1. Differential Equations MTH 242 Lecture # 11 Dr. Manshoor Ahmed

2. Summary (Recall) Construction of 2 nd solution from a known solution. Reduction of order. How to solve a homogeneous linear differential with constant coefficients? Auxiliary or characteristics equation. How to write the general of solution of linear differential with constant coefficients

3. Solution of non-homogeneous DE

4. Solution of non-homogeneous DE

5. The Method of Undetermined Coefficients

8. Caution! In addition to the form of the input function, the educated guess for must take into consideration the functions that make up the complementary function. No function in the assumed must be a solution of the associated homogeneous differential equation. This means that the assumed should not contain terms that duplicate terms in.

9. Trial particular solutions No TermsChoice for 1. 2. 3. 4.

10. Some functions and their particular solutions 1. Form of 2. 1(Any constant ) 3. 4. 5. 6. 7. 8.

11. 9. 10. 11. 12.

12. If equals a sum? Suppose that The input function consists of a sum of terms of the kind listed in the above table i.e. The trial forms corresponding to be. Then, the particular solution of the given non-homogeneous differential equation is In other words, the form of is a linear combination of all the Linearly independent functions generated by repeated differentiation of the input function.

13. Example 1 Solve the non homogenous equation by using Undetermined Coefficient Method. Solution: For complementary solution consider homogenous equation Auxiliary equation is

14. For particular solution we assume Substituting into the given differential equation Equating the coefficients of and constant we have Constant: Solving these equations we get the values of undetermined coefficients

15. Thus Hence the general solution is

16. Example 2 Solve the non homogenous equation by using Undetermined Coefficient Method. Solution:

19. Duplication between and ? If a function in the assumed is also present in then this function is a solution of the associated homogeneous differential equation. In this case the obvious assumption for the form of is not correct. In this case we suppose that the input function is made up of terms of kinds i.e. and corresponding to this input function the assumed particular solution is

20. If a contain terms that duplicate terms in, then that must be multiplied with, being the least positive integer that eliminates the duplication. Example 3 Find a particular solution of the following non-homogeneous differential Equation Solution: To find, we solve the associated homogeneous differential equation We put in the given equation, so that the auxiliary equation is

21. Thus Since Therefore Substituting in the given non-homogeneous differential equation, we obtain or Clearly we have made a wrong assumption for, as we did not remove the duplication.

22. Since is present in.Therefore, it is a solution of the associated homogeneous differential equation To avoid this we find a particular solution of the form We notice that there is no duplication between and this new Assumption for. Now Substituting in the given differential equation, we obtain or

23. So that a particular solution of the given equation is given by Hence, the general solution of the given equation is Example 4 Find a particular solution of Solution: Consider the associated homogeneous equation Put

24. Then the auxiliary equation is Roots of the auxiliary equation are real and equal. Therefore, Since Therefore, we assume that This assumption fails because of duplication between and. We multiply with. Therefore, we now assume

25. However, the duplication is still there. Therefore, we again multiply with and assume Since there is no duplication, this is acceptable form of the trial. Example 5 Solve the initial value problem Solution Associated homogeneous DE

26. Put Then the auxiliary equation is The roots of the auxiliary equation are complex. Therefore, the complementary function is Since Therefore, we assume that

27. So that Clearly, there is duplication of the functions and. To remove this duplication we multiply with. Therefore, we assume that So that Substituting into the given non-homogeneous differential equation, we have Equating constant terms and coefficients of,, we obtain

28. So that Thus Hence the general solution of the differential equation is We now apply the initial conditions to find and Since Therefore

29. Now Therefore Hence the solution of the initial value problem is Example 6 Solve the differential equation

30. Solution: The associated homogeneous differential equation is Put Then the auxiliary equation is Thus the complementary function is Since We assume that Corresponding to:

31. Thus the assumed form of the particular solution is The function in is duplicated between and. Multiplication with does not remove this duplication. However, if we multiply with, this duplication is removed. Thus the operative from of a particular solution is Then And Substituting into the given differential equation and collecting like term, we obtain

32. Equating constant terms and coefficients of and yields Solving these equations, we have the values of the unknown coefficients Thus,. Hence, the general solution

33. Higher –Order Equation The method of undetermined coefficients can also be used for higher order equations of the form with constant coefficients. The only requirement is that consists of the proper kinds of functions as discussed earlier. Example 7 Solve Solution: To find the complementary function we solve the associated homogeneous differential equation

34. Put Then the auxiliary equation is Or The auxiliary equation has equal and distinct real roots. Therefore, the complementary function is Since Therefore, we assume that Clearly, there is no duplication of terms between and.

37. Summary (Recall)