math differentiation equation

DEPARTMENT OF CIVIL ENGINEERING

Department of Civil Engineering Faculty of Engineering

SUBMITTED TO:SIR HASRAT HUSSAIN

H

SUBMITTED TO:SIR HASRAT HUSSAIN

H

FALL 2014 (CIVIL) 2ND


SHAHZAIB NASAR

HISTORICAL BACKGROUND OF EXACT DE i.e HISTORICAL BACKGROUND OF DE

In mathematics, the history of differential equations traces the development of "differential equations" from calculus, which

A snapshot overview of the historical origin of differential equations, a mathematical tool invented independently by Isaac Newton (1676) and Gottfried Leibniz (1693).


itself was independently invented by English physicist Isaac Newton and German mathematician Gottfried Leibniz. The history of the subject of differential equations, in concise form, from a synopsis of the recent article “The History of Differential Equations, 1670-1950”, reads: “Differential equations began with Leibniz, the Bernoulli brothers, and others from the 1680s, not long after Newton’s ‘fluxional equations’ in the 1670s.” Differential equations differ from ordinary equations of mathematics in that in addition to variables and constants they also contain derivatives of one or more of the variables involved. Newton-Leibniz years

The exact chronological origin and history to the subject of differential equations is a bit of a murky subject; for what seems to be a number of reasons: one being secretiveness, two being private publication issues (private works published only decades latter), and three being the nature of the battle of mathematical and scientific discovery, which is a type of intellectual "war" (in the words of English polymath Thomas Young).


In circa 1671, English physicist Isaac Newton wrote his then-unpublished The Method of Fluxions and Infinite Series(published in 1736), in which he classified first order differential equations, known to him as fluxional equations, into three classes, as follows (using modern notation):

Ordinary differential equations Partial differential equations

CLASS 1 ∂Y/∂X =F(Y)

Class2 ∂Y/∂X=F(X,Y)

Class 3 X∂U/∂X +Y∂U/∂Y= U


The first two classes contain only ordinary derivatives of one or more dependent variables, with respect to a single independent variable, and are known today as "ordinary differential equations"; the third class involves the partial derivatives of one dependent variable and today are called "partial differential equations". The study of "differential equations", according to British mathematician Edward Ince, is said to have began in 1675, when German mathematician Gottfried Leibniz wrote the following equation (date of introduction of integral sign; see:symbols):

In 1676, Newton solved his first differential equation. That same year, Leibniz introduced the term “differential equations” (aequatio differentialis, Latin) or to denote a relationship between the differentials dx and dy of two variables x and y.


In 1693, Leibniz solved his first differential equation and that same year Newton published the results of previous differential equation solution methods—a year that is said to mark the inception for the differential equations as a distinct field in mathematics


(Fahad Khan Afridi) INTRODUCTION TO AN EXACT DIFFERENTIAL EQUATION Given a simply connected and open sub-set D of R2 and two functions “M” and “N” which are continous on “”D” then an implicit first-order ordinary differential equation of the form. M(x,y) dx + N (x,y) dy = 0 Is called an exact equation if there exists a continuously differentiable function g, called the p9otential function, so that

∂g(x,y) = M & ∂g(x,y) = N ∂x ∂y

Note: For an exact DE to be solved, the potentiail function must be continuously differentiable i.e. the potential function must be exist and the potential function exists if and only if

∂M(x,y) =

∂N(x,y) Is satisfied

∂y ∂x


The existence of the potential function is therefore, must as all the exact differential equation are solved. HOW TO CHECK EXACTNESS To check exactness i.e. the existence of potential function. Example: 2xydx + (1 + x2) dy = 0 Solution: We have 2xydx + (1 + x2) dy = 0 (1) Compare with M(x,y)dx + N(x,y) dy = 0 M = 2xy , N = 1+x2


∂M = ∂(2x = 2x , ∂N = ∂(1+x2)

= 2x ∂y ∂y ∂x ∂x

My = 2x , Nx = 2x Since there is equality, so equation (1) is exact. If My ≠ Nx, then the equation would not be exact.


(Kamranullah Khan)

How to solve an exact DE Problem: (2x + y + 1) dx + ( 2y + x + 1) dy = 0 Solution: We have (2x + y + 1) dx + ( 2y + x + 1) dy = 0 (A) 1st: equation (A) exact? i. e. does My = Nx? Compare equation (A) with M(x, y) dx + N(x,y) dy = 0 M = 2x + y + 1 , N = 2y + x+ 1 ∂M =

My = 1

∂N = Nx =

1

∂y ∂x


We have equality so Equation (A) is exact Again we have (2x + y + 1) dx + ( 2y + x + 1) dy = 0 Compare with ∂g

Dx

+

∂g

dy

=

dg

∂x ∂y ∂g

=

2x + y + 1

(a) ,

∂g

=

2y + x + 1

(b)

∂x ∂y Integrating (a) w.r.t. x

∫

∂g

dx=

∫(2x + y + 1)dx

∂x

g = x2 + xy + x + h(y) (c) Partially differentiating (c) w.r.t. y

∂g

= 0 + x + 0 + h’(y)

∂y

∂g

= x + h’(y)

(d)

∂y Compare (b) and (d) 2y + x + 1 = x + h’(y) OR h’(y) = 2y + 1 ∂h (y) = 2y + 1 ∂y Integrating w.r.t y ∫∂h (y) dy = ∫ (2y + 1) dy ∂y


∫∂h (y) dy = ∫ (2y + 1) dy ∂y h(y) = y2 + y + c1 Putting in (c) g = x2 + xy + x + y2 + y + c1 Let g – c1 = C C = x2 + xy + x + y2 + y or x2 + x + y2 + y + xy = C Which is in implicit form?


(Akbar Khan)

How to solve an exact DE Problem: (2x + y + 1) dx + ( 2y + x + 1) dy = 0 Solution: We have (2x + y + 1) dx + ( 2y + x + 1) dy = 0 (A) 1st: equation (A) exact? i. e. does My = Nx? Compare equation (A) with M(x, y) dx + N(x,y) dy = 0 M = 2x + y + 1 , N = 2y + x+ 1 ∂M = My = 1 ∂N

= Nx =

1 ∂y ∂x


We have equality so Equation (A) is exact Again we have (2x + y + 1) dx + ( 2y + x + 1) dy = 0 Compare with ∂g

dx

+

∂g

Dy

=

dg

∂x ∂y ∂g

=

2x + y + 1

(a) ,

∂g

=

2y + x + 1

(b)

∂x ∂y Integrating (a) and (b) w.r.t. x and y, respectively

∫

∂g

dx=

∫(2x + y + 1)dx

, ∫

∂g

=

∫(2y + x + 1) dy

∂x ∂y


g = x2 + xy + x + h(y) (c) , g = y2 + xy + y + f(x) (d) Comparing (c) and (d) We see f(x) = x2 + x , h(y) = y2 + y g = x2 + x + y2 + y + xy as g(x,y) = c c = x2 + x + y2 + y + xy or x 2 + x + y2 + y + xy = c


(ISRAR ULLAH)

Problem: ydx - xdy =0 Sol: We have ydx - xdy =0 (A) As eq (A) in differential form so Compare with M(x,y)dx +N(x,y)dy =0 ( defferentail form) 1st eq (A) exist? I .e Does My=Nx ?


M= y , N= -x My = 1 (a) , Nx = -1 (b) As Eq(a) and Eq(b) are not equal i.e My ≠ Nx Thus Eq (A) is not exact So eq (A) has to be reduced to an exact DE 1st and then solved.


(Sher Khan)

REDUCING A NON-EXCAT DE TO AN EXACT DE :

INTEGRATING FACTOR:

IN 1739 SWISS MATHEMATICIAN LEAONHARD EULER GEGAN USING THE INTEGRATING FACTOR AS AN AID TO DRIVE DIFFERENTIAL EQUATIONS THAT WERE INTEGRATABLE IN FINITE FORM Consider a first-order ODE in the slightly different form

(1) Such an equation is said to be exact if

(2)


This statement is equivalent to the requirement that a conservative field exists, so that a scalar potential can be defined. For an exact equation, the solution is

(3)

where is a constant.

A first-order ODE (◇ ) is said to be inexact if

(4)

For a non exact equation, the solution may be obtained by defining an integrating factor of (◇ ) so that the new equation

(5)

satisfies


(6)

or, written out explicitly,

(7)

This transforms the nonexact equation into an exact one. Solving (7) for gives

(8)

Therefore, if a function satisfying (8) can be found, then writing

(9) (10)

in equation (◇ ) then gives

(11)

which is then an exact ODE. Special cases in which can be found include -dependent, -dependent, and -dependent integrating factors.


(6)

Given an inexact first-order ODE, we can also look for an integrating factors. factor so that

(12)

For the equation to be exact in and , the equation for a first-order nonexact ODE

(13)

becomes

(14)

Solving for gives

(15)

(16)

which will be integrable if

(17)

(18)


in which case

(19)

so that the equation is integrable

(20)

and the equation

(21)

with known is now exact and can be solved as an exact ODE.

Given an inexact first-order ODE, assume there exists an integrating factor

(22)

so . For the equation to be exact in and , equation (◇ ) becomes

(23)

(6)


(6)

Now, if

(24)

then

(25)

so that

(26)

and the equation

(27)

is now exact and can be solved as an exact ODE.


( Shazaib sabir) Every separable DE is an exact DE PROOF: We have M(x, y)dx +N(x,y)dy =0 (A) Eq (A) will be a separable DE If and only If M(x,y)= A(x) and N(x,y)=B(y) Eq (A) becomes A(x)dx + B(y)dy = 0 ( D.F)

(6)


Now if A(x) is partially differentiated w.r.t ‘ y ’ We get “0” i.e ∂A(x) =0 (a) ∂y

Similarly If B(x) is partially differentiated w.r.t ‘x’ , we get “0”i.e ∂B(x) =0 (b) ∂x

As Eq (a) and Eq (b) are equal so A(x) dx +B(y) dy =o I s an Exact DE

Hence every separable DE is an exact DE But every exact DE IS NOT SEPARABLE I-e an exact may or may not be separable.

(6)


Let the set A represent separable DE and the set B represent exact DE. (a) Every separable DE is exact DE:

Let A= {a, b, c} AND B= {a, b, c, d, e} We see A is the sub-set of B so every separable DE is an exact DE. (b) Exact DE may or may not be separable DE: (i) Let A= {a, b, c} and B= {a, b, c, d, e} In this case B is not the sub-set of A so in this case exact DE eq is not separable. (ii) AGAIN Let A={a,b,c} AND B={a,b,c} In this case B is the sub-set of A so in this case the exact DE is separable.

(6)

math differentiation equation

Education

order differential equations

newtons fluxional equations

newtonleibniz years

year newton

english physicist isaac

ordinary derivatives

partial derivatives

dependent variables