Download Exact Differential Equation

Mathematics: RC6901B46
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Mathematics : MTH - 102
TERM PAPER
Give a short summary of the method of solution of exact
differential equation. Using equation of your own choice. Show
that differential equations can be solved by more than one
method. Compare the amount of work involved in each case.
Submitted By
:-
saurabh srivastava
Acknowledgment
The completion of any term paper requires
coordination ,dedication and a combined effect of
various sources. First of I Thank God for making me
capable of completing this term paper. Thereafter I
Mathematics: RC6901B46
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thank Mr. Gurpindar our Math lecturer for spapring
his valuable precious time in guiding me in this term
paper.
CONTENTS
Differential Equations
Order & Degree
Mathematics: RC6901B46
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Exact Differential Equation
Method For Solving Exact Differential equation
Equations Reducible To Exact Equations
Other Methods Of Solving Differential equations
 Linear Differential equation
 Homogeneous differential equation
 Reducible To Homogeneous Form
Reference
Differential Equations :A differential equation is an equation which involves differential coefficients
or differentials.
Ex :- 𝑒 𝑥 𝑑𝑥 + 𝑒 𝑦 𝑑𝑦 = 0
d2x/dx2 + n2x = 0
Differntial equation in which all the differential coefficients have reference to a
single independent variable is known as an ordinary differential equation.
Order & Degree :-
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The order of a differential equatio is the order of thee highest derivative
appearing .
The degree of a differential equation is the degree of the highest derivative
occurring in it, after the equation has been expressed in a form free of radicalsand
fractions.
Ex :- 𝑒 𝑥 𝑑𝑥 + 𝑒 𝑦 𝑑𝑦 = 0
Order :- 1
Degree :- 1
d2x/dx2 + n2x = 0
Order :- 2
Degree :- 1
Exact Differential Equation :An exact differential equation is formed directly differentiating its
primitive(solution) without any other process.
Mdx + Ndy = 0
Is said to be an exact differential equation if it satisfies the following condition
𝝏𝑴
𝝏𝒚
Where
𝝏𝑴
𝝏𝒚
constant and
=
𝝏𝑵
𝝏𝒙
denotes the differential coefficient of M with respect to y keeping x
𝝏𝑵
𝝏𝒙
, the differential coefficient of N with respect to x, keeping y
constant.
Method For Solving Exact Differential Equation :Step 1 : Integrate M w.r.t. x keeping y constant.
Mathematics: RC6901B46
Step 2 : Integrate y, only those terms of N which do not contain x.
Step 3 : Result of 1 + Result of 2 = constant.
∫ 𝑀𝑑𝑥 + ∫(𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑁 𝑛𝑜𝑡 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑥)𝑑𝑦 = 𝐶
(y constant)
Ex :- Solve
(5𝑥 4 + 3𝑥 2 𝑦 2 − 2𝑥𝑦 3 )𝑑𝑥 + (2𝑥 3 𝑦 − 3𝑥 2 𝑦 2 − 5𝑦 4 )𝑑𝑦 = 0
Solution :- Here M = 5𝑥 4 + 3𝑥 2 𝑦 2 − 2𝑥𝑦 3
N = 2𝑥 3 𝑦 − 3𝑥 2 𝑦 2 − 5𝑦 4
𝜕𝑀
𝜕𝑦
Since
Now
𝜕𝑁
= 6𝑥 2 𝑦 − 6𝑥𝑦 2
∂M
∂y
=
∂N
∂x
𝜕𝑥
= 6𝑥 2 𝑦 − 6𝑥𝑦 2
, the given equation is exact.
∫ 𝑀𝑑𝑥 + ∫(𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑁 𝑛𝑜𝑡 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑥)𝑑𝑦 = 𝐶
(y constant)
∫(5𝑥 4 + 3𝑥 2 𝑦 2 − 2𝑥𝑦 3 )𝑑𝑥 + ∫ −5 𝑦 4 𝑑𝑦 = 𝐶
𝑥 5 + 𝑥 3𝑦2 − 𝑥 2𝑦3 − 𝑦5 = 𝐶
Ans.
Equations reducible to The Exact Equations :Sometimes a differential equation which is not exact may become so, on
multiplication by a suitable function known as the integrating factor.
RULE 1 :If (
𝜕𝑀
𝜕𝑦
−
𝜕𝑁
𝜕𝑥
)/N is a function of x alone, say f(x), then
I F = 𝒆∫ 𝒇(𝒙)𝒅𝒙
RULE 2 :-
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Mathematics: RC6901B46
If (
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
6
)/𝑀𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 𝑎𝑙𝑜𝑛𝑒, 𝑠𝑎𝑦 𝑓(𝑦), 𝑡ℎ𝑒𝑛
I F = 𝒆∫ 𝒇(𝒚)𝒅𝒚
RULE 3 :If M is of the form M = 𝑦𝑓 1(𝑥𝑦) and N is of the form N = 𝑥𝑓 2(𝑥𝑦) then,
IF =
𝟏
𝑴𝒙−𝑵𝒚
RULE 4 :Homogeneous equation . If 𝑀𝑑𝑥 + 𝑁𝑑𝑦 = 0 be a homogeneous equation in x
and y, then
IF =
𝟏
𝑴𝒙+𝑵𝒚
Other Methods Of Solving Differential Equations :Linear Differential Equation:A differential equation is said to be linear if the dependent variable and its
differential coefficients occur only in the first degree and not multiplied together.
Thus the standard form of a linear equation of the first order, commonly
known as Leibnitz’s linear equations, is
𝑑𝑦
𝑑𝑥
+ 𝑃𝑦 = 𝑄
Where P, Q are the functions of x.
To solve the equation, multiply both sides by
IF = 𝒆∫ 𝑷𝒅𝒙
𝐝𝐲
𝐝𝐱
i.e
𝐞∫ 𝐏𝐝𝐱 + 𝐲𝐞∫ 𝐏𝐝𝐱 = 𝐐𝐞∫ 𝐏𝐝𝐱
𝐝
𝐝𝐱
(𝐲𝐞∫ 𝐏𝐝𝐱 ) = 𝐐𝐞∫ 𝐏𝐝𝐱
Mathematics: RC6901B46
Integrating both sides , we get
y𝒆∫ 𝑷𝒅𝒙 = ∫ 𝑸𝒆∫ 𝑷𝒅𝒙 𝒅𝒙 + 𝑪
as the required solution.
Ex :- Solve
(𝑥 + 1)
𝑑𝑦
𝑑𝑥
− 𝑦 = 𝑒 3𝑥 (𝑥 + 1)2
Sol. Dividing throughout by (𝑥 + 1), given equation becomes
𝑑𝑦
𝑑𝑥
−
𝑦
𝑥+1
Here P = −
= 𝑒 3𝑥 (𝑥 + 1) which is Leibnitz’s equation.
1
𝑥+1
and ∫ 𝑃𝑑𝑥 = − log(𝑥 + 1) = log(𝑥 + 1)-1
Therefore,
IF = 𝐞∫ 𝐏𝐝𝐱
1
= 𝑒 𝑙𝑜𝑔𝑥+1
=
1
𝑥+1
Thus the solution of above equation is
𝑦(𝐼𝐹) = ∫[𝑒 3𝑥 (𝑥 + 1)](𝐼𝐹) 𝑑𝑥 + 𝑐
Or
𝑦
𝑥+1
= ∫ 𝑒 3𝑥 𝑑𝑥
1
= 𝑒 3𝑥 + 𝑐
3
Or
1
𝑦 = ( 𝑒 3𝑥 + 𝑐) (𝑥 + 1)
3
Homogeneous Equations :-
Ans.
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Mathematics: RC6901B46
These are of the form
𝑑𝑦
𝑑𝑥
=
8
𝑓(𝑥,𝑦)
∅(𝑥,𝑦)
Where 𝑓(𝑥, 𝑦)𝑎𝑛𝑑 ∅(𝑥, 𝑦)are homogeneous functions of the same degree in x and
y.
To solve homogeneous equations
(i) Put y=vx
Then
𝑑𝑦
𝑑𝑥
=𝑣+𝑥
𝑑𝑣
𝑑𝑥
(ii) Separate the variables v and x, and integrate.
Equations Reducible To Homogeneous Form:The equation of the form
𝑑𝑦
𝑑𝑥
=
𝑎𝑥+𝑏𝑦+𝑐
𝑎′ 𝑥+𝑏′ 𝑦+𝑐′
Can be reduced to the homogeneous form as follows:Case 1 :𝑎
𝑎′
=
𝑏
𝑏′
Putting
𝑥 = 𝑋 + ℎ , 𝑦 = 𝑌+k (h,k being constants)
So that
𝑑𝑥 = 𝑑𝑋
𝑑𝑦 = 𝑑𝑌
Now the given equation becomes
𝑑𝑌
𝑑𝑋
=
𝑎𝑋+𝑏𝑌+(𝑎ℎ+𝑏𝑘+𝑐)
𝑎′ 𝑋+𝑏′ 𝑌+(𝑎′ ℎ+𝑏′ 𝑘+𝑐 ′ )
----------------------(1)
Choose h,k so that above equation may become homogeneous
Put
𝑎ℎ + 𝑏𝑘 + 𝑐 = 0
Mathematics: RC6901B46
𝑎′ ℎ + 𝑏 ′ 𝑘 + 𝑐 ′ = 0
ℎ
So that
Or
=
𝑏𝑐 ′ −𝑏′𝑐
𝑘
𝑐𝑎′ −𝑐′𝑎
=
1
𝑎𝑏′ −𝑏𝑎′
𝑏𝑐 ′ −𝑏′𝑐
ℎ=
𝑎𝑏′ −𝑏𝑎′
𝑐𝑎′ −𝑐′𝑎
𝑘=
𝑎𝑏′ −𝑏𝑎′
Thus when 𝑎𝑏 ′ − 𝑏𝑎′ ≠ 0 (1) becomes
𝑑𝑋
=
𝑑𝑌
𝑎𝑋+𝑏𝑌
𝑎′ 𝑋+𝑏′𝑌
Which is homogeneous in X,Y and can be solved by putting
𝑌 = 𝑣𝑋
Case 2 :When
𝑎
𝑎′
≠
𝑏
𝑏′
i.e. 𝑎𝑏 ′ − 𝑏𝑎′ = 0, the above method fails as h and k become infinite or
indeterminate.
Now,
𝑎
𝑎′
=
𝑏
𝑏′
=
1
𝑚
(say)
𝑎′ = 𝑎𝑚
𝑏 ′ = 𝑏𝑚
Then given equation becomes
𝑑𝑦
𝑑𝑥
(𝑎𝑥=𝑏𝑦)+𝑐
= (𝑎𝑥+𝑏𝑦)𝑚+𝑐′
Put 𝑎𝑥 + 𝑏𝑦 = 𝑡, so that
𝑎+𝑏
𝑑𝑦
𝑑𝑥
=
𝑑𝑡
𝑑𝑥
-----------------------------------(2)
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Mathematics: RC6901B46
𝑑𝑦
Or
1
10
𝑑𝑡
= ( − 𝑎)
𝑏 𝑑𝑥
𝑑𝑥
Therefore (2) becomes
1
𝑑𝑡
𝑡+𝑐
( − 𝑎) = 𝑚𝑡=𝑐′
𝑏 𝑑𝑥
𝑑𝑡
Or
=𝑎+
𝑑𝑥
=
𝑏𝑡+𝑏𝑐
𝑚𝑡+𝑐 ′
(𝑎𝑚+𝑏)𝑡+𝑎𝑐 ′ +𝑏𝑐
𝑚𝑡+𝑐′
So that the variables are separable. In this solution , putting
𝑡 = 𝑎𝑥 + 𝑏𝑦, we get the required solution for the given differential equation.
Ex :- Solve
(3𝑦 + 2𝑥 + 4)𝑑𝑥 − (4𝑥 + 6𝑦 + 5)𝑑𝑦 = 0
Sol.
Given equation is
𝑑𝑦
𝑑𝑥
=
Putting 2𝑥 + 3𝑦 = 𝑡
So that
2+3
𝑑𝑦
=
𝑑𝑥
𝑑𝑡
𝑑𝑥
Therefore (1) becomes
1 𝑑𝑡
𝑡+4
( − 2) = 2𝑡+5
3 𝑑𝑥
Or
𝑑𝑡
𝑑𝑥
=2+
=
Or
2𝑡+5
7𝑡+22
3𝑡+12
2𝑡+5
7𝑡+22
2𝑡+5
𝑑𝑡 = 𝑑𝑥
Integrating both sides,
(2𝑥+3𝑦)+4
2(2𝑥+3𝑦)+5
------------------------(1)
Mathematics: RC6901B46
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2𝑡+5
∫ 7𝑡+22 𝑑𝑡 = ∫ 𝑑𝑥 + 𝑐
2
9
1
Or ∫ ( − .
) 𝑑𝑡 = 𝑥 + 𝑐
7
7 7𝑡+22
Or
2
7
𝑡−
9
49
log(7𝑡 + 22) = 𝑥 + 𝑐
Putting 𝑡 = 2𝑥 + 3𝑦, we have
14(2𝑥 + 3𝑦) − 9𝑙𝑜𝑓(14𝑥 + 21𝑦 + 22) = 49𝑥 + 49𝑐
Or 21𝑥 − 42𝑦 + 9 log(14𝑥 + 21𝑦 + 22) = 𝑐′ which is the required solution.
Work Comparison:There is no much difference in the work done in all the methods. Just the
main task is to find out the solution. In Exact differential equation type mainly the
its type should be satisfied. While in others also their type should be fulfilled first
then only the appropriate method can be applied. In Linear equations main
concentration is on Integrating Factor while in Homogeneous Equation substituting
of the variables with certain appropriate term is done so that we can solve it very
easily. For equations reducible to Homogeneous form its two cases should be
accomplished first then only the method of homogeneous equation is applied.
Reference: http://en.wikipedia.org/exactdifferentialequation
 http://www.mathforu.com
 http://www.mathsworld.com
Mathematics: RC6901B46
Books: Engineering mathematics ----B.S. Grewal
 Engineering mathematins-----H. K. Dass
 Differential Calculus-------Narayana shanty
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