derivative as parameter for solving differential equations

derivative as parameter for solving differential equations

The solution of some

differential equations

of the forms

x

=

f

⁢

(

d

⁢

y

d

⁢

x

)

and

y

=

f

⁢

(

d

⁢

y

d

⁢

x

)

may be expressed in a parametric form by taking for the

parameter

the

derivative

p

:=

d

⁢

y

d

⁢

x

.

(1)

I.

Consider first the equation

x

=

f

⁢

(

d

⁢

y

d

⁢

x

)

,

(2)

for which we suppose that

p

↦

f

⁢

(

p

)

and its derivative

p

↦

f

′

⁢

(

p

)

are

continuous

and

f

′

⁢

(

p

)

≠

0

on an

interval

[

p

1

,

p

2

]

.  It follows that on the interval, the

function

p

↦

f

⁢

(

p

)

changes monotonically from

f

⁢

(

p

1

)

:=

x

1

to

f

⁢

(

p

2

)

:=

x

2

, whence conversely the equation

x

=

f

⁢

(

p

)

(3)

defines from

[

p

1

,

p

2

]

onto

[

x

1

,

x

2

]

a bijection

p

=

g

⁢

(

x

)

(4)

which is

continuously differentiable

.  Thus on the interval

[

x

1

,

x

2

]

,  the differential equation (2) can be replaced by the equation

d

⁢

y

d

⁢

x

=

g

⁢

(

x

)

,

(5)

and therefore, the solution of (2) is

y

=

∫

g

⁢

(

x

)

⁢

𝑑

x

+

C

.

(6)

If we cannot express

g

⁢

(

x

)

in a , we take

p

as an

independent variable

through the substitution (3), which maps

[

x

1

,

x

2

]

bijectively onto

[

p

1

,

p

2

]

.  Then (6) becomes a function of

p

, and by the

chain rule

,

d

⁢

y

d

⁢

p

=

g

⁢

(

f

⁢

(

p

)

)

⁢

f

′

⁢

(

p

)

=

p

⁢

f

′

⁢

(

p

)

.

Accordingly, the solution of the given differential equation may be presented on

[

p

1

,

p

2

]

as

{

x

=

f

⁢

(

p

)

,

y

=

∫

p

⁢

f

′

⁢

(

p

)

⁢

𝑑

p

+

C

.

(7)

II.

With corresponding considerations, one can write the solution of the differential equation

y

=

f

⁢

(

d

⁢

y

d

⁢

x

)

:=

f

⁢

(

p

)

,

(8)

where

p

changes on some interval

[

p

1

,

p

2

]

where

f

⁢

(

p

)

and

f

′

⁢

(

p

)

are continuous and

p

⋅

f

′

⁢

(

p

)

≠

0

,  in the parametric presentation

{

x

=

∫

f

′

⁢

(

p

)

p

⁢

𝑑

p

+

C

,

y

=

f

⁢

(

p

)

.

(9)

III.

The procedures of

I

and

II

may be generalised for the differential equations of

x

=

f

⁢

(

y

,

p

)

and

y

=

f

⁢

(

x

,

p

)

;  let’s consider the former one.

In

x

=

f

⁢

(

y

,

p

)

(10)

we regard

y

as the independent variable and

differentiate

with respect to it:

d

⁢

x

d

⁢

y

=

1

p

=

f

y

′

⁢

(

y

,

p

)

+

f

p

′

⁢

(

y

,

p

)

⁢

d

⁢

p

d

⁢

y

.

Supposing that the

partial derivative

f

p

′

⁢

(

y

,

p

)

does not vanish identically, we get

d

⁢

p

d

⁢

y

=

1

p

-

f

y

′

⁢

(

y

,

p

)

f

p

′

⁢

(

y

,

p

)

:=

g

⁢

(

y

,

p

)

.

(11)

If

p

=

p

⁢

(

y

,

C

)

is the

general solution

of (11), we obtain the general solution of (10):

x

=

f

⁢

(

y

,

p

⁢

(

y

,

C

)

)

(12)

Title

derivative as parameter for solving differential equations

Canonical name

DerivativeAsParameterForSolvingDifferentialEquations

Date of creation

2013-03-22 18:28:39

Last modified on

2013-03-22 18:28:39

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Topic

Classification

msc 34A05

Related topic

InverseFunctionTheorem

Related topic

ImplicitFunctionTheorem

Related topic

DAlembertsEquation

Related topic

ClairautsEquation

Generated on Fri Feb 9 20:39:23 2018 by

LaTeXML