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
π π« π = π(π)
when πΉ π₯ is
ο‘ an exponential function: (π ππ₯ )
ο‘ a polynomial: (π0 π₯ π + π1 π₯ πβ1 + β― + π π)
ο‘ sinusoidal function (π πππΌπ₯ ππ πππ πΌπ₯)
ο‘ the more general case in which πΉ(π₯) I sum of a
product of terms of the above types, such as
πΉ π₯ = π ππ₯ (π0 π₯ π + π1 π₯ πβ1 + β― β¦ β¦ + π π)
π πππΌπ₯
πππ πΌπ₯
UNDETERMINED CO-EFFICIENT
7. ο‘ In π₯ π , k is the smallest +ve integer which will ensure that
no terms in π π is already in the C.F.
ο‘ If πΉ(π₯) is sum of several terms, write π π for each term
individually and then add up all of them.
ο‘ The π π and its derivative will be substituted into the
equation π π· π¦ = πΉ π₯ and coefficients of like terms on
the left hand and right hand sides will be equated to
determine the U.C. π΄ π, π΄1, β¦ β¦ π΄ π , π΅0, π΅1, β¦ β¦ π΅ π.
PROPERTIES
8. ο‘ Real and Distinct
π¦ π = π1 π π1π₯ + π2 π π2π₯ + β― π π π ππ π₯
ο‘ Repeated Real Roots
π¦ π = (π1 + π2 π₯ + β― π π)π₯ π π1π₯
ο‘ Complex Roots
π¦ π = π π π₯
[π1 sin ππ₯ + π2 cos ππ₯]
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
11. For P.I.
Using the table, π¦ π is given by
π¦ π = π₯ π
(π΄π₯2
+ π΅π₯ + πΆ)π π₯
Since π π₯ is already present in C.F. so we take k=1 so that no
term of C.F. is in π¦ π.Hence the modified P.I. is
π¦ π = (π΄π₯3
+ π΅π₯2
+ πΆπ₯)π π₯
Differentiate w.r.t x
π¦ π
β² = π΄π₯3 + π΅ π₯2 + πΆπ₯ π π₯ + (3π΄π₯2+2π΅π₯ + πΆ)π π₯
Differentiate again w.r.t x
π¦ π
β²β²
= π΄π₯3 + π΅π₯2 + πΆπ₯ π π₯ + 2(3π΄π₯2 + 2π΅π₯ + πΆ)π π₯ + (6π΅π₯ + 2π΅)π π₯
CONTDβ¦.