aarjaneiro/planetmath huggingface dataset:
natural symmetry of the Lorenz equation

natural symmetry of the Lorenz equation

The

Lorenz equation

has a natural

symmetry

defined by

(

x

,

y

,

z

)

↦

(

-

x

,

-

y

,

z

)

.

(1)

To verify that (

1

) is a symmetry of an

ordinary differential equation

(Lorenz equation) there must exist a

3

×

3

matrix which commutes with the

differential equation

. This can be easily verified by observing that the symmetry is associated with the matrix

R

defined as

R

=

[

-

1

0

0

0

-

1

0

0

0

1

]

.

(2)

Let

𝐱

˙

=

f

⁢

(

𝐱

)

=

[

σ

⁢

(

y

-

x

)

x

⁢

(

τ

-

z

)

-

y

x

⁢

y

-

β

⁢

z

]

(3)

where

f

⁢

(

𝐱

)

is the Lorenz equation and

𝐱

T

=

(

x

,

y

,

z

)

. We proceed by showing that

R

⁢

f

⁢

(

𝐱

)

=

f

⁢

(

R

⁢

𝐱

)

. Looking at the

left hand side

R

⁢

f

⁢

(

𝐱

)

=

[

-

1

0

0

0

-

1

0

0

0

1

]

⁢

[

σ

⁢

(

y

-

x

)

x

⁢

(

τ

-

z

)

-

y

x

⁢

y

-

β

⁢

z

]

=

[

σ

⁢

(

x

-

y

)

x

⁢

(

z

-

τ

)

+

y

x

⁢

y

-

β

⁢

z

]

and now looking at the right hand side

f

⁢

(

R

⁢

𝐱

)

=

f

⁢

(

[

-

1

0

0

0

-

1

0

0

0

1

]

⁢

[

x

y

z

]

)

=

f

⁢

(

[

-

x

-

y

z

]

)

=

[

σ

⁢

(

x

-

y

)

x

⁢

(

z

-

τ

)

+

y

x

⁢

y

-

β

⁢

z

]

.

Since the left hand side is equal to the right hand side then (

1

) is a symmetry of the Lorenz equation.

Title

natural symmetry of the Lorenz equation

Canonical name

NaturalSymmetryOfTheLorenzEquation

Date of creation

2013-03-22 13:44:12

Last modified on

2013-03-22 13:44:12

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

5

Author

Daume (40)

Entry type

Result

Classification

msc 34-00

Classification

msc 65P20

Classification

msc 65P30

Classification

msc 65P40

Classification

msc 65P99

Generated on Fri Feb 9 20:38:54 2018 by

LaTeXML
rectification theorem

rectification theorem

Let

U

be an open subset of

ℝ

n

and let

f

∈

C

1

⁢

(

U

)

be a

continuous

differentiable

vector field

f

:

U

→

ℝ

n

.

If there exists

x

0

∈

U

such that

f

⁢

(

x

0

)

≠

0

then there exists

U

0

⊂

U

an open neighborhood of

x

0

such that there exists a

diffeomorphism

of class

C

1

F

:

U

0

→

V

where

V

is an open subset of

ℝ

n

such that

[

D

⁢

F

⁢

(

x

)

]

⁢

f

⁢

(

x

)

=

e

1

for all

x

∈

U

0

where

[

D

⁢

F

⁢

(

x

)

]

is the

Jacobian

of the diffeomorphism

F

evaluated at

x

and

e

1

=

(

1

,

0

,

…

,

0

)

is the first vector of the canonical basis of

ℝ

n

. More generally if the vector field

f

is of class

C

r

then so is the diffeomorphism

F

.

[

AVI

]

References

AVI

Arnold, V.I.:

Ordinary Differential Equations

(translated by R.A. Silverman)

. The MIT Press, Cambridge, 1973.

Title

rectification theorem

Canonical name

RectificationTheorem

Date of creation

2013-03-22 14:57:15

Last modified on

2013-03-22 14:57:15

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

12

Author

Daume (40)

Entry type

Theorem

Classification

msc 34-00

Classification

msc 34A12

Related topic

ImplicitFunctionTheorem

Related topic

TangentMap

Generated on Fri Feb 9 20:38:56 2018 by

LaTeXML
some qualitative analysis of FitzHugh-Nagumo equation using nullclines

some qualitative analysis of FitzHugh-Nagumo equation using nullclines

The

FitzHugh-Nagumo model

which we will use for the analysis is:

u

˙

=

(

v

+

u

-

u

3

/

3

)

ϵ

v

˙

=

ϵ

⁢

(

u

+

β

-

γ

⁢

v

)

The following formulas

v

=

1

3

⁢

u

3

-

u

v

=

1

γ

⁢

u

+

β

γ

are the

u

and

v

nullcline

respectively.

(See graph of the nullcline below.)

Graph of the nullcline with

parameter

(

ϵ

=

0.2

,

β

=

0.7

,

γ

=

0.5

)

The nullclines

v

=

1

3

⁢

u

3

-

u

and

v

=

1

γ

⁢

u

+

β

γ

represent the set of points when the change with repect to time of

u

and

v

is null. Hence the solution will cross the ‘cubic’ nullcline

(i.e.

u

nullcline)

when travelling vertically while the solution will cross the ‘line’ nullcline

(i.e.

v

nullcline)

when travelling horizontally. Suppose that we fix the parameters of the FitzHugh-Nagumo model to

ϵ

=

0.2

,

β

=

0.7

,

γ

=

0.5

, then suppose that you let a solution start at

x

0

=

(

1

,

-

0.5

)

. Then we notice that

d

⁢

u

d

⁢

t

>

0

and that

d

⁢

v

d

⁢

t

>

0

, so the solution will travel upwards and to the right. It will only cross the ‘cubic’ nullcline vertically. Therefore once the solution will be close to the nullcline it will travel vertically and cross the nullcline. Once the solution has crossed the nullcline it will move upwards and to the left since

d

⁢

u

d

⁢

t

<

0

and that

d

⁢

v

d

⁢

t

>

0

. It will then cross the ‘line’ nullcline horizontally and the solution will move downwards and will remain moving towards the left since

d

⁢

u

d

⁢

t

<

0

and

d

⁢

v

d

⁢

t

<

0

, until the solution reaches the ‘cubic’ nullcline which it will cross vertically. Once the solution has crossed the last nullcline it will continue to move downwards but will now progressively start moving to the right since

d

⁢

u

d

⁢

t

>

0

and

d

⁢

v

d

⁢

t

<

0

. This is all that can be said about this specific case using the analysis of the nullcline. Below is an example of the solution. As we can see it intially behaves like the above description.

Graph of the nullcline with parameter

(

ϵ

=

0.2

,

β

=

0.7

,

γ

=

0.5

)

and solution starting at

x

0

=

(

1

,

-

0.5

)

.

Title

some qualitative analysis of

FitzHugh-Nagumo equation

using nullclines

Canonical

name

SomeQualitativeAnalysisOfFitzHughNagumoEquationUsingNullclines

Date of creation

2013-03-22 14:32:06

Last modified on

2013-03-22 14:32:06

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

4

Author

Daume (40)

Entry type

Result

Classification

msc 34-00

Generated on Fri Feb 9 20:38:58 2018 by

LaTeXML
symmetry of a solution of an ordinary differential equation

symmetry of a solution of an ordinary differential equation

Let

γ

be a symmetry of the

ordinary differential equation

(

http://planetmath.org/SymmetryOfAnOrdinaryDifferentialEquation

) and

x

0

be a

steady state solution

of

x

˙

=

f

⁢

(

x

)

. If

γ

⁢

x

0

=

x

0

then

γ

is called a

symmetry of the solution of

x

0

.

Let

γ

be a symmetry of the ordinary differential equation and

x

0

⁢

(

t

)

be a

periodic solution

of

x

˙

=

f

⁢

(

x

)

. If

γ

⁢

x

0

⁢

(

t

-

t

0

)

=

x

0

⁢

(

t

)

for a certain

t

0

then

(

γ

,

t

0

)

is called a

symmetry of the periodic solution of

x

0

⁢

(

t

)

.

lemma:

If

γ

is a symmetry of the ordinary differential equation and let

x

0

⁢

(

t

)

be a solution

(either steady state or periodic)

of

x

˙

=

f

⁢

(

x

)

. Then

γ

⁢

x

0

⁢

(

t

)

is a solution of

x

˙

=

f

⁢

(

x

)

.

proof:

If

x

0

⁢

(

t

)

is a solution of

d

⁢

x

d

⁢

t

=

f

⁢

(

x

)

implies

d

⁢

x

0

⁢

(

t

)

d

⁢

t

=

f

⁢

(

x

0

⁢

(

t

)

)

. Let’s now verify that

γ

⁢

x

0

⁢

(

t

)

is a solution, with a substitution into

d

⁢

x

d

⁢

t

=

f

⁢

(

x

)

. The left hand side of the equation becomes

d

⁢

γ

⁢

x

0

⁢

(

t

)

d

⁢

t

=

γ

⁢

d

⁢

x

0

⁢

(

t

)

d

⁢

t

and the right hand side of the equation becomes

f

⁢

(

γ

⁢

x

0

⁢

(

t

)

)

=

γ

⁢

f

⁢

(

x

0

⁢

(

t

)

)

since

γ

is a symmetry of the

differential equation

. Therefore we have that the left hand side equals the right hand side since

d

⁢

x

0

⁢

(

t

)

d

⁢

t

=

f

⁢

(

x

0

⁢

(

t

)

)

.
qed

References

GSS

Golubitsky, Martin. Stewart, Ian. Schaeffer, G. David: Singularities and Groups in

Bifurcation Theory

(Volume II)

. Springer-Verlag, New York, 1988.

Title

symmetry of a solution of an ordinary differential equation

Canonical

name

SymmetryOfASolutionOfAnOrdinaryDifferentialEquation

Date of creation

2013-03-22 13:42:26

Last modified on

2013-03-22 13:42:26

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

11

Author

Daume (40)

Entry type

Definition

Classification

msc 34-00

Synonym

symmetry of a periodic solution solution of an ordinary differential equation

Generated on Fri Feb 9 20:39:01 2018 by

LaTeXML
symmetry of an ordinary differential equation

symmetry of an ordinary differential equation

Let

f

:

ℝ

n

→

ℝ

n

be a

smooth function

and let

x

˙

=

f

⁢

(

x

)

be a

system of ordinary differential equations

, in

addition

let

γ

be an invertible matrix. Then

γ

is a

of the

ordinary differential equation

if

f

⁢

(

γ

⁢

x

)

=

γ

⁢

f

⁢

(

x

)

.

Example:

•

Natural symmetry of the

Lorenz equation

is a example of a symmetry of a

differential equation

.

References

GSS

Golubitsky, Martin. Stewart, Ian. Schaeffer, G. David: Singularities and Groups in Bifurcation Theory

(Volume II)

. Springer-Verlag, New York, 1988.

Title

symmetry of an ordinary differential equation

Canonical

name

SymmetryOfAnOrdinaryDifferentialEquation

Date of creation

2013-03-22 13:42:24

Last modified on

2013-03-22 13:42:24

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

10

Author

Daume (40)

Entry type

Definition

Classification

msc 34-00

Synonym

symmetry of an differential equation

Generated on Fri Feb 9 20:39:03 2018 by

LaTeXML
order n constant coefficient differential equations and matrix exponential

order n constant coefficient differential equations and matrix exponential

Let

P

be a degree

n

>

0

monic complex polynomial in one indeterminate, let

f

be a

continuous function

on the real line, let

k

be an

integer

varying from 0 to

n

-

1

, and let

y

k

be a

complex number

. The solution to the ODE

P

⁢

(

d

/

d

⁢

t

)

⁢

y

=

f

⁢

(

t

)

,

y

(

k

)

⁢

(

0

)

=

y

k

(1)

is

y

⁢

(

t

)

=

∑

y

k

⁢

g

k

⁢

(

t

)

+

∫

0

t

g

n

-

1

⁢

(

t

-

x

)

⁢

f

⁢

(

x

)

⁢

𝑑

x

,

(2)

where

g

k

⁢

(

t

)

is the

coefficient

of

z

k

in the product of

P

⁢

(

z

)

by the

singular

part of

e

t

⁢

z

P

⁢

(

z

)

.

Moreover, if

A

is a

complex

square matrix

annihilated by

P

, then

e

t

⁢

A

=

∑

g

k

⁢

(

t

)

⁢

A

k

.

(3)

(

1

) into

Y

′

-

B

⁢

Y

=

f

⁢

(

t

)

⁢

v

,

Y

⁢

(

0

)

=

Y

0

(4)

by putting

Y

k

:=

y

(

k

)

,

Y

0

⁢

k

:=

y

k

, and by letting

B

be the

transpose

companion matrix

of

P

, and

v

the last vector of the

canonical basis

of

ℂ

n

. The solution to (

4

) is

Y

⁢

(

t

)

=

e

t

⁢

B

⁢

Y

0

+

∫

0

t

f

⁢

(

x

)

⁢

e

(

t

-

x

)

⁢

B

⁢

v

⁢

𝑑

x

.

There is a unique

n

-tuple of

functions

h

k

such that

e

t

⁢

A

is the sum of the

h

k

⁢

(

t

)

⁢

A

k

whenever

A

is a complex square matrix annihilated by

P

. The first line of

B

k

being the

(

k

+

1

)

-th vector of the canonical basis of

ℂ

n

(for

0

≤

k

<

n

), we obtain

y

⁢

(

t

)

=

∑

y

k

⁢

h

k

⁢

(

t

)

+

∫

0

t

h

n

-

1

⁢

(

t

-

x

)

⁢

f

⁢

(

x

)

⁢

𝑑

x

,

so that the proof of (

2

) and (

3

) boils down to verifying

h

k

⁢

(

t

)

=

g

k

⁢

(

t

)

.

a real value of

t

, let

G

∈

ℂ

⁢

[

X

]

be the sum of the

g

k

⁢

(

t

)

⁢

X

k

, form the

entire function

φ

⁢

(

z

)

=

e

t

⁢

z

-

G

⁢

(

z

)

P

⁢

(

z

)

,

multiply the above equality by

P

⁢

(

z

)

, and replace

z

by

A

.

Title

order n constant coefficient differential equations and matrix exponential

Canonical name

OrderNConstantCoefficientDifferentialEquationsAndMatrixExponential

Date of creation

2013-03-22 19:01:00

Last modified on

2013-03-22 19:01:00

Owner

gaillard (1824)

Last modified by

gaillard (1824)

Numerical id

7

Author

gaillard (1824)

Entry type

Definition

Classification

msc 34-01

Generated on Fri Feb 9 20:39:09 2018 by

LaTeXML
d’Alembert’s equation

d’Alembert’s equation

The first

differential equation

y

=

φ

⁢

(

d

⁢

y

d

⁢

x

)

⋅

x

+

ψ

⁢

(

d

⁢

y

d

⁢

x

)

is called

d’Alembert’s differential equation

; here

φ

and

ψ

some known

differentiable

real functions.

If we denote

d

⁢

y

d

⁢

x

:=

p

, the equation is

y

=

φ

⁢

(

p

)

⋅

x

+

ψ

⁢

(

p

)

.

We take

p

as a new variable and derive the equation with respect to

p

, getting

p

-

φ

⁢

(

p

)

=

[

x

⁢

φ

′

⁢

(

p

)

+

ψ

′

⁢

(

p

)

]

⁢

d

⁢

p

d

⁢

x

.

If the equation

p

-

φ

⁢

(

p

)

=

0

has the roots

p

=

p

1

,

p

2

, …,

p

k

, then we have

d

⁢

p

ν

d

⁢

x

=

0

for all

ν

’s, and therefore there are the special solutions

y

=

p

ν

x

+

ψ

(

p

ν

)

(

ν

=

1

,

2

,

…

,

k

)

for the original equation.  If

φ

⁢

(

p

)

≢

p

, then the derived equation may be written as

d

⁢

x

d

⁢

p

=

φ

′

⁢

(

p

)

p

-

φ

⁢

(

p

)

⁢

x

+

ψ

′

⁢

(

p

)

p

-

φ

⁢

(

p

)

,

which linear differential equation has the solution

x

=

x

⁢

(

p

,

C

)

with the integration constant

C

.  Thus we get the

general solution

of d’Alembert’s equation as a parametric

{

x

=

x

⁢

(

p

,

C

)

,

y

=

φ

⁢

(

p

)

⁢

x

⁢

(

p

,

C

)

+

ψ

⁢

(

p

)

of the integral curves.

Title

d’Alembert’s equation

Canonical name

DAlembertsEquation

Date of creation

2013-03-22 14:31:05

Last modified on

2013-03-22 14:31:05

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

16

Author

pahio (2872)

Entry type

Derivation

Classification

msc 34A05

Synonym

Lagrange equation

Related topic

ClairautsEquation

Related topic

ContraharmonicProportion

Related topic

DerivativeAsParameterForSolvingDifferentialEquations

Generated on Fri Feb 9 20:39:20 2018 by

LaTeXML
exact differential equation

exact differential equation

Let

R

be a region in

ℝ

2

and let the functions

X

:

R

→

ℝ

,

Y

:

R

→

ℝ

have

continuous

partial derivatives

in

R

.  The first order

differential equation

X

⁢

(

x

,

y

)

+

Y

⁢

(

x

,

y

)

⁢

d

⁢

y

d

⁢

x

=

0

or

X

⁢

(

x

,

y

)

⁢

d

⁢

x

+

Y

⁢

(

x

,

y

)

⁢

d

⁢

y

=

0

(1)

is called an

exact differential equation

, if the condition

∂

⁡

X

∂

⁡

y

=

∂

⁡

Y

∂

⁡

x

(2)

is true in

R

.

By (2), the left hand side of (1) is the

total differential

of a function, there is a function

f

:

R

→

ℝ

such that the equation (1) reads

d

⁢

f

⁢

(

x

,

y

)

=

0

,

whence its

general integral

is

f

⁢

(

x

,

y

)

=

C

.

The solution function

f

can be calculated as the

line integral

f

⁢

(

x

,

y

)

:=

∫

P

0

P

[

X

⁢

(

x

,

y

)

⁢

d

⁢

x

+

Y

⁢

(

x

,

y

)

⁢

d

⁢

y

]

(3)

along any curve

γ

connecting an arbitrarily chosen point

P

0

=

(

x

0

,

y

0

)

and the point

P

=

(

x

,

y

)

in the region

R

(the

integrating factor

is now

≡

1

).

Example.

Solve the differential equation

2

⁢

x

y

3

⁢

d

⁢

x

+

y

2

-

3

⁢

x

2

y

4

⁢

d

⁢

y

=

0

.

This equation is exact, since

∂

∂

⁡

y

⁢

2

⁢

x

y

3

=

-

6

⁢

x

y

4

=

∂

∂

⁡

x

⁢

y

2

-

3

⁢

x

2

y

4

.

If we use as the integrating way the broken line from

(

0

,

1

)

to

(

x

,

1

)

and from this to

(

x

,

y

)

,  the

integral

(3) is simply

∫

0

x

2

⁢

x

1

3

⁢

𝑑

x

+

∫

1

y

y

2

-

3

⁢

x

2

y

4

⁢

𝑑

y

=

x

2

y

3

-

1

y

+

1

=

x

2

-

1

y

+

x

2

y

3

+

1

-

x

2

=

x

2

y

3

-

1

y

+

1

.

Thus we have the general integral

x

2

y

3

-

1

y

=

C

of the given differential equation.

Title

exact differential equation

Canonical name

ExactDifferentialEquation

Date of creation

2013-03-22 18:06:17

Last modified on

2013-03-22 18:06:17

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Result

Classification

msc 34A05

Related topic

HarmonicConjugateFunction

Related topic

Differential

Related topic

TotalDifferential

Defines

exact differential equation

Generated on Fri Feb 9 20:39:26 2018 by

LaTeXML
example of derivative as parameter

example of derivative as parameter

For solving the (nonlinear)

differential equation

x

=

y

3

⁢

p

-

2

⁢

p

⁢

y

2

(1)

with

p

=

d

⁢

y

d

⁢

x

,  according to

III

in the parent entry (

http://planetmath.org/DerivativeAsParameterForSolvingDifferentialEquations

), we

differentiate

both sides in regard to

y

, getting first

1

p

=

1

3

⁢

p

-

(

y

3

⁢

p

2

+

2

⁢

y

2

)

⁢

d

⁢

p

d

⁢

y

-

4

⁢

p

⁢

y

.

Removing the denominators, we obtain

2

⁢

p

+

(

y

+

6

⁢

p

2

⁢

y

2

)

⁢

d

⁢

p

d

⁢

y

+

12

⁢

p

3

⁢

y

=

0

.

The left hand side can be factored:

(

y

⁢

d

⁢

p

d

⁢

y

+

2

⁢

p

)

⁢

(

1

+

6

⁢

p

2

⁢

y

)

=

0

(2)

Now we may use the zero rule of product; the first factor of the product in (2) yields

y

⁢

d

⁢

p

d

⁢

y

=

-

2

⁢

p

, i.e.

2

⁢

∫

d

⁢

y

y

=

-

∫

d

⁢

p

p

+

ln

⁡

C

,

whence

y

2

=

C

p

,  i.e.

p

=

C

y

2

.  Substituting this into the original equation (1) we get

x

=

y

3

3

⁢

C

-

2

⁢

C

.  Hence the

general solution

of (1) may be written

y

3

=

3

⁢

C

⁢

x

+

6

⁢

C

2

.

The second factor in (2) yields

6

⁢

p

2

⁢

y

=

-

1

,  which is substituted into (1) multiplied by

3

⁢

p

:

3

⁢

p

⁢

x

=

y

-

(

-

y

)

Thus we see that

p

=

2

⁢

y

3

⁢

x

, which is again set into (1), giving

x

=

y

⋅

3

⁢

x

3

⋅

2

⁢

y

-

4

⁢

y

3

3

⁢

x

.

Finally, we can write it

3

⁢

x

2

=

-

8

⁢

y

3

,

which (a variant of the so-called semicubical parabola) is the

singular solution

of (1).

Title

example of

derivative

as parameter

Canonical name

ExampleOfDerivativeAsParameter

Date of creation

2013-03-22 18:29:03

Last modified on

2013-03-22 18:29:03

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Example

Classification

msc 34A05

Synonym

example of solving an ODE

Generated on Fri Feb 9 20:39:29 2018 by

LaTeXML
finding another particular solution of linear ODE

finding another particular solution of linear ODE

Consider the

homogeneous

(

http://planetmath.org/HomogeneousLinearDifferentialEquation

)

second-order

linear ordinary differential equation

y

′′

+

P

⁢

(

x

)

⁢

y

′

+

Q

⁢

(

x

)

⁢

y

=

0

.

(1)

If one knows one
particular solution (

http://planetmath.org/SolutionsOfOrdinaryDifferentialEquation

)

y

=

y

1

⁢

(

x

)

≢

0

of (1), it’s possible to derive
from it via two

quadratures

another solution

y

2

⁢

(

x

)

,

linearly independent

on

y

1

⁢

(

x

)

;  thus one can write the

general solution

y

=

C

1

⁢

y

1

⁢

(

x

)

+

C

2

⁢

y

2

⁢

(

x

)

of that

homogeneous differential equation

.

We will now show the

derivation

procedure.

We put

y

=

u

⁢

v

(2)

which renders (1) to

(

v

′′

+

P

⁢

v

′

+

Q

⁢

v

)

⁢

u

+

(

2

⁢

v

′

+

P

⁢

v

)

⁢

u

′

+

u

′′

⁢

v

=

0

.

(3)

Here one can choose

v

:=

y

1

⁢

(

x

)

, whence the first addend

vanishes

, and (3) gets the form

(

2

⁢

y

1

′

+

P

⁢

y

1

)

⁢

u

′

+

y

1

⁢

u

′′

=

0

.

(4)

This equation may be written as

u

′′

u

′

=

-

2

⁢

y

1

′

y

1

-

P

, which is integrated to

ln

⁡

|

d

⁢

u

d

⁢

x

|

=

ln

⁡

1

y

1

2

-

∫

P

⁢

𝑑

x

+

constant

,

i.e.

d

⁢

u

d

⁢

x

=

C

y

1

2

⁢

e

-

∫

P

⁢

𝑑

x

.

A new integration results from this the general solution of (4):

u

=

C

⁢

∫

e

-

∫

P

⁢

𝑑

x

y

1

2

⁢

𝑑

x

+

C

′

.

Thus by (2), we have obtained the wanted other solution

y

2

⁢

(

x

)

=

y

1

⁢

(

x

)

⁢

∫

e

-

∫

P

⁢

𝑑

x

y

1

2

⁢

𝑑

x

which is clearly linearly independent on y_1(x).

Consequently, we can express the general solution of the

differential equation

(1) as

y

=

y

1

⁢

(

x

)

⁢

u

=

C

1

⁢

y

1

⁢

(

x

)

+

C

2

⁢

y

1

⁢

(

x

)

⁢

∫

e

-

∫

P

⁢

𝑑

x

y

1

2

⁢

𝑑

x

,

where

C

1

and

C

2

are arbitrary

constants

.

Remark.

The

substitution

y

:=

e

-

1

2

⁢

∫

P

⁢

(

x

)

⁢

𝑑

x

⁢

u

converts the equation (1) into the form

d

2

⁢

u

d

⁢

x

2

+

(

Q

-

P

2

4

-

P

′

2

)

⁢

u

=

0

not containing the

derivative

d

⁢

u

d

⁢

x

.

References

1

Ernst Lindelöf

:

Differentiali- ja integralilasku
ja sen sovellutukset III.1

.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1935).

Title

finding another particular solution of linear ODE

Canonical name

FindingAnotherParticularSolutionOfLinearODE

Date of creation

2014-02-28 14:31:42

Last modified on

2014-02-28 14:31:42

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Algorithm

Classification

msc 34A05

Generated on Fri Feb 9 20:39:34 2018 by

LaTeXML
homogeneous linear differential equation

homogeneous linear differential equation

The

linear differential equation

a

n

⁢

(

x

)

⁢

y

(

n

)

+

a

n

-

1

⁢

(

x

)

⁢

y

(

n

-

1

)

+

…

+

a

1

⁢

(

x

)

⁢

y

′

+

a

0

⁢

(

x

)

⁢

y

=

b

⁢

(

x

)

(1)

is called

homogeneous

(

http://planetmath.org/HomogeneousLinearDifferentialEquation

)
iff

b

⁢

(

x

)

≡

0

.  If

b

⁢

(

x

)

≢

0

,
the equation (1) is

inhomogeneous

.

If (1) is homogeneous (

http://planetmath.org/HomogeneousLinearDifferentialEquation

),
then the sum of any solutions is a solution and any solution
multiplied by a

constant

is a solution.

The special case

c

n

⁢

x

n

⁢

y

(

n

)

+

c

n

-

1

⁢

x

n

-

1

⁢

y

(

n

-

1

)

+

…

+

c

1

⁢

x

⁢

y

′

+

c

0

⁢

y

=

0

of (1), where the

c

i

’s are constants, can via the
substitution

x

=

e

t

be transformed into a

homogeneous
linear differential equation

of the same order but with
constant coefficients.

Title

homogeneous linear differential equation

Canonical name

HomogeneousLinearDifferentialEquation

Date of creation

2014-02-27 10:07:04

Last modified on

2014-02-27 10:07:04

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

3

Author

pahio (2872)

Entry type

Definition

Classification

msc 34A05

Generated on Fri Feb 9 20:39:38 2018 by

LaTeXML
Landau-Ramanujan Constant

Landau-Ramanujan Constant

Let

N

be a

natural number

. Consider the equation

x

2

+

y

2

=

N

Let

L

⁢

(

N

)

denote the number of naturals

z

≤

N

such that the equation

x

2

+

y

2

=

z

has at least one

integer

solution

(

x

,

y

)

.
The Laudau-Ramanujan

constant

is defined as the limit

lim

N

→

∞

⁡

L

⁢

(

N

)

⁢

log

⁡

(

N

)

N

≈

0.76422365358922066299069873125

Title

Landau-Ramanujan Constant

Canonical name

LandauRamanujanConstant

Date of creation

2014-03-06 12:39:16

Last modified on

2014-03-06 12:39:16

Owner

Filipe (28191)

Last modified by

Filipe (28191)

Numerical id

2

Author

Filipe (28191)

Entry type

Definition

Generated on Fri Feb 9 20:39:46 2018 by

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