Lecture 6 new math3dadoiojai idajdiaiji (1).pptx

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1. Lecture 6 Differential equations of second – order. Reduction of order

2. Some second-order equation can be reduced to first-order equation, rendering then susceptible to the simple methods of solving equations of the first order. The following are three types of such second-order equations: Type 1 : Let (1) Integrating this equation (1), we obtain Integrating this equation once again we get

3. where Example 1 . Solve the differential equation Solution . This equation relates to the first type Then Type 2 : Second-order equations with the dependent variable mussing.

4. Example : ; Solution: = =  

5. ; = = { ; } = = .  

6. This is an equation of the type (2) The dependent variable y does not explicitly appear in the equation. This type of second-order equation is easily reduced to a first-order equation by the transformation . This substitution obviously implies , and the original equation becomes a first-order equation for p Suppose we have managed to integrate this equation and have obtained its general solution .

7. Then we have and therefore the general solution of equation (2) is thus obtained: Example 2 . Solve the differential equation Solution . Since the dependent variable y is missing, let and . These substitutions transform the given second-order equation into the first-order equation

8. ; , It is a separable equation. Separating the variables, we obtain Integrating , we have   Then or and hence So ;  

9. Example: Solution: ; ; ; ; ;  

10. . ; ; ; ;  

11. ; ; ; ; ; ; ;  

12. ; ;  

13. Type 3 . Second-order nonlinear equations with the independent variable missing. Let it be of the form ( 3) The independent variable x does not explicitly appear in the equation. The method for reducing the order of these second-order equations begins with the same substitution as for Type 2 equations, namely, replacing y / as p . But instead of simple writing y // as p / , the trick here is to express y // in terms of a first derivative with respect to y .

14. Therefore we write ; Now we deduce from equation (3) the equation   which is a first-order equation . If we manage to integrate it and to find its general solution then we have and therefore the general solution of equation (3) can be directly written in the form  

15. Example 3 . Solve the differential equation Solution. The substitutions and transform this second-order equation for y into the following first-order equation for p : Separating the variables, we obtain Integrating, we have It follows that or  

16. Now, since , this last result becomes Separating the variables, we get So or

17. Example: ; Solution: ; ; ; ; ; ; ; ; ;  

18. , , ; ; ;  

19. Second – Order Linear Differential Equations  A second-order linear differential equation has the form where and are continuous functions. If equation (1) is nonhomogeneous . If (2) equation (2) is called homogeneous linear equation.

20. Theorem 1 . If and are both solutions of the linear homogeneous equation (2) and C 1 and C 2 are any constants, then the linear combination is also a solution of equation (2) Proof. Since and are solution of (2), we have and

21. Therefore, using the basic rules for differentiation, we have Thus, is a solution of equation (2). The other fact we need is given by the following theorem. It says that the general solution is a linear combination of two linearly independent solutions and .

22. This means that neither and is a constant multiple of the other. For instance, the functions and are linearly dependent, but and are linearly independent. Theorem 2 . If and are linearly independent solutions of equation (2), then the general solution is given by where and are arbitrary constants. Definition. Let and be two differentiable functions. The Wronskian , associated to and , is the function

23. Theorem 3 . If and are linearly dependent, then Wronskian . Theorem 4 . If and are linearly independent solution of equation (2), then . Theorem 5 . If and are two solutions of equation (2), then . Proof . Since and are solutions of (2) we have (4)

24. Example: are linearly dependent or linearly independent ? Solution: .  

25. Example: ? Solution: is general solution of homogenous equation. ; Type equation here. ; ; or =0;  

26. .  

27. Example: are linearly dependent or linearly independent? Solution: linearly independent  

28. Example: are linearly dependent or linearly independent? Solution: linearly dependent  

29. Example: If are solutions, then find differential equation? Solution:I )  

30. ;.  

31. Example: ? Solution: =0 30  

32. Example: ;. ;  

33. (5) Now we multiply (4) by and multiply (5) by . Subtract the resulting two equations to obtain. (6) Recall the definition (3) and observe that Hence (6) is the equation   (7) Separating the variables, we get

34. Integrating, we have or   So   where .