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Linear homogeneous ODEn with constant coefficients

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Presentation on theme: "Linear homogeneous ODEn with constant coefficients"— Presentation transcript:

1 Linear homogeneous ODEn with constant coefficients
where Let be solutions to (H), we say that they are independent if implies

2 Let be n solutions to (H), the function
is called a Wronskian. It has some interesting and useful properties

3 Let be n solutions to (H) on an interval [a,b].
Then they are independent exactly when there is an such that

4 Linear non-homogeneous ODEn with constant coefficients
where associated homogeneous equation (H) If y(x) is a solution to (H) and u(x) is a solution to (N) then y(x) + u(x) is a solution to (N)

5 General solution to (H)
Assume substituting into (H) yields (E) If all the n roots of the above equation are real and simple, we obtain the following general solution

6 If  is a root of (E) of multiplicity k, then it can be proved that there are k independent solutions ... to the equation

7 If is a complex root of equation (E), then, since the coefficients are real, there must be also a root and the following two independent functions are solutions to (H). If is a root of multiplicity k, then again we have the following 2k independent solutions of (H):

8 To find a particular solution to
we can use a variation-of-constants method. Let be a general solution to (H) suppose that the constants c1, ... , cn become functions and let us substitute into (N)

9 if we put we have if we put we have if we put we have and finally

10

11 Thus, to find the functions we have to solve the following system of n algebraic equations
This system has a solution since the determinant of the system is the Wronskian which is non-zero due to the fact that are independent.

12 ... If are the functions solving the system (S), we get
and a general solution to (H)

13 Example where u is a particular solution to the original non-homogeneous solution Let

14 We have to solve the following system of algebraic equations
If we denote by A the system matrix then, clearly,

15 Calculating A-1, we get so that and Calculating the integrals, we can now write a general solution of the original non-homogeneous differential equation

16 Note that we could also find a particular solution u to the original non-homogeneous equation by assuming it in the form Assuming just would not do since 0 is a solution to the equation Substituting u, u’, u’’, and u’’’ into the non-homogeneous equation yields which means that and we have which is another particular solution to

17 This method can be generalized for some special types of the right-hand sides
If where Pk(x) is a polynomial of degree k we can assume a solution of the form where Qk(x) is a polynomial of degree k or if 0 is a root of (E) of multiplicity m

18 If where Pk(x) is a polynomial of degree k, Ql(x) is a polynomial of degree l, we can assume a solution of the form where Rm(x) and Sm(x) are polynomials of degree m with or if is a root of (E) of multiplicity m


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