UNDETERMINED COEFFICIENT

UNDETERMIEND COEFFICIENT AND ITS EXAMPLES AND APPLICATIONS>

1. Presented By: Amenah Gondal Presented To: Ma’am Mehwish Class: BS.ED(IV ) ORDINARY DIFFERENTIAL EQUATIONS

2. We will discuss: What is method of Undetermined Coefficient? Its detail The table to find solution of P.I. Three Roots to find solution of C.F. Applications of Differential Equation Example Objectives:

3. 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

4. The method of undetermined coefficients which can prove simpler in finding the particular integral of the equation when is an exponential function: ( a polynomial: sinusoidal function the more general case in which I sum of a product of terms of the above types, such as   Undetermined co-efficient

5. is of the form Take as +ve integer) +ve integer) C C   C C     The P.I . will be constructed according to the following table:

6. In , k is the smallest + ve integer which will ensure that no terms in is already in the C.F . If is sum of several terms, write for each term individually and then add up all of them . The and its derivative will be substituted into the equation and coefficients of like terms on the left hand and right hand sides will be equated to determine the U.C.   properties

7. Real and Distinct Repeated Real Roots Complex Roots   Three types of roots

8. In  medicine  for modelling cancer growth or the spread of disease In  engineering  for describing the movement of electricity In  chemistry  for modelling chemical reactions. In   economics  to find optimum investment strategies 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

9. Solve: ---(1) Solution: The characteristic eq. is For C.F. So is given by,   Example:

10. For P.I. Using the table, is given by Since is already present in C.F. so we take k=1 so that no term of C.F. is in .Hence the modified P.I. is Differentiate w.r.t x Differentiate again w.r.t x   CONTD….

11. Substituting for , , into (1), we have 2( Equating coefficients of like terms, we obtain Coeff . of : Coeff . of x: Coeff . of :   Contd ……

12. So, The general solution is given by,         Contd ….