Calculus 3 5: power series

Infinite Series and Differential Equations
Nguyen Thieu Huy
Hanoi University of Science and Technology

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Differential Equations: Motivations
Falling Body Problems
Vertically falling body of mass m with unknown velocity v :

Two forces: 1) Gravitation mg ; 2) Air resistance: −kv , where k: positive
constant.

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Differential Equations: Motivations
Falling Body Problems
Vertically falling body of mass m with unknown velocity v :

Two forces: 1) Gravitation mg ; 2) Air resistance: −kv , where k: positive
constant.

dv
v
Newton’s law: Ftotal = ma ⇔ mg − kv = m dv
dt ⇔ dt + k m = g
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Electrical Circuits
RL-circuit consisting of a resistance R, inductor L, electromotive force E .
Let find the current I :

dI
Kirchoff’s Voltage Law: VR + VL = E . Also, VR = RI ; VL = L dt
.

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Electrical Circuits
RL-circuit consisting of a resistance R, inductor L, electromotive force E .
Let find the current I :

dI
Kirchoff’s Voltage Law: VR + VL = E . Also, VR = RI ; VL = L dt
.
dI
dI
RI + L = E ⇔
+ RL I = E
dt
dt

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Electrical Circuits
RL-circuit consisting of a resistance R, inductor L, electromotive force E .
Let find the current I :

dI
Kirchoff’s Voltage Law: VR + VL = E . Also, VR = RI ; VL = L dt
.
dI
dI
RI + L = E ⇔
+ RL I = E
dt
dt

2.1 Definition
A differential equation is an equation involving an unknown function and
its derivatives.
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Basic Notions
Unknown function y :
• A differential equation is an
ordinary differential equation
if the unknown function depends on
only one independent variable.

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Basic Notions
Unknown function y :
• A differential equation is an
ordinary differential equation
if the unknown function depends on

only one independent variable.
• If the unknown function depends
on two or more ind. variables,
then the differential equation is a
partial differential equation.

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Basic Notions
Unknown function y :
• A differential equation is an
ordinary differential equation
if the unknown function depends on
only one independent variable.
• If the unknown function depends
on two or more ind. variables,
then the differential equation is a
partial differential equation.
• The order of a differential equation
is the order of the highest derivative
appearing in the equation.

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Basic Notions
Unknown function y :
• A differential equation is an
ordinary differential equation
if the unknown function depends on
only one independent variable.
• If the unknown function depends
on two or more ind. variables,
then the differential equation is a
partial differential equation.
• The order of a differential equation
is the order of the highest derivative
appearing in the equation.
Abbreviation:
DE: Differential Equation
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2.2 Definition: Solutions
A solution of a DE in the unknown function y and the ind. variable x on
the interval J ⊂ R is a function y (x) that satisfies the DE identically for
all x ∈ J. To solve a DE is to find all its solutions.

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2.2 Definition: Solutions
A solution of a DE in the unknown function y and the ind. variable x on
the interval J ⊂ R is a function y (x) that satisfies the DE identically for
all x ∈ J. To solve a DE is to find all its solutions.
Example: The function y (x) = 2 sin 2x + 3 cos 2x is a solution to
y ” + 4y = 0 in (−∞, ∞).

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2.2 Definition: Solutions
A solution of a DE in the unknown function y and the ind. variable x on
the interval J ⊂ R is a function y (x) that satisfies the DE identically for
all x ∈ J. To solve a DE is to find all its solutions.
Example: The function y (x) = 2 sin 2x + 3 cos 2x is a solution to
y ” + 4y = 0 in (−∞, ∞).
2.3 Definition: Particular and general solutions.
A particular solution of a DE is any one solution. The general solution of a

DE is the set of all its solutions.

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2.2 Definition: Solutions
A solution of a DE in the unknown function y and the ind. variable x on
the interval J ⊂ R is a function y (x) that satisfies the DE identically for
all x ∈ J. To solve a DE is to find all its solutions.
Example: The function y (x) = 2 sin 2x + 3 cos 2x is a solution to
y ” + 4y = 0 in (−∞, ∞).
2.3 Definition: Particular and general solutions.
A particular solution of a DE is any one solution. The general solution of a
DE is the set of all its solutions.
2.4 Initial-Value and Boundary-Value Problems.
1 A DE with subsidiary conditions on the unknown function and its
derivatives, all given at the same value of the independent variable,
constitutes an initial-value problem.

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2.2 Definition: Solutions
A solution of a DE in the unknown function y and the ind. variable x on
the interval J ⊂ R is a function y (x) that satisfies the DE identically for
all x ∈ J. To solve a DE is to find all its solutions.
Example: The function y (x) = 2 sin 2x + 3 cos 2x is a solution to
y ” + 4y = 0 in (−∞, ∞).
2.3 Definition: Particular and general solutions.
A particular solution of a DE is any one solution. The general solution of a
DE is the set of all its solutions.
2.4 Initial-Value and Boundary-Value Problems.
1 A DE with subsidiary conditions on the unknown function and its
derivatives, all given at the same value of the independent variable,
constitutes an initial-value problem. The sub. conditions are initial
conditions.

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2.2 Definition: Solutions
A solution of a DE in the unknown function y and the ind. variable x on
the interval J ⊂ R is a function y (x) that satisfies the DE identically for
all x ∈ J. To solve a DE is to find all its solutions.
Example: The function y (x) = 2 sin 2x + 3 cos 2x is a solution to
y ” + 4y = 0 in (−∞, ∞).
2.3 Definition: Particular and general solutions.
A particular solution of a DE is any one solution. The general solution of a

DE is the set of all its solutions.
2.4 Initial-Value and Boundary-Value Problems.
1 A DE with subsidiary conditions on the unknown function and its
derivatives, all given at the same value of the independent variable,
constitutes an initial-value problem. The sub. conditions are initial
conditions.
2 If the sub. conditions are given at more than one value of the
independent variable, the problem is a boundary-value problem and
the sub. conditions are boundary conditions.
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Example:
1

The problem y ” + 2y = x; y (π) = 1, y (π) = 2 is an initial value
problem, because the two subsidiary conditions are both given at
x = π.

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Example:
1

The problem y ” + 2y = x; y (π) = 1, y (π) = 2 is an initial value
problem, because the two subsidiary conditions are both given at
x = π.

2

The problem y ” + 2y = x; y (0) = 1, y (1) = 1 is a boundary-value
problem, because the two subsidiary conditions are given at x = 0 and
x = 1.

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Example:
1

The problem y ” + 2y = x; y (π) = 1, y (π) = 2 is an initial value
problem, because the two subsidiary conditions are both given at
x = π.

2

The problem y ” + 2y = x; y (0) = 1, y (1) = 1 is a boundary-value

problem, because the two subsidiary conditions are given at x = 0 and
x = 1.

2.5 Standard and Differential Forms
1 Standard form for a first-order DE in the unknown function y (x) is
y = f (x, y )

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(1)

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Example:
1

The problem y ” + 2y = x; y (π) = 1, y (π) = 2 is an initial value
problem, because the two subsidiary conditions are both given at
x = π.

2

The problem y ” + 2y = x; y (0) = 1, y (1) = 1 is a boundary-value
problem, because the two subsidiary conditions are given at x = 0 and
x = 1.

2.5 Standard and Differential Forms

1 Standard form for a first-order DE in the unknown function y (x) is
y = f (x, y )

2

(1)

The right side of (1) can always be written as a quotient of two other
M(x,y )
functions −M(x, y ) and N(x, y ). Then (1) becomes dy
dx = − N(x,y )
⇔ Differential form M(x, y )dx + N(x, y )dy = 0

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(2)
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FIRST-ORDER DIFFERENTIAL EQUATIONS
1. Separable Equations
1.1 Definition
The first-order separable differential equation has the form
A(x)dx + B(y )dy = 0

(3)

1.2 Solutions: The solution to (3) is

A(x)dx +

B(y )dy = c

(4)

where c represents an arbitrary constant.

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FIRST-ORDER DIFFERENTIAL EQUATIONS
1. Separable Equations
1.1 Definition
The first-order separable differential equation has the form
A(x)dx + B(y )dy = 0

(3)

1.2 Solutions: The solution to (3) is
A(x)dx +

B(y )dy = c

(4)

where c represents an arbitrary constant.
x 2 +2
Example. Solve the equation: dy
dx = y .
Rewrite in the differential form (x 2 + 2)dx − ydy = 0 which is separable
with A(x) = x 2 + 2 and B(y ) = −y . Its solution is
(x 2 + 2)dx − ydy = c or 31 x 3 + 2x − 12 y 2 = c for any constant c.

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FIRST-ORDER DIFFERENTIAL EQUATIONS
1. Separable Equations
1.1 Definition
The first-order separable differential equation has the form
A(x)dx + B(y )dy = 0

(3)

1.2 Solutions: The solution to (3) is
A(x)dx +

B(y )dy = c

(4)

where c represents an arbitrary constant.
x 2 +2
Example. Solve the equation: dy
dx = y .
Rewrite in the differential form (x 2 + 2)dx − ydy = 0 which is separable
with A(x) = x 2 + 2 and B(y ) = −y . Its solution is
(x 2 + 2)dx − ydy = c or 31 x 3 + 2x − 12 y 2 = c for any constant c.
Sometimes, it may not be algebraically possible to solve for y explicitly in
terms of x. In that case, the solution is left in implicit form.
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1.3 Solutions to the Initial-Value Problem:
The solution to the initial-value problem A(x)dx + B(y )dy = 0; y (x0 ) = y0
can be obtained first by formula (4) and then substitute y (x0 ) = y0 to find
x
y
constant c, or another way is using x0 A(t)dt + y0 B(s)ds = 0.

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1.3 Solutions to the Initial-Value Problem:
The solution to the initial-value problem A(x)dx + B(y )dy = 0; y (x0 ) = y0
can be obtained first by formula (4) and then substitute y (x0 ) = y0 to find
x
y
constant c, or another way is using x0 A(t)dt + y0 B(s)ds = 0.
2. Homogeneous Equations:
2.1 Definition:
A DE in standard form

dy
= f (x, y )
dx
is homogeneous if f (tx, ty ) = f (x, y ) for every real number t = 0.

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(5)

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