Calculus 3 5: power series

Basic NotionsUnknown function y : •A differential equation is an ordinary differential equation if the unknown function depends ononly one independent variable.. Basic NotionsUnknown fun

Infinite Series and Differential Equations

Nguyen Thieu Huy

Hanoi University of Science and Technology

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

Newton’s law: ~Ftotal = m~a ⇔ mg − kv = mdvdt ⇔ dv

dt + kmv = g

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: positiveconstant

Electrical Circuits

RL-circuit consisting of a resistance R, inductor L, electromotive force E

Let find the current I :

Kirchoff’s Voltage Law: VR + VL= E Also, VR = RI ; VL= LdIdt

Electrical Circuits

RL-circuit consisting of a resistance R, inductor L, electromotive force E

Let find the current I :

Kirchoff’s Voltage Law: VR + VL= E Also, VR = RI ; VL= LdIdt

Basic Notions

Unknown function y :

•A differential equation is an

ordinary differential equation

if the unknown function depends ononly one independent variable

• If the unknown function depends

on two or more ind variables,then the differential equation is a

partial differential equation

• Theorderof a differential equation

is the order of thehighest derivative

appearing in the equation

Abbreviation:

DE: Differential Equation

Basic Notions

Unknown function y :

•A differential equation is an

ordinary differential equation

if the unknown function depends ononly one independent variable

•If the unknown function depends

on two or more ind variables,then the differential equation is a

partial differential equation

• Theorderof a differential equation

is the order of thehighest derivative

appearing in the equation

Abbreviation:

DE: Differential Equation

Basic Notions

Unknown function y :

•A differential equation is an

ordinary differential equation

if the unknown function depends ononly one independent variable

•If the unknown function depends

on two or more ind variables,then the differential equation is a

partial differential equation

•The orderof a differential equation

is the order of thehighest derivative

appearing in the equation

Abbreviation:

DE: Differential Equation

Basic Notions

Unknown function y :

•A differential equation is an

ordinary differential equation

if the unknown function depends ononly one independent variable

•If the unknown function depends

on two or more ind variables,then the differential equation is a

partial differential equation

•The orderof a differential equation

is the order of thehighest derivative

appearing in the equation

Abbreviation:

DE: Differential Equation

2.2 Definition: Solutions

A solutionof 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 solvea 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 withsubsidiary conditions on the unknown function and itsderivatives, all given at the same value of the independent variable,constitutes aninitial-value problem

The sub conditions are initialconditions

2 If the sub conditions are given at more than one value of theindependent variable, the problem is aboundary-value problem andthe sub conditions are boundary conditions

2.2 Definition: Solutions

A solutionof 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 solvea 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 withsubsidiary conditions on the unknown function and itsderivatives, all given at the same value of the independent variable,constitutes aninitial-value problem

The sub conditions are initialconditions

2 If the sub conditions are given at more than one value of theindependent variable, the problem is aboundary-value problem andthe sub conditions are boundary conditions

2.2 Definition: Solutions

A solutionof 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 solvea 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 withsubsidiary conditions on the unknown function and itsderivatives, all given at the same value of the independent variable,constitutes aninitial-value problem

The sub conditions are initialconditions

2 If the sub conditions are given at more than one value of theindependent variable, the problem is aboundary-value problem andthe sub conditions are boundary conditions

2.2 Definition: Solutions

A solutionof 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 solvea 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 withsubsidiary conditions on the unknown function and its

derivatives, all given at the same value of the independent variable,

constitutes aninitial-value problem

The sub conditions are initialconditions

2 If the sub conditions are given at more than one value of theindependent variable, the problem is aboundary-value problem andthe sub conditions are boundary conditions

2.2 Definition: Solutions

A solutionof 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 solvea 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 withsubsidiary conditions on the unknown function and its

derivatives, all given at the same value of the independent variable,

constitutes aninitial-value problem The sub conditions are initial

conditions

If the sub conditions are given at more than one value of theindependent variable, the problem is aboundary-value problem andthe sub conditions are boundary conditions

2.2 Definition: Solutions

A solutionof a DE in the unknown function y and the ind variable x onthe interval J ⊂ R is a function y (x) that satisfies the DE identically forall x ∈ J To solvea 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 withsubsidiary conditions on the unknown function and itsderivatives, all given at the same value of the independent variable,constitutes aninitial-value problem The sub conditions are initialconditions

2 If the sub conditions are given at more than one value of the

independent variable, the problem is aboundary-value problem andthe sub conditions are boundary conditions

1 The problem y ” + 2y = x ; y (π) = 1, y0(π) = 2 is an initial value

problem, because the two subsidiary conditions are both given at

x = π

2 The problem y ” + 2y0 = x ; y (0) = 1, y (1) = 1 is a boundary-valueproblem, 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

1 The problem y ” + 2y = x ; y (π) = 1, y0(π) = 2 is an initial value

problem, because the two subsidiary conditions are both given at

x = π

2 The problem y ” + 2y0 = 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

1 The problem y ” + 2y = x ; y (π) = 1, y0(π) = 2 is an initial value

problem, because the two subsidiary conditions are both given at

x = π

2 The problem y ” + 2y0 = 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

2.5 Standard and Differential Forms

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

FIRST-ORDER DIFFERENTIAL EQUATIONS

1 Separable Equations

1.1 Definition

The first-order separable differential equation has the form

1.2 Solutions: The solution to (3) is

ZA(x )dx +

Z

where c represents an arbitrary constant

Example Solve the equation: dydx = xy+2.Rewrite in the differential form (x2+ 2)dx − ydy = 0 which is separablewith A(x ) = x2+ 2 and B(y ) = −y Its solution is

R (x2+ 2)dx −R ydy = c or 1

3x3+ 2x − 12y2 = c for any constant c.Sometimes, it may not be algebraically possible to solve for y explicitly interms of x In that case, the solution is left in implicit form

FIRST-ORDER DIFFERENTIAL EQUATIONS

1 Separable Equations

1.1 Definition

The first-order separable differential equation has the form

1.2 Solutions: The solution to (3) is

ZA(x )dx +

Z

where c represents an arbitrary constant

Example Solve the equation: dydx = x2y+2

Rewrite in the differential form (x2+ 2)dx − ydy = 0 which is separable

with A(x ) = x2+ 2 and B(y ) = −y Its solution is

R (x2+ 2)dx −R ydy = c or 1

3x3+ 2x − 12y2 = c for any constant c

Sometimes, it may not be algebraically possible to solve for y explicitly interms of x In that case, the solution is left in implicit form

FIRST-ORDER DIFFERENTIAL EQUATIONS

1 Separable Equations

1.1 Definition

The first-order separable differential equation has the form

1.2 Solutions: The solution to (3) is

ZA(x )dx +

Z

where c represents an arbitrary constant

Example Solve the equation: dydx = x2y+2

Rewrite in the differential form (x2+ 2)dx − ydy = 0 which is separablewith A(x ) = x2+ 2 and B(y ) = −y Its solution is

R (x2+ 2)dx −R ydy = c or 1x3+ 2x − 1y2 = c for any constant c

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

constant c, or another way is using Rx

is homogeneous if f (tx , ty ) = f (x , y ) for every real number t 6= 0

Consider x 6= 0 Then, f (x , y ) = f (x · 1, x · yx) = f (1,yx) := g (yx) for afunction g depending only on the ratio yx

2.2 Solution: The homogeneous differential equation can be transformedinto a separable equation by making the substitution: y = xv with itsderivative: dydx = v + xdvdx Then, v + xdvdx = g (v ) ⇔ g (v )−vdv = dxx for

g (v ) 6= v , which is a separable DE and can be solved as previously For

g (v ) = v it yields another solution of the form y = kx for any constant k

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

constant c, or another way is using Rx

is homogeneous if f (tx , ty ) = f (x , y ) for every real number t 6= 0

Consider x 6= 0 Then, f (x , y ) = f (x · 1, x · yx) = f (1,yx) := g (yx) for afunction g depending only on the ratio yx

2.2 Solution: The homogeneous differential equation can be transformedinto a separable equation by making the substitution: y = xv with itsderivative: dydx = v + xdvdx Then, v + xdvdx = g (v ) ⇔ g (v )−vdv = dxx for

g (v ) 6= v , which is a separable DE and can be solved as previously For

g (v ) = v it yields another solution of the form y = kx for any constant k

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

constant c, or another way is using Rx

is homogeneous if f (tx , ty ) = f (x , y ) for every real number t 6= 0

Consider x 6= 0 Then, f (x , y ) = f (x · 1, x · yx) = f (1,yx) := g (yx) for a

function g depending only on the ratio yx

2.2 Solution: The homogeneous differential equation can be transformedinto a separable equation by making the substitution: y = xv with itsderivative: dydx = v + xdvdx Then, v + xdvdx = g (v ) ⇔ g (v )−vdv = dxx for

g (v ) 6= v , which is a separable DE and can be solved as previously For

g (v ) = v it yields another solution of the form y = kx for any constant k

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 findconstant c, or another way is using Rx

is homogeneous if f (tx , ty ) = f (x , y ) for every real number t 6= 0

Consider x 6= 0 Then, f (x , y ) = f (x · 1, x · yx) = f (1,yx) := g (yx) for afunction g depending only on the ratio yx

2.2 Solution: The homogeneous differential equation can be transformedinto a separable equation by making the substitution: y = xv with itsderivative: dydx = v + xdvdx Then, v + xdvdx = g (v ) ⇔ g (v )−vdv = dxx for

3.3 Solution: To solve Equation (6), first solve the equations(∂g (x ,y )

3.3 Solution: To solve Equation (6), first solve the equations(∂g (x ,y )

3.2 Test for exactness:If M(x , y ) and N(x , y ) are continuous functions and

have continuous first partial derivatives on some rectangle of the xy -plane,

then Equation (6) is exact if and only if ∂M(x ,y )∂y = ∂N(x ,y )∂x in this domain

3.3 Solution: To solve Equation (6), first solve the equations(∂g (x ,y )

3.3 Solution: To solve Equation (6), first solve the equations

(∂g (x ,y )

∂x = M(x , y )

∂g (x ,y )

∂y = N(x , y ) for g (x , y ).