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UNDETERMINED COEFFICIENT
- 2. Presented By: Amenah Gondal Presented To: Maβam 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 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
- 6. π π is of the form Take π π as 1. π π΄π₯ π 1. ππ₯ π (π ππ π +ve integer) π₯ π (π΄0 π₯ π + π΄1 π₯ πβ1 + β― β¦ β¦ . π΄ πβ1 π₯ + π΄ π)π ππ₯ 1. ππ₯ π π ππ₯ (π ππ π +ve integer) π₯ π(π΄0 π₯ π + π΄1 π₯ πβ1 + β― β¦ β¦ . π΄ πβ1 π₯ + π΄ π)π ππ₯ 1. Cπ₯ π πππ ππ₯ 2. Cπ₯ π π ππππ₯ π₯ π(π΄πππ ππ₯ + π΅π ππππ₯) 1. Cπ₯ π π ππ₯ πππ ππ₯ 2. Cπ₯ π π ππ₯ π ππππ₯ π₯ π π΄0 π₯ π + π΄1 π₯ πβ1 + β― β¦ β¦ . π΄ πβ1 π₯ + π΄ π π ππ₯ πππ + (π΅0 π₯ π + π΅1 π₯ πβ1 + β― β¦ β¦ . π΅ πβ1 π₯ + π΅π)π ππ₯ π ππππ₯ 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 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
- 10. Solve: πβ²β² β ππβ² + ππ = π π π π ---(1) Solution: The characteristic eq. is π·2 β 3π· + 2 = π₯3 π π₯ For C.F. π·2 β 2π· β π· + 2 = 0 π· π· β 2 β 1 π· β 2 = 0 π· β 1 π· β 2 = 0 π· = 2 , π· = 1 So π¦ π is given by, π¦ π = π1 π π₯ + π2 π2π₯ EXAMPLE:
- 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¦.
- 12. Substituting for π¦ π, π¦ π β² , π¦ π β²β² into (1), we have π΄π₯3 + π΅π₯2 + πΆπ₯ π π₯ + 2(3π΄π₯2 + 2π΅π₯ + πΆ)π π₯ + 6π΅π₯ + 2π΅ π π₯ β 3 π΄π₯3 + π΅π₯2 + πΆπ₯ π π₯ β 3(3π΄π₯2 + 2π΅π₯ + πΆ)π π₯ +2(π΄π₯3 + π΅π₯2 + πΆπ₯)π π₯ = π₯3 π π₯ [β3π΄π₯2 6π΄ β 2π΅ π₯ + πΆ] = π₯3 π π₯ Equating coefficients of like terms, we obtain Coeff. of π₯3: β3π΄ = 1 ππ π΄ = β 1 3 Coeff. of x: 6π΄ β 2π΅ = 0 ππ π΅ = β1 Coeff. of π₯0: πΆ = 0 CONTDβ¦β¦
- 13. So, π¦ π = β 1 3 π₯3 π π₯ β π₯2 π π₯ The general solution is given by, π¦ = π¦ π + π¦ π π = π π π π + π π π ππ β π π π π π π β π π π π CONTDβ¦.