The document provides a historical overview of differential equations, which were independently invented by Isaac Newton and Gottfried Leibniz. It discusses that differential equations began being studied in the 1680s by Leibniz, Bernoulli brothers, and others. Differential equations contain variables, constants, and derivatives of the variables. The document then discusses Newton and Leibniz's early work in more detail in the 1670s and 1680s, including Newton's classification of differential equations into three classes. It notes that the study of differential equations is generally considered to have begun in 1675 when Leibniz wrote his first differential equation.

2. DEPARTMENT OF CIVIL ENGINEERING

3. Department of Civil Engineering Faculty of Engineering
SUBMITTED TO:
SIR HASRAT HUSSAIN
H
SUBMITTED TO:
SIR HASRAT HUSSAIN
H
FALL 2014 (CIVIL) 2ND

4. Department of Civil Engineering Faculty of Engineering
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).

5. Department of Civil Engineering Faculty of Engineering
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).

6. Department of Civil Engineering Faculty of Engineering
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

7. Department of Civil Engineering Faculty of Engineering
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.

8. Department of Civil Engineering Faculty of Engineering
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

9. Department of Civil Engineering Faculty of Engineering
(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

10. Department of Civil Engineering Faculty of Engineering
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

11. Department of Civil Engineering Faculty of Engineering
∂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.

12. Department of Civil Engineering Faculty of Engineering
(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

13. Department of Civil Engineering Faculty of Engineering
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

14. 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

15. Department of Civil Engineering Faculty of Engineering
∫∂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?

16. Department of Civil Engineering Faculty of Engineering
(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

17. Department of Civil Engineering Faculty of Engineering
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

18. Department of Civil Engineering Faculty of Engineering
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

19. Department of Civil Engineering Faculty of Engineering
(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 ?

20. Department of Civil Engineering Faculty of Engineering
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.

21. Department of Civil Engineering Faculty of Engineering
(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)

22. Department of Civil Engineering Faculty of Engineering
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

23. Department of Civil Engineering Faculty of Engineering
(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.

24. Department of Civil Engineering Faculty of Engineering
(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)

25. Department of Civil Engineering Faculty of Engineering
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)

26. Department of Civil Engineering Faculty of Engineering
(6)
Now, if
(24)
then
(25)
so that
(26)
and the equation
(27)
is now exact and can be solved as an exact ODE.

27. Department of Civil Engineering Faculty of Engineering
( 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)

28. Department of Civil Engineering Faculty of Engineering
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)

29. Department of Civil Engineering Faculty of Engineering
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)

30. Department of Civil Engineering Faculty of Engineering