INTRODUCTION

It was pointed out in the Remarks at the end of Section 3.4 that the method of undetermined coefficients has an inherent weakness that limits its wider application to linear equations: The DE must have constant coefficients, and the input function g(x) must be of the type listed in Table 3.4.1. In this section we examine a method for determining a particular solution y_p of a nonhomogeneous linear DE that has, in theory, no such restrictions on it. This method, due to the eminent Italian astronomer and mathematician Joseph-Louis Lagrange (1736–1813), is known as variation of parameters.

Some Assumptions

To adapt the method of variation of parameters to a linear second-order differential equation

a₂(x)y″ + a₁(x)y′ + a₀(x)y = g(x), (1)

we begin as we did in Section 3.2—we put (1) in the standard form

y″ + P(x)y′ + Q(x)y = f(x) (2)

by dividing through by the lead coefficient a₂(x). Equation (2) is the second-order analogue of the linear first-order equation dy/dx + P(x)y = f(x). In (2) we shall assume P(x), Q(x), and f(x) are continuous on some common interval I. As we have already seen in Section 3.3, there is no difficulty in obtaining the complementary function y_c of (2) when the coefficients are constants.

Method of Variation of Parameters

Corresponding to the substitution y_p = u₁(x)y₁(x) that we used in Section 2.3 to find a particular solution y_p of dy/dx + P(x)y = f(x), for the linear second-order DE (2) we seek a solution of the form

y_p(x) = u₁(x)y₁(x) + u₂(x)y₂(x), (3)

where y₁ and y₂ form a fundamental set of solutions on I of the associated homogeneous form of (1). Using the product rule to differentiate y_p twice, we get

Substituting (3) and the foregoing derivatives into (2) and grouping terms yields

(4)

Because we seek to determine two unknown functions u₁ and u₂, reason dictates that we need two equations. We can obtain these equations by making the further assumption that the functions u₁ and u₂ satisfy . This assumption does not come out of the blue but is prompted by the first two terms in (4), since, if we demand that , then (4) reduces to . We now have our desired two equations, albeit two equations for determining the derivatives and . By Cramer’s rule, the solution of the system

can be expressed in terms of determinants:

(5)

where

(6)

The functions u₁ and u₂ are found by integrating the results in (5). The determinant W is recognized as the Wronskian of y₁ and y₂. By linear independence of y₁ and y₂ on I, we know that W(y₁(x), y₂(x)) ≠ 0 for every x in the interval.

Summary of the Method

Usually it is not a good idea to memorize formulas in lieu of understanding a procedure. However, the foregoing procedure is too long and complicated to use each time we wish to solve a differential equation. In this case it is more efficient simply to use the formulas in (5). Thus to solve a₂y″ + a₁y′ + a₀y = g(x), first find the complementary function y_c = c₁y₁ + c₂y₂ and then compute the Wronskian W(y₁(x), y₂(x)). By dividing by a₂, we put the equation into the standard form y″ + Py′ + Qy = f(x) to determine f(x). We find u₁ and u₂ by integrating = W₁/W and = W₂/W, where W₁ and W₂ are defined as in (6). A particular solution is y_p = u₁y₁ + u₂ y₂. The general solution of equation (1) is then y = y_c + y_p.

EXAMPLE 1 General Solution Using Variation of Parameters

EXAMPLE 2 General Solution Using Variation of Parameters

Constants of Integration

When computing the indefinite integrals of and , we need not introduce any constants. This is because

Integral-Defined Functions

We have seen several times in the preceding sections and chapters that when a solution method involves integration we may encounter nonelementary integrals. As the next example shows, sometimes the best we can do in constructing a particular solution (3) of a linear second-order differential equation is to use the integral-defined functions.

Here we assume that the integrand is continuous on the interval [x₀, x]. See Problems 27–30 in Exercises 3.5.

EXAMPLE 3 General Solution Using Variation of Parameters

In Example 3 we can integrate on any interval [x₀, x] not containing the origin. Also see Examples 2 and 3 in Section 3.10.

Higher-Order Equations

The method we have just examined for nonhomogeneous second-order differential equations can be generalized to linear nth-order equations that have been put into the standard form

y⁽ⁿ⁾ + P_n−1(x)y⁽ⁿ⁻¹⁾ + ... + P₁(x)y′ + P₀(x)y = f(x). (8)

If y_c = c₁y₁ + c₂y₂ + ... + c_ny_n is the complementary function for (8), then a particular solution is

y_p(x) = u₁(x)y₁(x) + u₂(x)y₂(x) + ... + u_n(x)y_n(x),

where the , k = 1, 2, ..., n, are determined by the n equations

(9)

The first n − 1 equations in this system, like y₁ + y₂ = 0 in (4), are assumptions made to simplify the resulting equation after y_p = u₁(x)y₁(x) + ... + u_n(x)y_n(x) is substituted in (8). In this case, Cramer’s rule gives

= k = 1, 2, ..., n,

where W is the Wronskian of y₁, y₂, ..., y_n and W_k is the determinant obtained by replacing the kth column of the Wronskian by the column consisting of the right-hand side of (9), that is, the column (0, 0, ..., f(x)). When n = 2 we get (5). When n = 3, the particular solution is y_p = u₁y₁ + u₂y₂ + u₃y₃, where y₁, y₂, and y₃ constitute a linearly independent set of solutions of the associated homogeneous DE, and u₁, u₂, u₃ are determined from

(10)

See Problems 33 and 34 in Exercises 3.5.

Particular Solution May Not Be Unique

In the problems in Exercises 3.5, do not hesitate to simplify, if possible, the form of the particular solution y_p found by variation of parameters. Depending on how the antiderivatives of and are found, you might not obtain the same y_p given in the answer section. For example, in Problem 3 in Exercises 3.5, both and are particular solutions of the differential equation If we use the first particular solution, then a general solution of the differential equation is

But note that by writing and defining another perfectly valid general solution is

3.5 Exercises Answers to selected odd-numbered problems begin on page ANS-6.

In Problems 1–20, solve each differential equation by variation of parameters.

1. y″ + y = sec x
2. y″ + y = tan x
3. y″ + y = sin x
4. y″ + y = sec θ tan θ
5. y″ + y = cos²x
6. y″ + y = sec²x
7. y″ − y = cosh x
8. y″ − y = sinh 2x
9. y″ − 9y =
10. 4y″ − y = e^x/2 + 3
11. y″ + y′ = 3
12. y″ − 4y′ + 3y = e^x
13. y″ + 3y′ + 2y =
14. y″ − 2y′ + y =
15. y″ + 3y′ + 2y = sin e^x
16. y″ − 2y′ + y = e^t arctan t
17. y″ + 2y′ + y = e^−t ln t
18. 2y″ + y′ = 6x
19. 3y″ − 6y′ + 6y = e^x sec x
20. 4y″ − 4y′ + y =

In Problems 21–26, solve each differential equation by variation of parameters subject to the initial conditions y(0) = 1, y′(0) = 0.

21. 4y″ − y = xe^x/2
22. y″ + y′ = e^−x
23. y″ − 2y′ + y = e^x sec² x
24. 2y″ + y′ − y = x + 1
25. y″ + 2y′ − 8y = 2e^−2x − e^−x
26. y″ − 4y′ + 4y = (12x² − 6x)e^2x

In Problems 27–30, proceed as in Example 3 and solve each differential equation by variation of parameters.

In Problems 31 and 32, the indicated functions are known linearly independent solutions of the associated homogeneous differential equation on the interval (0, ∞). Find the general solution of the given nonhomogeneous equation.

31. x²y″ + xy′ +(x² − )y = ; y₁ = cos x,
    y₂ = sin x
32. x²y″ + xy′ + y = sec(ln x); y₁ = cos(ln x), y₂ = sin(ln x)

In Problems 33–36, solve the given third-order differential equation by variation of parameters.

33. y‴ + y′ = tan x
34. y‴ + 4y′ = sec 2x
35. y‴ −2y″ −y′ + 2y = e^4x
36. 

Discussion Problems

In Problems 37 and 38, discuss how the methods of undetermined coefficients and variation of parameters can be combined to solve the given differential equation. Carry out your ideas.

37. 3y″ − 6y′ + 30y = 15 sin x + e^x tan 3x
38. y″ − 2y′ + y = 4x² − 3 + x⁻¹e^x
39. What are the intervals of definition of the general solutions in Problems 1, 7, 17, and 20? Discuss why the interval of definition of the general solution in Problem 32 is not (0, ∞).
40. Find the general solution of x⁴y″ + x³y′ − 4x²y = 1 given that y₁ = x² is a solution of the associated homogeneous equation.

Computer Lab Assignments

In Problems 41 and 42, the indefinite integrals of the equations in (5) are nonelementary. Use a CAS to find the first four nonzero terms of a Maclaurin series of each integrand and then integrate the result. Find a particular solution of the given differential equation.

41. y″ + y =
42.