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Math 3120 Differential Equations with Boundary Value Problems

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1 Math 3120 Differential Equations with Boundary Value Problems
Chapter 4: Higher-Order Differential Equations Section 4-6: Variation of Parameters

2 Ch 3.7: Variation of Parameters
Recall the nonhomogeneous equation where p, q, g are continuous functions on an open interval I. The associated homogeneous equation is In this section we will learn the variation of parameters method to solve the nonhomogeneous equation. As with the method of undetermined coefficients, this procedure relies on knowing solutions to homogeneous equation. Variation of parameters is a general method, and requires no detailed assumptions about solution form. However, certain integrals need to be evaluated, and this can present difficulties.

3 Example: Variation of Parameters (1 of 6)
We seek a particular solution to the equation below. We cannot use method of undetermined coefficients since g(t) is a quotient of sin t or cos t, instead of a sum or product. Recall that the solution to the homogeneous equation is To find a particular solution to the nonhomogeneous equation, we begin with the form Then or

4 Example: Derivatives, 2nd Equation (2 of 6)
From the previous slide, Note that we need two equations to solve for u1 and u2. The first equation is the differential equation. To get a second equation, we will require Then Next,

5 Example: Two Equations (3 of 6)
Recall that our differential equation is Substituting y'' and y into this equation, we obtain This equation simplifies to Thus, to solve for u1 and u2, we have the two equations:

6 Example: Solve for u1' (4 of 6)
To find u1 and u2 , we need to solve the equations From second equation, Substituting this into the first equation,

7 Example : Solve for u1 and u2 (5 of 6)
From the previous slide, Then Thus

8 Example: General Solution (6 of 6)
Recall our equation and homogeneous solution yC: Using the expressions for u1 and u2 on the previous slide, the general solution to the differential equation is

9 Summary Suppose y1, y2 are fundamental solutions to the homogeneous equation associated with the nonhomogeneous equation above, where we note that the coefficient on y'' is 1. To find u1 and u2, we need to solve the equations Doing so, and using the Wronskian, we obtain Thus

10 Theorem 3.7.1 Consider the equations
If the functions p, q and g are continuous on an open interval I, and if y1 and y2 are fundamental solutions to Eq. (2), then a particular solution of Eq. (1) is and the general solution is

11 Ch 4.4: Variation of Parameters
The variation of parameters method can be used to find a particular solution of the nonhomogeneous nth order linear differential equation provided g is continuous. As with 2nd order equations, begin by assuming y1, y2 …, yn are fundamental solutions to homogeneous equation. Next, assume the particular solution Y has the form where u1, u2,… un are functions to be solved for. In order to find these n functions, we need n equations.

12 Variation of Parameters Derivation (2 of 5)
First, consider the derivatives of Y: If we require then Thus we next require Continuing in this way, we require and hence

13 Variation of Parameters Derivation (3 of 5)
From the previous slide, Finally, Next, substitute these derivatives into our equation Recalling that y1, y2 …, yn are solutions to homogeneous equation, and after rearranging terms, we obtain

14 Variation of Parameters Derivation (4 of 5)
The n equations needed in order to find the n functions u1, u2,… un are Using Cramer’s Rule, for each k = 1, …, n, and Wk is determinant obtained by replacing k th column of W with (0, 0, …, 1).

15 Variation of Parameters Derivation (5 of 5)
From the previous slide, Integrate to obtain u1, u2,… un: Thus, a particular solution Y is given by where t0 is arbitrary.

16 Example (1 of 3) Consider the equation below, along with the given solutions of corresponding homogeneous solutions y1, y2, y3: Then a particular solution of this ODE is given by It can be shown that

17 Example (2 of 3) Also,

18 Example (3 of 3) Thus a particular solution is
Choosing t0 = 0, we obtain More simply,


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