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Presented By:
Amenah Gondal
Presented To:
Ma’am Mehwish
Class:
BS.ED(IV)
ORDINARY DIFFERENTIAL
EQUATIONS
We will discuss:
ο‚‘What is method of Undetermined
Coefficient?
ο‚‘Its detail
ο‚‘The table to find solution of P.I.
ο‚‘Three Roots to find solution of C.F.
ο‚‘Applications of Differential Equation
ο‚‘Example
OBJECTIVES:
In mathematics, the method of
undetermined coefficient is an approach to
finding a particular solution to certain
nonhomogeneous ordinary differential
equations.
METHOD OF UNDETERMINED
COEFFICIENT
The method of undetermined coefficients which
can prove simpler in finding the particular integral
of the equation
𝒇 𝑫 π’š = 𝑭(𝒙)
when 𝐹 π‘₯ is
ο‚‘ an exponential function: (𝑒 π‘Žπ‘₯ )
ο‚‘ a polynomial: (𝑏0 π‘₯ 𝑛 + 𝑏1 π‘₯ π‘›βˆ’1 + β‹― + 𝑏 𝑛)
ο‚‘ sinusoidal function (𝑠𝑖𝑛𝛼π‘₯ π‘œπ‘Ÿ π‘π‘œπ‘ π›Όπ‘₯)
ο‚‘ the more general case in which 𝐹(π‘₯) I sum of a
product of terms of the above types, such as
𝐹 π‘₯ = 𝑒 π‘Žπ‘₯ (𝑏0 π‘₯ 𝑛 + 𝑏1 π‘₯ π‘›βˆ’1 + β‹― … … + 𝑏 𝑛)
𝑠𝑖𝑛𝛼π‘₯
π‘π‘œπ‘ π›Όπ‘₯
UNDETERMINED CO-EFFICIENT
𝑭 𝒙 is of the
form
Take π’š 𝒑 as
1. π‘Ž 𝐴π‘₯ π‘˜
1. π‘Žπ‘₯ 𝑛
(𝑛 𝑖𝑠 π‘Ž +ve integer)
π‘₯ π‘˜
(𝐴0 π‘₯ 𝑛
+ 𝐴1 π‘₯ π‘›βˆ’1
+ β‹― … … . 𝐴 π‘›βˆ’1 π‘₯ + 𝐴 𝑛)𝑒 π‘Ÿπ‘₯
1. π‘Žπ‘₯ 𝑛 𝑒 π‘Ÿπ‘₯
(𝑛 𝑖𝑠 π‘Ž +ve integer)
π‘₯ π‘˜(𝐴0 π‘₯ 𝑛 + 𝐴1 π‘₯ π‘›βˆ’1 + β‹― … … . 𝐴 π‘›βˆ’1 π‘₯ + 𝐴 𝑛)𝑒 π‘Ÿπ‘₯
1. Cπ‘₯ 𝑛 π‘π‘œπ‘ π‘Žπ‘₯
2. Cπ‘₯ 𝑛
π‘ π‘–π‘›π‘Žπ‘₯
π‘₯ π‘˜(π΄π‘π‘œπ‘ π‘Žπ‘₯ + π΅π‘ π‘–π‘›π‘Žπ‘₯)
1. Cπ‘₯ 𝑛
𝑒 π‘Ÿπ‘₯
π‘π‘œπ‘ π‘Žπ‘₯
2. Cπ‘₯ 𝑛
𝑒 π‘Ÿπ‘₯
π‘ π‘–π‘›π‘Žπ‘₯
π‘₯ π‘˜
𝐴0 π‘₯ 𝑛
+ 𝐴1 π‘₯ π‘›βˆ’1
+ β‹― … … . 𝐴 π‘›βˆ’1 π‘₯ + 𝐴 𝑛 𝑒 π‘Ÿπ‘₯
π‘π‘œπ‘ 
+ (𝐡0 π‘₯ 𝑛 + 𝐡1 π‘₯ π‘›βˆ’1 + β‹― … … . 𝐡 π‘›βˆ’1 π‘₯
+ 𝐡𝑛)𝑒 π‘Ÿπ‘₯
π‘ π‘–π‘›π‘Žπ‘₯
THE P.I. WILL BE CONSTRUCTED ACCORDING TO
THE FOLLOWING TABLE:
ο‚‘ In π‘₯ π‘˜ , k is the smallest +ve integer which will ensure that
no terms in π’š 𝒑 is already in the C.F.
ο‚‘ If 𝐹(π‘₯) is sum of several terms, write π’š 𝒑 for each term
individually and then add up all of them.
ο‚‘ The π’š 𝒑 and its derivative will be substituted into the
equation 𝑓 𝐷 𝑦 = 𝐹 π‘₯ and coefficients of like terms on
the left hand and right hand sides will be equated to
determine the U.C. 𝐴 π‘œ, 𝐴1, … … 𝐴 𝑛 , 𝐡0, 𝐡1, … … 𝐡 𝑛.
PROPERTIES
ο‚‘ Real and Distinct
𝑦 𝑐 = 𝑐1 𝑒 π‘š1π‘₯ + 𝑐2 𝑒 π‘š2π‘₯ + β‹― 𝑐 𝑛 𝑒 π‘šπ‘› π‘₯
ο‚‘ Repeated Real Roots
𝑦 𝑐 = (𝑐1 + 𝑐2 π‘₯ + β‹― 𝑐 𝑛)π‘₯ 𝑒 π‘š1π‘₯
ο‚‘ Complex Roots
𝑦 𝑐 = 𝑒 π‘Ž π‘₯
[𝑐1 sin 𝑏π‘₯ + 𝑐2 cos 𝑏π‘₯]
THREE TYPES OF ROOTS
ο‚‘ In medicine for modelling cancer growth or the
spread of disease
ο‚‘ In engineering for describing the movement of
electricity
ο‚‘ In chemistry for modelling chemical reactions.
ο‚‘ In economics to find optimum investment strategies
ο‚‘ In physics to describe the motion of waves,
pendulums or chaotic systems. It is also used in
physics with Newton's Second Law of Motion and the
Law of Cooling.
APPLICATIONS OF DIFFERENTIAL
EQUATIONS
Solve:
π’šβ€²β€²
βˆ’ πŸ‘π’šβ€²
+ πŸπ’š = 𝒙 πŸ‘
𝒆 𝒙
---(1)
Solution:
The characteristic eq. is
𝐷2 βˆ’ 3𝐷 + 2 = π‘₯3 𝑒 π‘₯
For C.F.
𝐷2 βˆ’ 2𝐷 βˆ’ 𝐷 + 2 = 0
𝐷 𝐷 βˆ’ 2 βˆ’ 1 𝐷 βˆ’ 2 = 0
𝐷 βˆ’ 1 𝐷 βˆ’ 2 = 0
𝐷 = 2 , 𝐷 = 1
So 𝑦 𝑐 is given by,
𝑦 𝑐 = 𝑐1 𝑒 π‘₯ + 𝑐2 𝑒2π‘₯
EXAMPLE:
For P.I.
Using the table, 𝑦 𝑝 is given by
𝑦 𝑝 = π‘₯ π‘˜
(𝐴π‘₯2
+ 𝐡π‘₯ + 𝐢)𝑒 π‘₯
Since 𝑒 π‘₯ is already present in C.F. so we take k=1 so that no
term of C.F. is in 𝑦 𝑝.Hence the modified P.I. is
𝑦 𝑝 = (𝐴π‘₯3
+ 𝐡π‘₯2
+ 𝐢π‘₯)𝑒 π‘₯
Differentiate w.r.t x
𝑦 𝑝
β€² = 𝐴π‘₯3 + 𝐡 π‘₯2 + 𝐢π‘₯ 𝑒 π‘₯ + (3𝐴π‘₯2+2𝐡π‘₯ + 𝐢)𝑒 π‘₯
Differentiate again w.r.t x
𝑦 𝑝
β€²β€²
= 𝐴π‘₯3 + 𝐡π‘₯2 + 𝐢π‘₯ 𝑒 π‘₯ + 2(3𝐴π‘₯2 + 2𝐡π‘₯ + 𝐢)𝑒 π‘₯ + (6𝐡π‘₯ + 2𝐡)𝑒 π‘₯
CONTD….
Substituting for 𝑦 𝑝, 𝑦 𝑝
β€² , 𝑦 𝑝
β€²β€² into (1), we have
𝐴π‘₯3
+ 𝐡π‘₯2
+ 𝐢π‘₯ 𝑒 π‘₯
+ 2(3𝐴π‘₯2
+ 2𝐡π‘₯ + 𝐢)𝑒 π‘₯
+ 6𝐡π‘₯ + 2𝐡 𝑒 π‘₯
βˆ’
3 𝐴π‘₯3 + 𝐡π‘₯2 + 𝐢π‘₯ 𝑒 π‘₯ βˆ’ 3(3𝐴π‘₯2 + 2𝐡π‘₯ + 𝐢)𝑒 π‘₯ +2(𝐴π‘₯3 + 𝐡π‘₯2 +
𝐢π‘₯)𝑒 π‘₯ = π‘₯3 𝑒 π‘₯
[βˆ’3𝐴π‘₯2 6𝐴 βˆ’ 2𝐡 π‘₯ + 𝐢] = π‘₯3 𝑒 π‘₯
Equating coefficients of like terms, we obtain
Coeff. of π‘₯3:
βˆ’3𝐴 = 1 π‘œπ‘Ÿ 𝐴 = βˆ’
1
3
Coeff. of x:
6𝐴 βˆ’ 2𝐡 = 0 π‘œπ‘Ÿ 𝐡 = βˆ’1
Coeff. of π‘₯0:
𝐢 = 0
CONTD……
So,
𝑦 𝑝 = βˆ’
1
3
π‘₯3 𝑒 π‘₯ βˆ’ π‘₯2 𝑒 π‘₯
The general solution is given by,
𝑦 = 𝑦 𝑐 + 𝑦 𝑝
π’š = 𝒄 𝟏 𝒆 𝒙 + 𝒄 𝟐 𝒆 πŸπ’™ βˆ’
𝟏
πŸ‘
𝒙 πŸ‘ 𝒆 𝒙 βˆ’ 𝒙 𝟐 𝒆 𝒙
CONTD….
UNDETERMINED COEFFICIENT

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UNDETERMINED COEFFICIENT

  • 1.
  • 2. Presented By: Amenah Gondal Presented To: Ma’am Mehwish Class: BS.ED(IV) ORDINARY DIFFERENTIAL EQUATIONS
  • 3. We will discuss: ο‚‘What is method of Undetermined Coefficient? ο‚‘Its detail ο‚‘The table to find solution of P.I. ο‚‘Three Roots to find solution of C.F. ο‚‘Applications of Differential Equation ο‚‘Example OBJECTIVES:
  • 4. In mathematics, the method of undetermined coefficient is an approach to finding a particular solution to certain nonhomogeneous ordinary differential equations. METHOD OF UNDETERMINED COEFFICIENT
  • 5. The method of undetermined coefficients which can prove simpler in finding the particular integral of the equation 𝒇 𝑫 π’š = 𝑭(𝒙) when 𝐹 π‘₯ is ο‚‘ an exponential function: (𝑒 π‘Žπ‘₯ ) ο‚‘ a polynomial: (𝑏0 π‘₯ 𝑛 + 𝑏1 π‘₯ π‘›βˆ’1 + β‹― + 𝑏 𝑛) ο‚‘ sinusoidal function (𝑠𝑖𝑛𝛼π‘₯ π‘œπ‘Ÿ π‘π‘œπ‘ π›Όπ‘₯) ο‚‘ the more general case in which 𝐹(π‘₯) I sum of a product of terms of the above types, such as 𝐹 π‘₯ = 𝑒 π‘Žπ‘₯ (𝑏0 π‘₯ 𝑛 + 𝑏1 π‘₯ π‘›βˆ’1 + β‹― … … + 𝑏 𝑛) 𝑠𝑖𝑛𝛼π‘₯ π‘π‘œπ‘ π›Όπ‘₯ UNDETERMINED CO-EFFICIENT
  • 6. 𝑭 𝒙 is of the form Take π’š 𝒑 as 1. π‘Ž 𝐴π‘₯ π‘˜ 1. π‘Žπ‘₯ 𝑛 (𝑛 𝑖𝑠 π‘Ž +ve integer) π‘₯ π‘˜ (𝐴0 π‘₯ 𝑛 + 𝐴1 π‘₯ π‘›βˆ’1 + β‹― … … . 𝐴 π‘›βˆ’1 π‘₯ + 𝐴 𝑛)𝑒 π‘Ÿπ‘₯ 1. π‘Žπ‘₯ 𝑛 𝑒 π‘Ÿπ‘₯ (𝑛 𝑖𝑠 π‘Ž +ve integer) π‘₯ π‘˜(𝐴0 π‘₯ 𝑛 + 𝐴1 π‘₯ π‘›βˆ’1 + β‹― … … . 𝐴 π‘›βˆ’1 π‘₯ + 𝐴 𝑛)𝑒 π‘Ÿπ‘₯ 1. Cπ‘₯ 𝑛 π‘π‘œπ‘ π‘Žπ‘₯ 2. Cπ‘₯ 𝑛 π‘ π‘–π‘›π‘Žπ‘₯ π‘₯ π‘˜(π΄π‘π‘œπ‘ π‘Žπ‘₯ + π΅π‘ π‘–π‘›π‘Žπ‘₯) 1. Cπ‘₯ 𝑛 𝑒 π‘Ÿπ‘₯ π‘π‘œπ‘ π‘Žπ‘₯ 2. Cπ‘₯ 𝑛 𝑒 π‘Ÿπ‘₯ π‘ π‘–π‘›π‘Žπ‘₯ π‘₯ π‘˜ 𝐴0 π‘₯ 𝑛 + 𝐴1 π‘₯ π‘›βˆ’1 + β‹― … … . 𝐴 π‘›βˆ’1 π‘₯ + 𝐴 𝑛 𝑒 π‘Ÿπ‘₯ π‘π‘œπ‘  + (𝐡0 π‘₯ 𝑛 + 𝐡1 π‘₯ π‘›βˆ’1 + β‹― … … . 𝐡 π‘›βˆ’1 π‘₯ + 𝐡𝑛)𝑒 π‘Ÿπ‘₯ π‘ π‘–π‘›π‘Žπ‘₯ THE P.I. WILL BE CONSTRUCTED ACCORDING TO THE FOLLOWING TABLE:
  • 7. ο‚‘ In π‘₯ π‘˜ , k is the smallest +ve integer which will ensure that no terms in π’š 𝒑 is already in the C.F. ο‚‘ If 𝐹(π‘₯) is sum of several terms, write π’š 𝒑 for each term individually and then add up all of them. ο‚‘ The π’š 𝒑 and its derivative will be substituted into the equation 𝑓 𝐷 𝑦 = 𝐹 π‘₯ and coefficients of like terms on the left hand and right hand sides will be equated to determine the U.C. 𝐴 π‘œ, 𝐴1, … … 𝐴 𝑛 , 𝐡0, 𝐡1, … … 𝐡 𝑛. PROPERTIES
  • 8. ο‚‘ Real and Distinct 𝑦 𝑐 = 𝑐1 𝑒 π‘š1π‘₯ + 𝑐2 𝑒 π‘š2π‘₯ + β‹― 𝑐 𝑛 𝑒 π‘šπ‘› π‘₯ ο‚‘ Repeated Real Roots 𝑦 𝑐 = (𝑐1 + 𝑐2 π‘₯ + β‹― 𝑐 𝑛)π‘₯ 𝑒 π‘š1π‘₯ ο‚‘ Complex Roots 𝑦 𝑐 = 𝑒 π‘Ž π‘₯ [𝑐1 sin 𝑏π‘₯ + 𝑐2 cos 𝑏π‘₯] THREE TYPES OF ROOTS
  • 9. ο‚‘ In medicine for modelling cancer growth or the spread of disease ο‚‘ In engineering for describing the movement of electricity ο‚‘ In chemistry for modelling chemical reactions. ο‚‘ In economics to find optimum investment strategies ο‚‘ In physics to describe the motion of waves, pendulums or chaotic systems. It is also used in physics with Newton's Second Law of Motion and the Law of Cooling. APPLICATIONS OF DIFFERENTIAL EQUATIONS
  • 10. Solve: π’šβ€²β€² βˆ’ πŸ‘π’šβ€² + πŸπ’š = 𝒙 πŸ‘ 𝒆 𝒙 ---(1) Solution: The characteristic eq. is 𝐷2 βˆ’ 3𝐷 + 2 = π‘₯3 𝑒 π‘₯ For C.F. 𝐷2 βˆ’ 2𝐷 βˆ’ 𝐷 + 2 = 0 𝐷 𝐷 βˆ’ 2 βˆ’ 1 𝐷 βˆ’ 2 = 0 𝐷 βˆ’ 1 𝐷 βˆ’ 2 = 0 𝐷 = 2 , 𝐷 = 1 So 𝑦 𝑐 is given by, 𝑦 𝑐 = 𝑐1 𝑒 π‘₯ + 𝑐2 𝑒2π‘₯ EXAMPLE:
  • 11. For P.I. Using the table, 𝑦 𝑝 is given by 𝑦 𝑝 = π‘₯ π‘˜ (𝐴π‘₯2 + 𝐡π‘₯ + 𝐢)𝑒 π‘₯ Since 𝑒 π‘₯ is already present in C.F. so we take k=1 so that no term of C.F. is in 𝑦 𝑝.Hence the modified P.I. is 𝑦 𝑝 = (𝐴π‘₯3 + 𝐡π‘₯2 + 𝐢π‘₯)𝑒 π‘₯ Differentiate w.r.t x 𝑦 𝑝 β€² = 𝐴π‘₯3 + 𝐡 π‘₯2 + 𝐢π‘₯ 𝑒 π‘₯ + (3𝐴π‘₯2+2𝐡π‘₯ + 𝐢)𝑒 π‘₯ Differentiate again w.r.t x 𝑦 𝑝 β€²β€² = 𝐴π‘₯3 + 𝐡π‘₯2 + 𝐢π‘₯ 𝑒 π‘₯ + 2(3𝐴π‘₯2 + 2𝐡π‘₯ + 𝐢)𝑒 π‘₯ + (6𝐡π‘₯ + 2𝐡)𝑒 π‘₯ CONTD….
  • 12. Substituting for 𝑦 𝑝, 𝑦 𝑝 β€² , 𝑦 𝑝 β€²β€² into (1), we have 𝐴π‘₯3 + 𝐡π‘₯2 + 𝐢π‘₯ 𝑒 π‘₯ + 2(3𝐴π‘₯2 + 2𝐡π‘₯ + 𝐢)𝑒 π‘₯ + 6𝐡π‘₯ + 2𝐡 𝑒 π‘₯ βˆ’ 3 𝐴π‘₯3 + 𝐡π‘₯2 + 𝐢π‘₯ 𝑒 π‘₯ βˆ’ 3(3𝐴π‘₯2 + 2𝐡π‘₯ + 𝐢)𝑒 π‘₯ +2(𝐴π‘₯3 + 𝐡π‘₯2 + 𝐢π‘₯)𝑒 π‘₯ = π‘₯3 𝑒 π‘₯ [βˆ’3𝐴π‘₯2 6𝐴 βˆ’ 2𝐡 π‘₯ + 𝐢] = π‘₯3 𝑒 π‘₯ Equating coefficients of like terms, we obtain Coeff. of π‘₯3: βˆ’3𝐴 = 1 π‘œπ‘Ÿ 𝐴 = βˆ’ 1 3 Coeff. of x: 6𝐴 βˆ’ 2𝐡 = 0 π‘œπ‘Ÿ 𝐡 = βˆ’1 Coeff. of π‘₯0: 𝐢 = 0 CONTD……
  • 13. So, 𝑦 𝑝 = βˆ’ 1 3 π‘₯3 𝑒 π‘₯ βˆ’ π‘₯2 𝑒 π‘₯ The general solution is given by, 𝑦 = 𝑦 𝑐 + 𝑦 𝑝 π’š = 𝒄 𝟏 𝒆 𝒙 + 𝒄 𝟐 𝒆 πŸπ’™ βˆ’ 𝟏 πŸ‘ 𝒙 πŸ‘ 𝒆 𝒙 βˆ’ 𝒙 𝟐 𝒆 𝒙 CONTD….