve it?
Given an equation, you would like to solve it.
Neela Nataraj Lecture 1: D3

66. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Neela Nataraj Lecture 1: D3

67. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
Neela Nataraj Lecture 1: D3

68. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
Neela Nataraj Lecture 1: D3

69. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
2 Does an equation always have a solution?
Neela Nataraj Lecture 1: D3

70. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
2 Does an equation always have a solution? If so, how many?
Neela Nataraj Lecture 1: D3

71. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
2 Does an equation always have a solution? If so, how many?
3 Can the solutions be expressed in a nice form?
Neela Nataraj Lecture 1: D3

72. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
2 Does an equation always have a solution? If so, how many?
3 Can the solutions be expressed in a nice form? If not, how to
get a feel for it?
Neela Nataraj Lecture 1: D3

73. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
2 Does an equation always have a solution? If so, how many?
3 Can the solutions be expressed in a nice form? If not, how to
get a feel for it?
4 How much can we proceed in a systematic manner?
Neela Nataraj Lecture 1: D3

74. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
2 Does an equation always have a solution? If so, how many?
3 Can the solutions be expressed in a nice form? If not, how to
get a feel for it?
4 How much can we proceed in a systematic manner?
order
Neela Nataraj Lecture 1: D3

75. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
2 Does an equation always have a solution? If so, how many?
3 Can the solutions be expressed in a nice form? If not, how to
get a feel for it?
4 How much can we proceed in a systematic manner?
order - first, second, ..., nth
, ...
Neela Nataraj Lecture 1: D3

76. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
2 Does an equation always have a solution? If so, how many?
3 Can the solutions be expressed in a nice form? If not, how to
get a feel for it?
4 How much can we proceed in a systematic manner?
order - first, second, ..., nth
, ...
linear
Neela Nataraj Lecture 1: D3

77. Can we solve it?
Given an equation, you would like to solve it. At least, try to solve
it.
Questions:
1 What is a solution?
2 Does an equation always have a solution? If so, how many?
3 Can the solutions be expressed in a nice form? If not, how to
get a feel for it?
4 How much can we proceed in a systematic manner?
order - first, second, ..., nth
, ...
linear or non-linear?
Neela Nataraj Lecture 1: D3

78. What is a solution?
Consider F(x, y, y , . . . , y(n)
) = 0.
Neela Nataraj Lecture 1: D3

79. What is a solution?
Consider F(x, y, y , . . . , y(n)
) = 0. We assume that it is always possible
to solve a differential equation for the highest derivative, obtaining
y(n)
= f (x, y, y , · · · , y(n−1)
)
and study equations of this form.
Neela Nataraj Lecture 1: D3

80. What is a solution?
Consider F(x, y, y , . . . , y(n)
) = 0. We assume that it is always possible
to solve a differential equation for the highest derivative, obtaining
y(n)
= f (x, y, y , · · · , y(n−1)
)
and study equations of this form. This is to avoid the ambiguity which
may arise because a single equation F(x, y, y , . . . , y(n)
) = 0 may
correspond to several equations of the form y(n)
= f (x, y, y , · · · , y(n−1)
).
Neela Nataraj Lecture 1: D3

81. What is a solution?
Consider F(x, y, y , . . . , y(n)
) = 0. We assume that it is always possible
to solve a differential equation for the highest derivative, obtaining
y(n)
= f (x, y, y , · · · , y(n−1)
)
and study equations of this form. This is to avoid the ambiguity which
may arise because a single equation F(x, y, y , . . . , y(n)
) = 0 may
correspond to several equations of the form y(n)
= f (x, y, y , · · · , y(n−1)
).
For example, the equation y 2
+ xy + 4y = 0 leads to the two equations
y =
−x + x2 − 16y
2
or y =
−x − x2 − 16y
2
.
Neela Nataraj Lecture 1: D3

82. What is a solution?
Consider F(x, y, y , . . . , y(n)
) = 0. We assume that it is always possible
to solve a differential equation for the highest derivative, obtaining
y(n)
= f (x, y, y , · · · , y(n−1)
)
and study equations of this form. This is to avoid the ambiguity which
may arise because a single equation F(x, y, y , . . . , y(n)
) = 0 may
correspond to several equations of the form y(n)
= f (x, y, y , · · · , y(n−1)
).
For example, the equation y 2
+ xy + 4y = 0 leads to the two equations
y =
−x + x2 − 16y
2
or y =
−x − x2 − 16y
2
.
Definition
A explicit solution of the ODE y(n)
= f (x, y, y , · · · , y(n−1)
) on the
interval α < x < β is a function φ(x) such that φ , φ , · · · , φ(n)
exist and
satisfy
φ(n)
(x) = f (x, φ, φ , · · · , φ(n−1)
),
for every x in α < x < β.
Neela Nataraj Lecture 1: D3

83. Implicit solution & Formal solution
Definition
A relation g(x, y) = 0 is called an implicit solution of
y(n)
= f (x, y, y , · · · , y(n−1)
) if this relation defines at least one function
φ(x) on an interval α < x < β, such that, this function is an explicit
solution of y(n)
= f (x, y, y , · · · , y(n−1)
) in this interval.
Neela Nataraj Lecture 1: D3

84. Implicit solution & Formal solution
Definition
A relation g(x, y) = 0 is called an implicit solution of
y(n)
= f (x, y, y , · · · , y(n−1)
) if this relation defines at least one function
φ(x) on an interval α < x < β, such that, this function is an explicit
solution of y(n)
= f (x, y, y , · · · , y(n−1)
) in this interval.
Examples :
1 x2
+ y2
− 25 = 0 is an implicit solution of x + yy = 0 in
−5 < x < 5, because it defines two functions
φ1(x) = 25 − x2, φ2(x) = − 25 − x2
which are solutions of the DE in the given interval.
Neela Nataraj Lecture 1: D3

85. Implicit solution & Formal solution
Definition
A relation g(x, y) = 0 is called an implicit solution of
y(n)
= f (x, y, y , · · · , y(n−1)
) if this relation defines at least one function
φ(x) on an interval α < x < β, such that, this function is an explicit
solution of y(n)
= f (x, y, y , · · · , y(n−1)
) in this interval.
Examples :
1 x2
+ y2
− 25 = 0 is an implicit solution of x + yy = 0 in
−5 < x < 5, because it defines two functions
φ1(x) = 25 − x2, φ2(x) = − 25 − x2
which are solutions of the DE in the given interval. Verify!
2 Consider x2
+ y2
+ 25 = 0 =⇒ x + yy = 0 =⇒ y = −
x
y
.
Neela Nataraj Lecture 1: D3

86. Implicit solution & Formal solution
Definition
A relation g(x, y) = 0 is called an implicit solution of
y(n)
= f (x, y, y , · · · , y(n−1)
) if this relation defines at least one function
φ(x) on an interval α < x < β, such that, this function is an explicit
solution of y(n)
= f (x, y, y , · · · , y(n−1)
) in this interval.
Examples :
1 x2
+ y2
− 25 = 0 is an implicit solution of x + yy = 0 in
−5 < x < 5, because it defines two functions
φ1(x) = 25 − x2, φ2(x) = − 25 − x2
which are solutions of the DE in the given interval. Verify!
2 Consider x2
+ y2
+ 25 = 0 =⇒ x + yy = 0 =⇒ y = −
x
y
. We say
x2
+ y2
+ 25 = 0 formally satisfies x + yy = 0. But it is NOT an
implicit solution of DE as this relation doesn’t yield φ which is an
explicit solution of the DE on any real interval I.
Neela Nataraj Lecture 1: D3

87. First order ODE & Initial Value Problem for first order
ODE
We now consider first order ODE of the form F(x, y, y ) = 0 or
y = f (x, y) .
Neela Nataraj Lecture 1: D3

88. First order ODE & Initial Value Problem for first order
ODE
We now consider first order ODE of the form F(x, y, y ) = 0 or
y = f (x, y) .
Consider a linear first order ODE of the form y + a(x)y = b(x) .
Neela Nataraj Lecture 1: D3

89. First order ODE & Initial Value Problem for first order
ODE
We now consider first order ODE of the form F(x, y, y ) = 0 or
y = f (x, y) .
Consider a linear first order ODE of the form y + a(x)y = b(x) .
If b(x) = 0, then we say that the equation is homogeneous.
Neela Nataraj Lecture 1: D3

90. First order ODE & Initial Value Problem for first order
ODE
We now consider first order ODE of the form F(x, y, y ) = 0 or
y = f (x, y) .
Consider a linear first order ODE of the form y + a(x)y = b(x) .
If b(x) = 0, then we say that the equation is homogeneous.
MA106 nostalgia : Do the solutions of a homogeneous differential
equation form a vector space under usual addition and scalar
multiplication?
Neela Nataraj Lecture 1: D3

91. First order ODE & Initial Value Problem for first order
ODE
We now consider first order ODE of the form F(x, y, y ) = 0 or
y = f (x, y) .
Consider a linear first order ODE of the form y + a(x)y = b(x) .
If b(x) = 0, then we say that the equation is homogeneous.
MA106 nostalgia : Do the solutions of a homogeneous differential
equation form a vector space under usual addition and scalar
multiplication?
Definition
Initial value problem (IVP) : A DE along with an initial condition is
an IVP.
y = f (x, y), y(x0) = y0.
Neela Nataraj Lecture 1: D3

92. Examples
Given an amount of a radioactive substance, say 1 gm, find the
amount present at any later time.
Neela Nataraj Lecture 1: D3

93. Examples
Given an amount of a radioactive substance, say 1 gm, find the
amount present at any later time.
The relevant ODE is
dy
dt
= −k · y.
Neela Nataraj Lecture 1: D3

94. Examples
Given an amount of a radioactive substance, say 1 gm, find the
amount present at any later time.
The relevant ODE is
dy
dt
= −k · y.
Initial amount given is 1 gm at time t = 0.
Neela Nataraj Lecture 1: D3

95. Examples
Given an amount of a radioactive substance, say 1 gm, find the
amount present at any later time.
The relevant ODE is
dy
dt
= −k · y.
Initial amount given is 1 gm at time t = 0. i.e.,
y(0) = 1.
Neela Nataraj Lecture 1: D3

96. Examples
Given an amount of a radioactive substance, say 1 gm, find the
amount present at any later time.
The relevant ODE is
dy
dt
= −k · y.
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.
Neela Nataraj Lecture 1: D3

97. Examples
Given an amount of a radioactive substance, say 1 gm, find the
amount present at any later time.
The relevant ODE is
dy
dt
= −k · y.
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 =
Neela Nataraj Lecture 1: D3

98. Examples
Given an amount of a radioactive substance, say 1 gm, find the
amount present at any later time.
The relevant ODE is
dy
dt
= −k · y.
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.
Neela Nataraj Lecture 1: D3

99. Examples
Find the curve through the point (1, 1) in the xy-plane having at
each of its points, the slope −
y
x
.
Neela Nataraj Lecture 1: D3

100. Examples
Find the curve through the point (1, 1) in the xy-plane having at
each of its points, the slope −
y
x
.
The relevant ODE is
y = −
y
x
.
Neela Nataraj Lecture 1: D3

101. Examples
Find the curve through the point (1, 1) in the xy-plane having at
each of its points, the slope −
y
x
.
The relevant ODE is
y = −
y
x
.
By inspection,
y =
c
x
is its general solution for an arbitrary constant c; that is, a family
of hyperbolas.
Neela Nataraj Lecture 1: D3

102. Examples
Find the curve through the point (1, 1) in the xy-plane having at
each of its points, the slope −
y
x
.
The relevant ODE is
y = −
y
x
.
By inspection,
y =
c
x
is its general solution for an arbitrary constant c; that is, a family
of hyperbolas.
The initial condition given is
y(1) = 1,
which implies c = 1.
Neela Nataraj Lecture 1: D3

103. Examples
Find the curve through the point (1, 1) in the xy-plane having at
each of its points, the slope −
y
x
.
The relevant ODE is
y = −
y
x
.
By inspection,
y =
c
x
is its general solution for an arbitrary constant c; that is, 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
y =
1
x
.
Neela Nataraj Lecture 1: D3

104. Examples
A first order IVP can have
1 NO solution :
Neela Nataraj Lecture 1: D3

105. Examples
A first order IVP can have
1 NO solution : |y | + |y| = 0, y(0) = 3.
Neela Nataraj Lecture 1: D3

106. Examples
A first order IVP can have
1 NO solution : |y | + |y| = 0, y(0) = 3.
2 Precisely one solution :
Neela Nataraj Lecture 1: D3

107. Examples
A first order IVP can have
1 NO solution : |y | + |y| = 0, y(0) = 3.
2 Precisely one solution : y = x, y(0) = 1.
Neela Nataraj Lecture 1: D3

108. Examples
A first order IVP can have
1 NO solution : |y | + |y| = 0, y(0) = 3.
2 Precisely one solution : y = x, y(0) = 1. What is the
solution?
3 Infinitely many solutions:
Neela Nataraj Lecture 1: D3

109. Examples
A first order IVP can have
1 NO solution : |y | + |y| = 0, y(0) = 3.
2 Precisely one solution : y = x, y(0) = 1. What is the
solution?
3 Infinitely many solutions: xy = y − 1, y(0) = 1
Neela Nataraj Lecture 1: D3

110. Examples
A first order IVP can have
1 NO solution : |y | + |y| = 0, y(0) = 3.
2 Precisely one solution : y = x, y(0) = 1. What is the
solution?
3 Infinitely many solutions: xy = y − 1, y(0) = 1 The solutions
are y = 1 + cx.
Neela Nataraj Lecture 1: D3

111. Examples
A first order IVP can have
1 NO solution : |y | + |y| = 0, y(0) = 3.
2 Precisely one solution : y = x, y(0) = 1. What is the
solution?
3 Infinitely many solutions: xy = y − 1, y(0) = 1 The solutions
are y = 1 + cx.
Motivation to study conditions under which the solution would
exist and the conditions under which it will be unique!
Neela Nataraj Lecture 1: D3

112. Examples
A first order IVP can have
1 NO solution : |y | + |y| = 0, y(0) = 3.
2 Precisely one solution : y = x, y(0) = 1. What is the
solution?
3 Infinitely many solutions: xy = y − 1, y(0) = 1 The solutions
are y = 1 + cx.
Motivation to study conditions under which the solution would
exist and the conditions under which it will be unique!
We first start with a few methods for finding out the solution of
first order ODEs, discuss the geometric meaning of solutions and
then proceed to study existence-uniqueness results.
Neela Nataraj Lecture 1: D3

113. Separable ODE’s
An ODE of the form
M(x) + N(y)y = 0
is called a separable ODE.
Neela Nataraj Lecture 1: D3

114. Separable ODE’s
An ODE of the form
M(x) + N(y)y = 0
is called a separable ODE.
Let H1(x) and H2(y) be any functions such that H1(x) = M(x)
and H2(y) = N(y).
Neela Nataraj Lecture 1: D3

115. Separable ODE’s
An ODE of the form
M(x) + N(y)y = 0
is called a separable ODE.
Let H1(x) and H2(y) be any functions such that H1(x) = M(x)
and H2(y) = N(y).
Substituting in the DE, we obtain
H1(x) + H2(y)y = 0.
Neela Nataraj Lecture 1: D3

116. Separable ODE’s
An ODE of the form
M(x) + N(y)y = 0
is called a separable ODE.
Let H1(x) and H2(y) be any functions such that H1(x) = M(x)
and H2(y) = N(y).
Substituting in the DE, we obtain
H1(x) + H2(y)y = 0.
Using chain rule,
d
dx
H2(y) = H2(y)
dy
dx
.
Neela Nataraj Lecture 1: D3

117. Separable ODE’s
An ODE of the form
M(x) + N(y)y = 0
is called a separable ODE.
Let H1(x) and H2(y) be any functions such that H1(x) = M(x)
and H2(y) = N(y).
Substituting in the DE, we obtain
H1(x) + H2(y)y = 0.
Using chain rule,
d
dx
H2(y) = H2(y)
dy
dx
.
Hence,
d
dx
(H1(x) + H2(y)) = 0.
Neela Nataraj Lecture 1: D3

118. Separable ODE’s
An ODE of the form
M(x) + N(y)y = 0
is called a separable ODE.
Let H1(x) and H2(y) be any functions such that H1(x) = M(x)
and H2(y) = N(y).
Substituting in the DE, we obtain
H1(x) + H2(y)y = 0.
Using chain rule,
d
dx
H2(y) = H2(y)
dy
dx
.
Hence,
d
dx
(H1(x) + H2(y)) = 0.
Integrating, H1(x) + H2(y) = c, where c is an arbirtary
constant.
Note:
Neela Nataraj Lecture 1: D3

119. Separable ODE’s
An ODE of the form
M(x) + N(y)y = 0
is called a separable ODE.
Let H1(x) and H2(y) be any functions such that H1(x) = M(x)
and H2(y) = N(y).
Substituting in the DE, we obtain
H1(x) + H2(y)y = 0.
Using chain rule,
d
dx
H2(y) = H2(y)
dy
dx
.
Hence,
d
dx
(H1(x) + H2(y)) = 0.
Integrating, H1(x) + H2(y) = c, where c is an arbirtary
constant.
Note: This method many times gives us an implicit solution and
not necessarily an explicit one!
Neela Nataraj Lecture 1: D3

120. Separable ODE - Example 1
Solve the DE :
y = −2xy.
Neela Nataraj Lecture 1: D3

121. Separable ODE - Example 1
Solve the DE :
y = −2xy.
Separating the variables, we get :
Neela Nataraj Lecture 1: D3

122. Separable ODE - Example 1
Solve the DE :
y = −2xy.
Separating the variables, we get :
dy
y
= −2xdx.
Neela Nataraj Lecture 1: D3

123. Separable ODE - Example 1
Solve the DE :
y = −2xy.
Separating the variables, we get :
dy
y
= −2xdx.
Integrating both sides, we obtain :
Neela Nataraj Lecture 1: D3

124. Separable ODE - Example 1
Solve the DE :
y = −2xy.
Separating the variables, we get :
dy
y
= −2xdx.
Integrating both sides, we obtain :
ln |y| = −x2
+ c1.
Neela Nataraj Lecture 1: D3

125. Separable ODE - Example 1
Solve the DE :
y = −2xy.
Separating the variables, we get :
dy
y
= −2xdx.
Integrating both sides, we obtain :
ln |y| = −x2
+ c1.
Thus, the solutions are
Neela Nataraj Lecture 1: D3

126. Separable ODE - Example 1
Solve the DE :
y = −2xy.
Separating the variables, we get :
dy
y
= −2xdx.
Integrating both sides, we obtain :
ln |y| = −x2
+ c1.
Thus, the solutions are
y = ce−x2
.
Neela Nataraj Lecture 1: D3

127. Separable ODE - Example 1
Solve the DE :
y = −2xy.
Separating the variables, we get :
dy
y
= −2xdx.
Integrating both sides, we obtain :
ln |y| = −x2
+ c1.
Thus, the solutions are
y = ce−x2
.
How do they look?
Neela Nataraj Lecture 1: D3

128. If we are given an initial condition
y(x0) = y0,
Neela Nataraj Lecture 1: D3

129. If we are given an initial condition
y(x0) = y0,
then we get:
c = y0ex2
0
Neela Nataraj Lecture 1: D3

130. If we are given an initial condition
y(x0) = y0,
then we get:
c = y0ex2
0
and y = y0ex2
0 −x2
.
Neela Nataraj Lecture 1: D3

131. Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx
=
y cos x
1 + 2y2
; y(0) = 1.
Neela Nataraj Lecture 1: D3

132. Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx
=
y cos x
1 + 2y2
; y(0) = 1.
Assume y = 0.
Neela Nataraj Lecture 1: D3

133. Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx
=
y cos x
1 + 2y2
; y(0) = 1.
Assume y = 0. Then,
1 + 2y2
y
dy = cos x dx.
Neela Nataraj Lecture 1: D3

134. Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx
=
y cos x
1 + 2y2
; y(0) = 1.
Assume y = 0. Then,
1 + 2y2
y
dy = cos x dx.
Integrating,
Neela Nataraj Lecture 1: D3

135. Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx
=
y cos x
1 + 2y2
; y(0) = 1.
Assume y = 0. Then,
1 + 2y2
y
dy = cos x dx.
Integrating,
ln |y| + y2
= sin x + c.
Neela Nataraj Lecture 1: D3

136. Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx
=
y cos x
1 + 2y2
; y(0) = 1.
Assume y = 0. Then,
1 + 2y2
y
dy = cos x dx.
Integrating,
ln |y| + y2
= sin x + c.
As y(0) = 1, we get c = 1.
Neela Nataraj Lecture 1: D3

137. Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx
=
y cos x
1 + 2y2
; y(0) = 1.
Assume y = 0. Then,
1 + 2y2
y
dy = cos x dx.
Integrating,
ln |y| + y2
= sin x + c.
As y(0) = 1, we get c = 1. Hence a particular solution to the IVP
is
ln |y| + y2
= sin x + 1.
Neela Nataraj Lecture 1: D3

138. Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx
=
y cos x
1 + 2y2
; y(0) = 1.
Assume y = 0. Then,
1 + 2y2
y
dy = cos x dx.
Integrating,
ln |y| + y2
= sin x + c.
As y(0) = 1, we get c = 1. Hence a particular solution to the IVP
is
ln |y| + y2
= sin x + 1.
Note: y ≡ 0 is a solution to the DE
Neela Nataraj Lecture 1: D3

139. Separable ODE - Example 2
Find the solution to the initial value problem:
dy
dx
=
y cos x
1 + 2y2
; y(0) = 1.
Assume y = 0. Then,
1 + 2y2
y
dy = cos x dx.
Integrating,
ln |y| + y2
= sin x + c.
As y(0) = 1, we get c = 1. Hence a particular solution to the IVP
is
ln |y| + y2
= sin x + 1.
Note: y ≡ 0 is a solution to the DE but it is not a solution to the
given IVP.
Neela Nataraj Lecture 1: D3

140. Separable ODE - Example 3
Escape velocity.
Neela Nataraj Lecture 1: D3

141. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth.
Neela Nataraj Lecture 1: D3

142. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity.
Neela Nataraj Lecture 1: D3

143. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Neela Nataraj Lecture 1: D3

144. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Its weight at height x
Neela Nataraj Lecture 1: D3

145. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Its weight at height x (from the surface of the earth)
Neela Nataraj Lecture 1: D3

146. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Its weight at height x (from the surface of the earth) is given by,
w(x)
Neela Nataraj Lecture 1: D3

147. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Its weight at height x (from the surface of the earth) is given by,
w(x) = −
mgR2
(R + x)2
,
Neela Nataraj Lecture 1: D3

148. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Its weight at height x (from the surface of the earth) is given by,
w(x) = −
mgR2
(R + x)2
,
where R is the radius of the earth.
Neela Nataraj Lecture 1: D3

149. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Its weight at height x (from the surface of the earth) is given by,
w(x) = −
mgR2
(R + x)2
,
where R is the radius of the earth.
Neglect force due to air resistance and other celestial bodies.
Neela Nataraj Lecture 1: D3

150. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Its weight at height x (from the surface of the earth) is given by,
w(x) = −
mgR2
(R + x)2
,
where R is the radius of the earth.
Neglect force due to air resistance and other celestial bodies.
Therefore, the equation of motion is
Neela Nataraj Lecture 1: D3

151. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Its weight at height x (from the surface of the earth) is given by,
w(x) = −
mgR2
(R + x)2
,
where R is the radius of the earth.
Neglect force due to air resistance and other celestial bodies.
Therefore, the equation of motion is
m
dv
dt
= −
mgR2
(R + x)2
;
Neela Nataraj Lecture 1: D3

152. Separable ODE - Example 3
Escape velocity.
A projectile of mass m moves in a direction perpendicular to the
surface of the earth. Suppose v0 is its initial velocity. We want to
calculate the height the projectile reaches.
Its weight at height x (from the surface of the earth) is given by,
w(x) = −
mgR2
(R + x)2
,
where R is the radius of the earth.
Neglect force due to air resistance and other celestial bodies.
Therefore, the equation of motion is
m
dv
dt
= −
mgR2
(R + x)2
; v(0) = v0.
Neela Nataraj Lecture 1: D3

153. Separable ODE’s
By chain rule,
dv
dt
Neela Nataraj Lecture 1: D3

154. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
Neela Nataraj Lecture 1: D3

155. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Neela Nataraj Lecture 1: D3

156. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
Neela Nataraj Lecture 1: D3

157. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
Neela Nataraj Lecture 1: D3

158. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
This ODE is
Neela Nataraj Lecture 1: D3

159. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
This ODE is separable.
Neela Nataraj Lecture 1: D3

160. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
This ODE is separable. Linear or non-linear?
Neela Nataraj Lecture 1: D3

161. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
This ODE is separable. Linear or non-linear? (NL)
Neela Nataraj Lecture 1: D3

162. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
This ODE is separable. Linear or non-linear? (NL)
Separating the variables and integrating, we get:
Neela Nataraj Lecture 1: D3

163. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
This ODE is separable. Linear or non-linear? (NL)
Separating the variables and integrating, we get:
v2
2
=
gR2
R + x
+ c.
Neela Nataraj Lecture 1: D3

164. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
This ODE is separable. Linear or non-linear? (NL)
Separating the variables and integrating, we get:
v2
2
=
gR2
R + x
+ c.
For x = 0, we get
v2
0
2
= gR + c,
Neela Nataraj Lecture 1: D3

165. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
This ODE is separable. Linear or non-linear? (NL)
Separating the variables and integrating, we get:
v2
2
=
gR2
R + x
+ c.
For x = 0, we get
v2
0
2
= gR + c, hence, c =
v2
0
2
− gR,
Neela Nataraj Lecture 1: D3

166. Separable ODE’s
By chain rule,
dv
dt
=
dv
dx
·
dx
dt
= v ·
dv
dx
.
Thus,
v ·
dv
dx
= −
gR2
(R + x)2
.
This ODE is separable. Linear or non-linear? (NL)
Separating the variables and integrating, we get:
v2
2
=
gR2
R + x
+ c.
For x = 0, we get
v2
0
2
= gR + c, hence, c =
v2
0
2
− gR, and,
v = ± v2
0 − 2gR +
2gR2
R + x
.
Neela Nataraj Lecture 1: D3

167. Separable ODE’s
Suppose the body reaches the maximum height H.
Neela Nataraj Lecture 1: D3

168. Separable ODE’s
Suppose the body reaches the maximum height H. Then v =
Neela Nataraj Lecture 1: D3

169. Separable ODE’s
Suppose the body reaches the maximum height H. Then v = 0 at
this height.
Neela Nataraj Lecture 1: D3

170. Separable ODE’s
Suppose the body reaches the maximum height H. Then v = 0 at
this height.
v2
0 − 2gR +
2gR2
(R + H)
= 0.
Neela Nataraj Lecture 1: D3

171. Separable ODE’s
Suppose the body reaches the maximum height H. Then v = 0 at
this height.
v2
0 − 2gR +
2gR2
(R + H)
= 0.
Thus,
v2
0 = 2gR −
2gR2
R + H
= 2gR
H
R + H
.
Neela Nataraj Lecture 1: D3

172. Separable ODE’s
Suppose the body reaches the maximum height H. Then v = 0 at
this height.
v2
0 − 2gR +
2gR2
(R + H)
= 0.
Thus,
v2
0 = 2gR −
2gR2
R + H
= 2gR
H
R + H
.
The escape velocity is found by taking limit as
Neela Nataraj Lecture 1: D3

173. Separable ODE’s
Suppose the body reaches the maximum height H. Then v = 0 at
this height.
v2
0 − 2gR +
2gR2
(R + H)
= 0.
Thus,
v2
0 = 2gR −
2gR2
R + H
= 2gR
H
R + H
.
The escape velocity is found by taking limit as H → ∞.
Neela Nataraj Lecture 1: D3

174. Separable ODE’s
Suppose the body reaches the maximum height H. Then v = 0 at
this height.
v2
0 − 2gR +
2gR2
(R + H)
= 0.
Thus,
v2
0 = 2gR −
2gR2
R + H
= 2gR
H
R + H
.
The escape velocity is found by taking limit as H → ∞. Thus,
ve = 2gR
Neela Nataraj Lecture 1: D3

175. Separable ODE’s
Suppose the body reaches the maximum height H. Then v = 0 at
this height.
v2
0 − 2gR +
2gR2
(R + H)
= 0.
Thus,
v2
0 = 2gR −
2gR2
R + H
= 2gR
H
R + H
.
The escape velocity is found by taking limit as H → ∞. Thus,
ve = 2gR ∼ 11 km/sec.
Neela Nataraj Lecture 1: D3