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MTH102
EXACT DIFFERENTIAL
EQUATION
SUBMITTED TO :
MR.GURPREET SIR
SUBMITTED BY :
AVINASH RAJPUT
ACKNOWLEDGMENT
I would express my gratitude to all those who gave me the
possibility to complete this term work. I want to thank the
department of MTH102 for giving me permission to
commence this work in the first instance and to use the
research data.
I am deeply indebted to my teacher gurpreet sir whose
suggestions and encouragement helped me in all time of
research for writing this term work.
CONTANT:
EXACT DIFFERENTIAL EQUATIONS
EQUATION REDUCIBLE TO EXACT
EQUATION:

I.F FOUND BY INSPECTION
INTEGRATING FACTORS
SEPARABLE EQUATIONS(I.F OF
HOMOGENEOUS)
PROBLEM
BILIOGRPHY:
EXACT DIFFERENTIAL EQUATIONS
A differential equation obtained from its primitive directly by
differentiation without any operation of multiplication, elimination or
reduction is said to be an exact differential equation.
Thus a differential equation of the form M(X,Y)dx+N(X,Y)dy=0is an
exact differential equation if it can be obtained directly by differentiating
the equation u(x,y)=c which is primitive.
Du=Mdx+Ndy
 Mdx+Ndy=0is integ.Mdx+integ(terms of N not containing X)
Theorem: Solutions to Exact Differential Equations
Let M, N, My, and Nx be continuous with
My = Nx
Then there is a function f with
fx = M
and
fy = N
such that
f(x,y) = C
is a solution to the differential equation
M(x,y) + N(x,y)y' = 0
A first-order differential equation is one containing a first—but no
higher—derivative of the unknown function. For virtually every such
equation encountered in practice, the general solution will contain one
arbitrary constant, that is, one parameter, so a first-order IVP will
contain one initial condition. There is no general method that solves
every first-order equation, but there are methods to solve particular
typesGiven a function f( x, y) of two variables, its total differential df is
defined by the equation
The DE's that come up in Calculus are Separable. As we just saw this
means they can be
and
.
This means that
so that
.
Such a du is called an "Exact", "Perfect" or "Total" differential.
As we will see in Orthogonal Trajectories (1.8), the expression
represents
a one-parameter family of curves in the plane. For example,
is a family of circles of radius
parabolas.
Let us find the differential du for
.
Calculate
(solution)
and ended with
and
is a family of
1st order D. E. of the form
This D. E. is called exact if there is some function u(x, y) so that
and, of course,
. since
.
Since
and
,
and
we make the mild assumption that
a freebie that
,
If
are continuous then we get as
Now we can carry out two "partial" integrations:
so
so
Notice that the integration so-called constants each depend on one of the
variables.
Now we do some "criss-crossing" to get our solution
First, get
by solving for dk/dy in
and then carry out a ("partial') integration:
.
.
EQUATION REDUCIBLE TO EXACT
EQUATION:
Differential equation which are not exact can be made exact after
multiplying by a suitable factor called the integrating factor.
Ydx-xdy=0
,M=y
,N=-x
the equation is not exact.
 Multiplying the equation 1/y2 it ydx-xdy/y2 =0
 Multiplying the equation 1/x2 it d(y/x)=0
 Multiplying the equation 1/xy it dx/x-dy/y=0
 Which is exact -1/y2,1/x2,1/xy are integrating factors.
I.F found by inspection :in a number of problem a little
analyisis helps to find the ingrting factor .

M ( x , y)dx
y
1. Ydx+xdy=d(xy)
Set
F ( x, y ) =
F( x, y)
 M ( x , y) .
x


N ( x , y)dy
x
Integrate with respect to x to get F ( x, y )
M ( x , y)dx +  ( y)
Differentiate with respect to y to get
 ( y) ]= N ( x , y)  ( y)
N ( x , y)
F ( x, y ) = g(x,y) +
Integrating Factors
What if
=
M( x, y) N( x, y)

y
x
 ( y) ,

y
[

M ( x , y)dx +
if no boundary value is given.
Definition
If M(x, y)dx  N(x, y)dy  0 is not exact, but  (x, y)M(x, y)dx   (x, y) N(x, y)  0 is
exact, then  ( x , y) is called an integrating factor.
Example :
Show that
such that yn
i
 2
y  2xy  2 y  2x
y


y

 2xy dx  x 2 dy  0
is not exact, then find n
is an integrating factor.
2

( x 2 )  2 x therefore
x

the DE is
not exact.
ii.
Multiply the DE by yn, then
solve. y n ( y 2  2xy )dx  y n x 2 dy  0
equal

 y n x 2  2 y n x
x

2n  1xy n

must equal
 n2
y  2xy n 1  n  2y n 1  2n  1xy n
y

which means
 2xy n

must
n  2y n 1 must equal 0 and
for this to be so n must equal -2.
Separable Equations(I.F OF HOMOGENEOUS)
F( x )G ( y)dx  f ( x )g( x )dy  0 This
type of DE is called separable because it can
be written in the form (variables can be separable) M(x )dx  N( y)dy  0
The first equation is usually not exact but multiplying it by the
appropriate integrating factor will make it exact, but use of an
integrating factor may eliminate solutions or may lead to extraneous
solutions.After multiplying by the integrating factor
equation becomes:
F( x )
g ( y)
dx 
dy  0 where
f (x)
G ( y)
M( x ) 
F( x )
f (x)
1
f ( x )G ( y )
and
the
N( x ) 
where
g ( y)
.
G ( y)
Solutions are of the form
 M(x)dx   N( y)dy  c
f ( x )  0 & G ( y)  0 .
Problem  4xy
3
 6 x  sin x sin y  dx   cos x cos y  12 y 2  6x2 y 2  dy  0 ,
y    0
Step 0: Put the equation into standard form.
 4xy
3
 6 x  sin x sin y    cos x cos y  12 y 2  6x 2 y 2  y  0
Then M  x, y   4 xy 3  6 x  sin x sin y and N  x, y   cos x cos y  12 y 2  6 x 2 y 2 .
Step 1: Compute M y  x, y  and N x  x, y  .
M y  x, y   12 xy 2  sin x cos y
N x  x, y    sin x cos y  12 xy 2
Thus M y  x, y   N x  x, y  .
Step 2: Make a choice between two methods of solving the equation.
Choose one of the two methods. Below each method is presented,
but you need only one of them. The purpose of presenting both methods
is simply to demonstrate that either one will provide the answer.
[First method:   x, y    M  x, y  dx ]
Step 3: Perform the integration according to the method chosen in Step
2.
Since the equation is exact, there is a function  for which
 x  x, y   M  x, y   4 xy 3  6 x  sin x sin y
and
 y  x, y   N  x, y   cos x cos y  12 y 2  6 x 2 y 2 .
Then   x, y    M  x, y  dx    4xy 3  6x  sin x sin y dx
 2 x 2 y 3  3x 2  cos x sin y  g  y  .
Now it is given that N  x, y   cos x cos y  12 y 2  6 x 2 y 2 ; and from above
 y  x, y  

2 x 2 y 3  3x 2  cos x sin y  g  y    6 x 2 y 2  cos x cos y  g   y  .

y
Since  y  x, y   N  x, y  ,
6 x 2 y 2  cos x cos y  g   y   cos x cos y  12 y 2  6 x 2 y 2
g  y   12 y 2
g  y   4 y3
So the one parameter family of functions that define the solutions of the
equations is
2 x2 y3  3x2  cos x sin y  4 y3  C .
[Second method:   x, y    N  x, y  dy ]
Step 3: Perform the integration according to the method chosen in Step
2.
Since the equation is exact, there is a function  for which
 x  x, y   M  x, y   4 xy 3  6 x  sin x sin y
and
 y  x, y   N  x, y   cos x cos y  12 y 2  6 x 2 y 2 .
Then   x, y    N  x, y  dy    cos x cos y  12 y 2  6 x2 y 2 dy
 cos x sin y  4 y 3  2 x 2 y 3  f  x  .
Now it is given that M  x, y   4 xy 3  6 x  sin x sin y ; and from above
 x  x, y  

cos x sin y  4 y 3  2 x 2 y 3  f  x     sin x sin y  4 xy 3  f   x 

x
Since  x  x, y   M  x, y  ,
 sin x sin y  4 xy 3  f   x   4 xy 3  6 x  sin x sin y
f   x   6 x
f  x   3x 2
So the one parameter family of functions that define the solutions of the
equations is
cos x sin y  4 y3  2 x2 y3  3x2  C .
Step 4: Evaluate C using the initial condition.
Using the initial condition y    0 ,
C  2 2  0  3 2  cos  sin0  4  0  3 2 .
3
3
So the solution of the is
2x2 y3  3x2  cos x sin y  4 y3  3 2
REFRANCE:
www.wolframe.com
www.math.fsu.edu/~fusaro/EngMath/Ch1/SEDE.htm
www.ltcconline.net/greenl/courses/204/.../exactDiffEQs.html
www.cliffsnotes.com/.../Exact-Equations.topicArticleId19736,articleId-19710.htm
chapter form n.p.bali/b.s Grewal.
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