5.3 Series Solutions near an Ordinary Point, Part II
3.4
Complex Roots of the Characteristic Equation
1
2
set of solutions on that interval.
1
2
a fundamental set of solutions on that interval.
1
2
0
0
3.4 Complex Roots of the Characteristic Equation
We continue our discussion of the equation ay + by + cy = 0,
(1)
where a, b, and c are given real numbers. In Section 3.1 we found that if we seek solutions of the form y = ert , then r must be a root of the characteristic equation ar 2 + br + c = 0.
(2)
If the roots r and r are real and different, which occurs whenever the discriminant
1
2
b2 − 4ac is positive, then the general solution of Eq. (1) is y = c er t
t
1
+ c er2 .
(3)
1
2
Suppose now that b2 − 4ac is negative. Then the roots of Eq. (2) are conjugate complex numbers; we denote them by r = λ + iµ,r = λ − iµ,
(4)
1
2
where λ and µ are real. The corresponding expressions for y are y (t) = exp[(λ + iµ)t],y (t) = exp[(λ − iµ)t].
(5)
1
2
Our first task is to explore what is meant by these expressions, which involve evaluating the exponential function for a complex exponent. For example, if λ = −1, µ = 2, and t = 3, then from Eq. (5) y (3) = e−3+6i .
(6)
1
What does it mean to raise the number e to a complex power? The answer is provided by an important relation known as Euler’s formula.
Chapter 3. Second Order Linear EquationsEuler’s Formula.
To assign a meaning to the expressions in Eqs. (5) we need to give a definition of the complex exponential function. Of course, we want the definition to reduce to the familiar real exponential function when the exponent is real. There are several ways to accomplish this extension of the exponential function. Here we use a method based on infinite series; an alternative is outlined in Problem 28.
Recall from calculus that the Taylor series for et about t = 0 is
∞
tn
et = , −∞ < t < ∞.
(7) n!
n=0
If we now assume that we can substitute i t for t in Eq. (7), then we have
∞ (it)neit = n!
n=0
∞
(−
∞
(−
=
1)nt2n + 1)n−1t2n−1 ,
(
i
(8) 2n)!
(2n − 1)! n=0
n=1
where we have separated the sum into its real and imaginary parts, making use of the fact that i 2 = −1, i3 = −i, i4 = 1, and so forth. The first series in Eq. (8) is precisely the Taylor series for cos t about t = 0, and the second is the Taylor series for sin t
about t = 0. Thus we have eit = cos t + i sin t.
(9)
Equation (9) is known as Euler’s formula and is an extremely important mathematical relationship. While our derivation of Eq. (9) is based on the unverified assumption that the series (7) can be used for complex as well as real values of the independent variable, our intention is to use this derivation only to make Eq. (9) seem plausible. We now put matters on a firm foundation by adopting Eq. (9) as the definition of eit . In other words, whenever we write eit , we mean the expression on the right side of Eq. (9).
There are some variations of Euler’s formula that are also worth noting. If we replace t by −t in Eq. (9) and recall that cos(−t) = cos t and sin(−t) = − sin t, then we have e−it = cos t − i sin t.
(10)
Further, if t is replaced by µt in Eq. (9), then we obtain a generalized version of Euler’s formula, namely, eiµt = cos µt + i sin µt.
(11)
Next, we want to extend the definition of the exponential function to arbitrary complex exponents of the form (λ + iµ)t. Since we want the usual properties of the exponential function to hold for complex exponents, we certainly want exp[(λ + iµ)t] to satisfy e(λ+iµ)t = eλt eiµt .
(12)
Then, substituting for eiµt from Eq. (11), we obtain e(λ+iµ)t = eλt (cos µt + i sin µt) = eλt cos µt + ieλt sin µt.
(13)
We now take Eq. (13) as the definition of exp[(λ + iµ)t]. The value of the exponential function with a complex exponent is a complex number whose real and imaginary parts are given by the terms on the right side of Eq. (13). Observe that the real and imaginary parts of exp[(λ + iµ)t] are expressed entirely in terms of elementary realvalued functions. For example, the quantity in Eq. (6) has the value e−3+6i = e−3 cos 6 + ie−3 sin 6 ∼ = 0.0478041 − 0.0139113i.
With the definitions (9) and (13) it is straightforward to show that the usual laws of exponents are valid for the complex exponential function. It is also easy to verify that the differentiation formula d (ert) = rert
(14)
dt
also holds for complex values of r .
Real-Valued Solutions.
The functions y (t) and y (t), given by Eqs. (5) and with
1
2
the meaning expressed by Eq. (13), are solutions of Eq. (1) when the roots of the characteristic equation (2) are complex numbers λ ± iµ. Unfortunately, the solutions y1 and y are complex-valued functions, whereas in general we would prefer to have real 2
valued solutions, if possible, because the differential equation itself has real coefficients.
Such solutions can be found as a consequence of Theorem 3.2.2, which states that if y and y are solutions of Eq. (1), then any linear combination of y and y is also a
1
2
1
2
solution. In particular, let us form the sum and then the difference of y and y . We
1
2
have y (t) + y (t) = eλt (cos µt + i sin µt) + eλt (cos µt − i sin µt)
1
2
= 2eλt cos µt
and y (t) − y (t) = eλt (cos µt + i sin µt) − eλt (cos µt − i sin µt)
1
2
= 2ieλt sin µt.
Hence, neglecting the constant multipliers 2 and 2i , respectively, we have obtained a pair of real-valued solutions u(t) = eλt cos µt,v(t) = eλt sin µt.
(15)
Observe that u and v are simply the real and imaginary parts, respectively, of y .
1
By direct computation you can show that the Wronskian of u and v is W (u, v)(t) = µe2λt .
(16)
Thus, as long as µ = 0, the Wronskian W is not zero, so u and v form a fundamental set of solutions. (Of course, if µ = 0, then the roots are real and the discussion in this section is not applicable.) Consequently, if the roots of the characteristic equation are complex numbers λ ± iµ, with µ = 0, then the general solution of Eq. (1) is y = c eλt cos µt + c eλt sin µt,
(17)
1
2
where c and c are arbitrary constants. Note that the solution (17) can be written down
1
2
as soon as the values of λ and µ are known.
Chapter 3. Second Order Linear Equations
Find the general solution of E X A M P L E
1
y + y + y = 0.
(18)
The characteristic equation is r 2 + r + 1 = 0, and its roots are
√
−1 ± (1 − 4)1/2
3
r = = −1 ± i.
2
2
2
√ Thus λ = −1/2 and µ = 3/2, so the general solution of Eq. (18) is
√
√
y = c e−t/2 cos( 3t/2) + c e−t/2 sin( 3t/2) .
(19)
1
2
Find the general solution of E X A M P L E
2
y + 9y = 0.
(20)
The characteristic equation is r 2 + 9 = 0 with the roots r = ±3i; thus λ = 0 and µ = 3. The general solution is y = c cos 3t + c sin 3t; (21)
1
2
note that if the real part of the roots is zero, as in this example, then there is no exponential factor in the solution.
Find the solution of the initial value problem E X A M P L E
3
16y − 8y + 145y = 0,y(0) = −2,y(0) = 1.
(22)
The characteristic equation is 16r 2 − 8r + 145 = 0 and its roots are r = 1/4 ± 3i.
Thus the general solution of the differential equation is y = c et/4 cos 3t + c et/4 sin 3t.
(23)
1
2
To apply the first initial condition we set t = 0 in Eq. (23); this gives y(0) = c = −2.
1
For the second initial condition we must differentiate Eq. (23) and then set t = 0. In this way we find that y(0) = 1 c + 3c = 1, 4 1
2
from which c = 1/2. Using these values of c and c in Eq. (23), we obtain
2
1
2
y = −2et/4 cos 3t + 1 et/4 sin 3t (24)
2
as the solution of the initial value problem (22).
We will discuss the geometrical properties of solutions such as these more fully in Section 3.8, so we will be very brief here. Each of the solutions u and v in represents an oscillation, because of the trigonometric factors, and also either grows or decays exponentially, depending on the sign of λ (unless λ = 0). In Example 1 we have λ = −1/2 < 0, so solutions are decaying oscillations. The graph of a typical solution of Eq. (18) is shown in Figure 3.4.1. On the other hand, λ = 1/4 > 0 in Example 3, so solutions of the differential equation (22) are growing oscillations. The graph of the solution (24) of the given initial value problem is shown in Figure 3.4.2. The intermediate case is illustrated in Example 2 in which λ = 0. In this case the solution neither grows nor decays exponentially, but oscillates steadily; a typical solution of Eq. (20) is shown in Figure 3.4.3.
y
2
1
4
6
2
8
t
FIGURE 3.4.1
A typical solution of y + y + y = 0.
y
10
y = –2et/4 cos 3t + 1 et/4 sin 3t
2
5
2
4
6
8 t
–5
–10
FIGURE 3.4.2
Solution of 16y − 8y + 145y = 0, y(0) = −2, y(0) = 1.
Chapter 3. Second Order Linear Equations
y
1
2
4
6
8
10
t
–1
FIGURE 3.4.3
A typical solution of y + 9y = 0.
PROBLEMS
In each of Problems 1 through 6 use Euler’s formula to write the given expression in the form a + ib.
In each of Problems 7 through 16 find the general solution of the given differential equation.
䉴 23. Consider the initial value problem
䉴 24. Consider the initial value problem
䉴 25. Consider the initial value problem
3.4Complex Roots of the Characteristic Equation
䉴 26. Consider the initial value problem
1
2
(i)
1
2
1
2
1
2
1
2
29. Using Euler’s formula, show that
t
t
1
2
1
2
dt
(i)
1
2
1
2
1
2
Change of Variables.
Often a differential equation with variable coefficients, (i)
into an equation with constant coefficients by a change of the independent variable. Let
Chapter 3. Second Order Linear Equations later.
(a) Show that
d y
2
dt
dt d x
dt
(b) Show that the differential equation (i) becomes
d y p(t) d x
(ii)
dt
dt
d x
be 1; hence
(iii)
(iv)
An equation of the form
equations are discussed in detail in Section 5.5.
3.5 Repeated Roots; Reduction of Order
In earlier sections we showed how to solve the equation ay + by + cy = 0
(1)
are either real and different, or are complex conjugates. Now we consider the third possibility, namely, that the two roots r and r are equal. This case occurs when the
1
2
discriminant b2 − 4ac is zero, and it follows from the quadratic formula that r = r = −b/2a.
(3)
1
2
The difficulty is immediately apparent; both roots yield the same solution y (t) = e−bt/2a
(4)
1
of the differential equation (1), and it is not obvious how to find a second solution.
Solve the differential equation E X A M P L E
1
y + 4y + 4y = 0.
(5)
The characteristic equation is r 2 + 4r + 4 = (r + 2)2 = 0, so r = r = −2. Therefore one solution of Eq. (5) is y (t) = e−2t . To find the general
1
2
1
solution of Eq. (5) we need a second solution that is not a multiple of y . This second
1
solution can be found in several ways (see Problems 20 through 22); here we use a method originated by D’Alembert 5 in the eighteenth century. Recall that since y (t) is
1
a solution of Eq. (1), so is cy (t) for any constant c. The basic idea is to generalize this
1
observation by replacing c by a function v(t) and then trying to determine v(t) so that the product v(t)y (t) is a solution of Eq. (1).
1
To carry out this program we substitute y = v(t)y (t) in Eq. (1) and use the resulting
1
equation to find v(t). Starting with y = v(t)y (t) = v(t)e−2t , (6)
1
we have y = v(t)e−2t − 2v(t)e−2t (7) and y = v(t)e−2t − 4v(t)e−2t + 4v(t)e−2t .
(8)
By substituting the expressions in Eqs. (6), (7), and (8) in Eq. (5) and collecting terms, we obtain [v(t) − 4v(t) + 4v(t) + 4v(t) − 8v(t) + 4v(t)]e−2t = 0, which simplifies to v(t) = 0.
(9)
5Jean d’Alembert (1717–1783), a French mathematician, was a contemporary of Euler and Daniel Bernoulli, and is known primarily for his work in mechanics and differential equations. D’Alembert’s principle in mechanics and d’Alembert’s paradox in hydrodynamics are named for him, and the wave equation first appeared in his paper on vibrating strings in 1747. In his later years he devoted himself primarily to philosophy and to his duties as science editor of Diderot’s Encyclope´die.
Chapter 3. Second Order Linear Equations
Therefore v(t) = c1 and v(t) = c t + c ,
(10)
1
2
where c and c are arbitrary constants. Finally, substituting for v(t) in Eq. (6), we
1
2
obtain y = c te−2t + c e−2t .
(11)
1
2
The second term on the right side of Eq. (11) corresponds to the original solution y (t) = exp(−2t), but the first term arises from a second solution, namely y (t) =
1
2
t exp(−2t). These two solutions are obviously not proportional, but we can verify that they are linearly independent by calculating their Wronskian:
e−2t
te−2t
W (y , y )(t) =
1
2
−2e−2t
(1 − 2t)e−2t = e−4t − 2te−4t + 2te−4t = e−4t = 0.
Therefore y (t) = e−2t ,y (t) = te−2t
(12)
1
2
form a fundamental set of solutions of Eq. (5), and the general solution of that equation is given by Eq. (11). Note that both y (t) and y (t) tend to zero as t → ∞; consequently,
1
2
all solutions of Eq. (5) behave in this way. The graph of a typical solution is shown in Figure 3.5.1.
y
2
1
0.5
1
1.5
2
t
FIGURE 3.5.1
A typical solution of y + 4y + 4y = 0.
The procedure used in Example 1 can be extended to a general equation whose characteristic equation has repeated roots. That is, we assume that the coefficients in Eq. (1) satisfy b2 − 4ac = 0, in which case y (t) = e−bt/2a
1
is a solution. Then we assume that y = v(t)y (t) = v(t)e−bt/2a
(13)
1
and substitute in Eq. (1) to determine v(t). We have y = v(t)e−bt/2a − b v(t)e−bt/2a
(14)
2a
and y = v(t)e−bt/2a − b v(t)e−bt/2a + b2 v(t)e−bt/2a.
(15)
a
4a2
Then, by substituting in Eq. (1), we obtain
a v(t) − b v(t) + b2 v(t) + b v(t) − b v(t) + cv(t) e−bt/2a = 0.
(16)
a
4a2
2a
Canceling the factor exp(−bt/2a), which is nonzero, and rearranging the remaining terms, we find that
b2 av(t) + (−b + b)v(t) + − b2 + c v(t) = 0.
(17)
4a
2a
The term involving v(t)is obviously zero. Further, the coefficient of v(t) is c − (b2/4a), which is also zero because b2 − 4ac = 0 in the problem that we are considering. Thus, just as in Example 1, Eq. (17) reduces to v(t) = 0; therefore, v(t) = c t + c .
1
2
Hence, from Eq. (13), we have y = c te−bt/2a + c e−bt/2a.
(18)
1
2
Thus y is a linear combination of the two solutions y (t) = e−bt/2a,y (t) = te−bt/2a.
(19)
1
2
The Wronskian of these two solutions is
e−bt/2a
te−bt/2a
W (y , y )(t) = = e−bt/a.
(20)
1
2
− b e−bt/2a 1 − bt
e−bt/2a
2a
2a
Since W (y , y )(t) is never zero, the solutions y and y given by Eq. (19) are a
1
2
1
2
fundamental set of solutions. Further, Eq. (18) is the general solution of Eq. (1) when the roots of the characteristic equation are equal. In other words, in this case, there is one exponential solution corresponding to the repeated root, while a second solution is obtained by multiplying the exponential solution by t.
Find the solution of the initial value problem E X A M P L E
2
y − y + 0.25y = 0,y(0) = 2,y(0) = 1 .
(21)
3
The characteristic equation is r 2 − r + 0.25 = 0,Chapter 3. Second Order Linear Equations so the roots are r = r = 1/2. Thus the general solution of the differential equation is
1
2
y = c et/2 + c tet/2.
(22)
1
2
The first initial condition requires that y(0) = c = 2.
1
To satisfy the second initial condition, we first differentiate Eq. (22) and then set t = 0.
This gives y(0) = 1 c + c = 1 , 2 1
2
3
so c = −2/3. Thus, the solution of the initial value problem is
2
y = 2et/2 − 2 tet/2.
(23)
3
The graph of this solution is shown in Figure 3.5.2.
y
4
y'(0) = 2: y = 2et/2 + tet/2
3
y'(0) = :
1
y = 2et/2 – 2 tet/2
3
3
2
1
1
2
3
t
–1
FIGURE 3.5.2
Solutions of y − y + 0.25y = 0, y(0) = 2, with y(0) = 1/3 and with y(0) = 2, respectively.
Let us now modify the initial value problem (21) by changing the initial slope; to be specific, let the second initial condition be y(0) = 2. The solution of this modified problem is y = 2et/2 + tet/2 and its graph is also shown in Figure 3.5.2. The graphs shown in this figure suggest that there is a critical initial slope, with a value between 1 and 2, that separates solutions
3
that grow positively from those that ultimately grow negatively. In Problem 16 you are asked to determine this critical initial slope.
3.5
Repeated Roots; Reduction of Order
The geometrical behavior of solutions is similar in this case to that when the roots are real and different. If the exponents are either positive or negative, then the magnitude of the solution grows or decays accordingly; the linear factor t has little influence.
A decaying solution is shown in Figure 3.5.1 and growing solutions in However, if the repeated root is zero, then the differential equation is y = 0 and the general solution is a linear function of t.
Summary.
We can now summarize the results that we have obtained for second order linear homogeneous equations with constant coefficients, ay + by + cy = 0.
(1)
Let r and r be the roots of the corresponding characteristic polynomial
1
2
ar 2 + br + c = 0.
(2)
If r and r are real but not equal, then the general solution of the differential
1
2
equation (1) is y = c er t
t
1
+ c er2 .
(24)
1
2
If r and r are complex conjugates λ ± iµ, then the general solution is
1
2
y = c eλt cos µt + c eλt sin µt.
(25)
1
2
If r = r , then the general solution is
1
2
y = c er t
t
1
+ c ter1 .
(26)
1
2
Reduction of Order.
It is worth noting that the procedure used earlier in this section for equations with constant coefficients is more generally applicable. Suppose we know one solution y (t), not everywhere zero, of
1
y + p(t)y + q(t)y = 0.
(27)
To find a second solution, let y = v(t)y (t); (28)
1
then y = v(t)y (t) + v(t)y (t)
1
1
and y = v(t)y (t) + 2v(t)y (t) + v(t)y(t).
1
1
1
Substituting for y, y, and y in Eq. (27) and collecting terms, we find that y v + (2y + py )v + (y + py + qy )v = 0.
(29)
1
1
1
1
1
1
Since y is a solution of Eq. (27), the coefficient of v in Eq. (29) is zero, so that Eq. (29)
1
becomes y v + (2y + py )v = 0.
(30)
1
1
1
Despite its appearance, Eq. (30) is actually a first order equation for the function v and can be solved either as a first order linear equation or as a separable equation. Once v has been found, then v is obtained by an integration. Finally, y is determined from Chapter 3. Second Order Linear Equations Eq. (28). This procedure is called the method of reduction of order because the crucial step is the solution of a first order differential equation for v rather than the original second order equation for y. Although it is possible to write down a formula for v(t), we will instead illustrate how this method works by an example.
Given that y (t) = t−1 is a solution of
1
E X A M P L E
3
2t2 y + 3t y − y = 0,t > 0, (31)
find a second linearly independent solution.
We set y = v(t)t−1; then y = vt−1 − vt−2,y = vt−1 − 2vt−2 + 2vt−3.
Substituting for y, y, and y in Eq. (31) and collecting terms, we obtain 2t2(vt−1 − 2vt−2 + 2vt−3) + 3t (vt−1 − vt−2) − vt−1 = 2tv + (−4 + 3)v + (4t−1 − 3t−1 − t−1)v
= 2tv − v = 0.
(32)
Note that the coefficient of v is zero, as it should be; this provides a useful check on our algebra.
Separating the variables in Eq. (32) and solving for v(t), we find that v(t) = ct1/2 ; then v(t) = 2ct3/2 + k.
3
It follows that y = 2 ct1/2 + kt−1,
(33)
3
where c and k are arbitrary constants. The second term on the right side of Eq. (33) is a multiple of y (t) and can be dropped, but the first term provides a new independent
1
solution. Neglecting the arbitrary multiplicative constant, we have y (t) = t1/2.
2
PROBLEMS
In each of Problems 1 through 10 find the general solution of the given differential equation.
3.5
Repeated Roots; Reduction of Order 䉴 15. Consider the initial value problem
(b) Determine where the solution has the value zero.
0
0
16. Consider the following modification of the initial value problem in Example 2:
䉴 17. Consider the initial value problem (a) Solve the initial value problem and plot the solution.
0
0
M
M
M
M
18. Consider the initial value problem (a) Solve the initial value problem.
0 can take on the value zero at most once.
1
2
1
2
1
2
1
1
1
2
1
2
1
2
1
2
1
2
2
1
2
1
2
1
1
1
2
2
1
1
(i)
1
Chapter 3. Second Order Linear Equations
1
1
∂
∂
=
).
ert
(ii)
r
∂r
1
1
1
1
1
1
1
1
1
1
√
1
1
31. The differential equation
1
2
2
2
1!
2!
2
1
32. The differential equation
1
an integral.
33. The method of Problem 20 can be extended to second order equations with variable
1
2
2
1
1
2
1
1
2
1
2
2
1
1
6 T. A. Newton, “On Using a Differential Equation to Generate Polynomials,” American Mathematical Monthly
81 (1974), pp. 592 – 601. Also see the references given there.
3.6
Nonhomogeneous Equations; Method of Undetermined Coefficients
1
1
0
0
Euler Equations.
Use the substitution introduced in Problem 38 in Section 3.4 to solve each of the equations in Problems 41 and 42.
3.6 Nonhomogeneous Equations; Method of Undetermined Coefficients
We now return to the nonhomogeneous equation L[y] = y + p(t)y + q(t)y = g(t), (1)
where p, q, and g are given (continuous) functions on the open interval I . The equation L[y] = y + p(t)y + q(t)y = 0, (2)
in which g(t) = 0 and p and q are the same as in Eq. (1), is called the homogeneous equation corresponding to Eq. (1). The following two results describe the structure of solutions of the nonhomogeneous equation (1) and provide a basis for constructing its general solution.
Chapter 3. Second Order Linear Equations
Theorem 3.6.1 If Y and Y are two solutions of the nonhomogeneous equation (1), then their
1
2
difference Y − Y is a solution of the corresponding homogeneous equation (2). If,
1
2
in addition, y and y are a fundamental set of solutions of Eq. (2), then
1
2
Y (t) − Y (t) = c y (t) + c y (t), (3)
1
2
1 1
2 2 where c and c are certain constants.
1
2
To prove this result, note that Y and Y satisfy the equations
1
2
L[Y ](t) = g(t),L[Y ](t) = g(t).
(4)
1
2
Subtracting the second of these equations from the first, we have L[Y ](t) − L[Y ](t) = g(t) − g(t) = 0.
(5)
1
2
However, L[Y ] − L[Y ] = L[Y − Y ], 1
2
1
2
so Eq. (5) becomes L[Y − Y ](t) = 0.
(6)
1
2
Equation (6) states that Y − Y is a solution of Eq. (2). Finally, since all solutions of
1
2
Eq. (2) can be expressed as linear combinations of a fundamental set of solutions by Theorem 3.2.4, it follows that the solution Y − Y can be so written. Hence Eq. (3)
1
2
holds and the proof is complete.
Theorem 3.6.2
The general solution of the nonhomogeneous equation (1) can be written in the form y = φ(t) = c y (t) + c y (t) + Y (t), (7)
1 1
2 2 where y and y are a fundamental set of solutions of the corresponding homogeneous
1
2
equation (2), c and c are arbitrary constants, and Y is some specific solution of the
1
2
nonhomogeneous equation (1).
The proof of Theorem 3.6.2 follows quickly from the preceding theorem. Note that Eq. (3) holds if we identify Y with an arbitrary solution φ of Eq. (1) and Y with the
1
2
specific solution Y . From Eq. (3) we thereby obtain φ(t) − Y (t) = c y (t) + c y (t), (8)
1 1
2 2 which is equivalent to Eq. (7). Since φ is an arbitrary solution of Eq. (1), the expression on the right side of Eq. (7) includes all solutions of Eq. (1); thus it is natural to call it the general solution of Eq. (1).
In somewhat different words, Theorem 3.6.2 states that to solve the nonhomogeneous equation (1), we must do three things: 1.
Find the general solution c y (t) + c y (t) of the corresponding homogeneous 1 1
2 2 equation. This solution is frequently called the complementary solution and may be denoted by y (t).
c
2.
Find some single solution Y (t) of the nonhomogeneous equation. Often this solution is referred to as a particular solution.
3.
Add together the functions found in the two preceding steps.
3.6
Nonhomogeneous Equations; Method of Undetermined Coefficients
We have already discussed how to find y (t), at least when the homogeneous equa c tion (2) has constant coefficients. Therefore, in the remainder of this section and in the next, we will focus on finding a particular solution Y (t) of the nonhomogeneous equation (1). There are two methods that we wish to discuss. They are known as the method of undetermined coefficients and the method of variation of parameters, respectively.
Each has some advantages and some possible shortcomings.
Method of Undetermined Coefficients.
The method of undetermined coefficients re quires that we make an initial assumption about the form of the particular solution Y (t), but with the coefficients left unspecified. We then substitute the assumed expression into Eq. (1) and attempt to determine the coefficients so as to satisfy that equation. If we are successful, then we have found a solution of the differential equation (1) and can use it for the particular solution Y (t). If we cannot determine the coefficients, then this means that there is no solution of the form that we assumed. In this case we may modify the initial assumption and try again.
The main advantage of the method of undetermined coefficients is that it is straight forward to execute once the assumption is made as to the form of Y (t). Its major limitation is that it is useful primarily for equations for which we can easily write down the correct form of the particular solution in advance. For this reason, this method is usually used only for problems in which the homogeneous equation has constant coefficients and the nonhomogeneous term is restricted to a relatively small class of functions. In particular, we consider only nonhomogeneous terms that consist of polynomials, exponential functions, sines, and cosines. Despite this limitation, the method of undetermined coefficients is useful for solving many problems that have important applications. However, the algebraic details may become tedious and a computer algebra system can be very helpful in practical applications. We will illustrate the method of undetermined coefficients by several simple examples and then summarize some rules for using it.
Find a particular solution of E X A M P L E
1
y − 3y − 4y = 3e2t .
(9)
We seek a function Y such that the combination Y (t) − 3Y (t) − 4Y (t) is equal to 3e2t . Since the exponential function reproduces itself through differentiation, the most plausible way to achieve the desired result is to assume that Y (t) is some multiple of e2t , that is, Y (t) = Ae2t , where the coefficient A is yet to be determined. To find A we calculate Y (t) = 2Ae2t ,Y (t) = 4Ae2t , and substitute for y, y, and y in Eq. (9). We obtain (4A − 6A − 4A)e2t = 3e2t.
Hence −6Ae2t must equal 3e2t , so A = −1/2. Thus a particular solution is Y (t) = − 1 e2t .
(10)
2
Chapter 3. Second Order Linear Equations
Find a particular solution of E X A M P L E
2
y − 3y − 4y = 2 sin t.
(11)
By analogy with Example 1, let us first assume that Y (t) = A sin t, where A is a constant to be determined. On substituting in Eq. (11) and rearranging the terms, we obtain −5A sin t − 3A cos t = 2 sin t, or (2 + 5A) sin t + 3A cos t = 0.
(12)
The functions sin t and cos t are linearly independent, so Eq. (12) can hold on an interval only if the coefficients 2 + 5A and 3A are both zero. These contradictory requirements mean that there is no choice of the constant A that makes Eq. (12) true for all t. Thus we conclude that our assumption concerning Y (t) is inadequate. The appearance of the cosine term in Eq. (12) suggests that we modify our original assumption to include a cosine term in Y (t), that is, Y (t) = A sin t + B cos t, where A and B are to be determined. Then Y (t) = A cos t − B sin t,Y (t) = −A sin t − B cos t.
By substituting these expressions for y, y, and y in Eq. (11) and collecting terms, we obtain (−A + 3B − 4A) sin t + (−B − 3A − 4B) cos t = 2 sin t.
(13) To satisfy Eq. (13) we must match the coefficients of sin t and cos t on each side of the equation; thus A and B must satisfy the equations −5A + 3B = 2, −3A − 5B = 0.
Hence A = −5/17 and B = 3/17, so a particular solution of Eq. (11) is Y (t) = − 5 sin t + 3 cos t.
17
17
The method illustrated in the preceding examples can also be used when the right side of the equation is a polynomial. Thus, to find a particular solution of y − 3y − 4y = 4t2 − 1, (14)
we initially assume that Y (t) is a polynomial of the same degree as the nonhomogeneous term, that is, Y (t) = At2 + Bt + C.
To summarize our conclusions up to this point: if the nonhomogeneous term g(t) in Eq. (1) is an exponential function eαt , then assume that Y (t) is proportional to the same exponential function; if g(t) is sin βt or cos βt, then assume that Y (t) is a linear combination of sin βt and cos βt; if g(t) is a polynomial, then assume that Y (t) is a polynomial of like degree. The same principle extends to the case where g(t) is a product of any two, or all three, of these types of functions, as the next example illustrates.
3.6
Nonhomogeneous Equations; Method of Undetermined Coefficients
Find a particular solution of E X A M P L E
3
y − 3y − 4y = −8et cos 2t.
(15)
In this case we assume that Y (t) is the product of et and a linear combination of cos 2t and sin 2t, that is, Y (t) = Aet cos 2t + Bet sin 2t.
The algebra is more tedious in this example, but it follows that Y (t) = (A + 2B)et cos 2t + (−2A + B)et sin 2t and Y (t) = (−3A + 4B)et cos 2t + (−4A − 3B)et sin 2t.
By substituting these expressions in Eq. (15), we find that A and B must satisfy 10 A + 2B = 8, 2 A − 10B = 0.
Hence A = 10/13 and B = 2/13; therefore a particular solution of Eq. (15) is Y (t) = 10 et cos 2t + 2 et sin 2t.
13
13
Now suppose that g(t) is the sum of two terms, g(t) = g (t) + g (t), and suppose
1
2
that Y and Y are solutions of the equations
1
2
ay + by + cy = g (t)
(16)
1
and ay + by + cy = g (t), (17)
2
respectively. Then Y + Y is a solution of the equation
1
2
ay + by + cy = g(t).
(18)
To prove this statement, substitute Y (t) + Y (t) for y in Eq. (18) and make use
1
2
of Eqs. (16) and (17). A similar conclusion holds if g(t) is the sum of any finite number of terms. The practical significance of this result is that for an equation whose nonhomogeneous function g(t) can be expressed as a sum, one can consider instead several simpler equations and then add the results together. The following example is an illustration of this procedure.
Find a particular solution of E X A M P L E
4
y − 3y − 4y = 3e2t + 2 sin t − 8et cos 2t.
(19)
By splitting up the right side of Eq. (19), we obtain the three equations y − 3y − 4y = 3e2t ,
y − 3y − 4y = 2 sin t, and y − 3y − 4y = −8et cos 2t.
Chapter 3. Second Order Linear Equations
Solutions of these three equations have been found in Examples 1, 2, and 3, respectively.
Therefore a particular solution of Eq. (19) is their sum, namely, Y (t) = − 1 e2t + 3 cos t − 5 sin t + 10 et cos 2t + 2 et sin 2t.
2
17
17
13
13
The procedure illustrated in these examples enables us to solve a fairly large class of problems in a reasonably efficient manner. However, there is one difficulty that sometimes occurs. The next example illustrates how it arises.
Find a particular solution of E X A M P L E
5
y + 4y = 3 cos 2t.
(20)
Proceeding as in Example 2, we assume that Y (t) = A cos 2t + B sin 2t. By substi tuting in Eq. (20), we then obtain (4A − 4A) cos 2t + (4B − 4B) sin 2t = 3 cos 2t.
(21)
Since the left side of Eq. (21) is zero, there is no choice of A and B that satisfies this equation. Therefore, there is no particular solution of Eq. (20) of the assumed form. The reason for this possibly unexpected result becomes clear if we solve the homogeneous equation y + 4y = 0 (22)
that corresponds to Eq. (20). A fundamental set of solutions of Eq. (22) is y (t) =
1
cos 2t and y (t) = sin 2t. Thus our assumed particular solution of Eq. (20) is actually
2
a solution of the homogeneous equation (22); consequently, it cannot possibly be a solution of the nonhomogeneous equation (20).
To find a solution of Eq. (20) we must therefore consider functions of a somewhat different form. The simplest functions, other than cos 2t and sin 2t themselves, that when differentiated lead to cos 2t and sin 2t are t cos 2t and t sin 2t. Hence we assume that Y (t) = At cos 2t + Bt sin 2t. Then, upon calculating Y (t) and Y (t), substituting them into Eq. (20), and collecting terms, we find that −4A sin 2t + 4B cos 2t = 3 cos 2t.
Therefore A = 0 and B = 3/4, so a particular solution of Eq. (20) is Y (t) = 3 t sin 2t.
4
The fact that in some circumstances a purely oscillatory forcing term leads to a solution that includes a linear factor t as well as an oscillatory factor is important in some applications; see Section 3.9 for a further discussion.
The outcome of Example 5 suggests a modification of the principle stated previously:
If the assumed form of the particular solution duplicates a solution of the corresponding homogeneous equation, then modify the assumed particular solution by multiplying it by t. Occasionally, this modification will be insufficient to remove all duplication with the solutions of the homogeneous equation, in which case it is necessary to multiply by t a second time. For a second order equation, it will never be necessary to carry the process further than this.
Summary.
We now summarize the steps involved in finding the solution of an initial value problem consisting of a nonhomogeneous equation of the form ay + by + cy = g(t), (23)
where the coefficients a, b, and c are constants, together with a given set of initial conditions:
1.
Find the general solution of the corresponding homogeneous equation.
2.
Make sure that the function g(t) in Eq. (23) belongs to the class of functions discussed in this section, that is, it involves nothing more than exponential functions, sines, cosines, polynomials, or sums or products of such functions. If this is not the case, use the method of variation of parameters (discussed in the next section).
3.
If g(t) = g (t) + · · · + g (t), that is, if g(t) is a sum of n terms, then form n
1
n
subproblems, each of which contains only one of the terms g (t), . . . , g (t). The
1
ni th subproblem consists of the equation ay + by + cy = g (t),
i where i runs from 1 to n.
4.
For the i th subproblem assume a particular solution Y (t) consisting of the ap i
propriate exponential function, sine, cosine, polynomial, or combination thereof.
If there is any duplication in the assumed form of Y (t) with the solutions of the i homogeneous equation (found in step 1), then multiply Y (t) by t, or (if necessary)
i by t2, so as to remove the duplication. See Table 3.6.1.
5.
Find a particular solution Y (t) for each of the subproblems. Then the sum Y (t) +
i
1
· · · + Y (t) is a particular solution of the full nonhomogeneous equation (23).
n
6.
Form the sum of the general solution of the homogeneous equation (step 1) and the particular solution of the nonhomogeneous equation (step 5). This is the general solution of the nonhomogeneous equation.
7.
Use the initial conditions to determine the values of the arbitrary constants remaining in the general solution.
TABLE 3.6.1 The Particular Solution of ay + by + cy = g (t)ig (t)
Y (t)
i
i
P (t) = a tn + a tn−1 + · · · + ats (A tn + A tn−1 + · · · + A )
n
0
1
n
0
1
n
P (t)eαtts (A tn + A tn−1 + · · · + A )eαt
n
0
1
n
sinβt
P (t)eαtts [(A tn + A tn−1 + · · · + A )eαt cos βt
n
cos βt
0
1
n
+(B tn + B tn−1 + · · · + B )eαt sin βt]
0
1
n
Notes. Here s is the smallest nonnegative integer (s = 0, 1, or 2) that will ensure that no term in Y (t) is a solution of the corresponding homogeneous equation. Equivalently, for the three cases,
i
s is the number of times 0 is a root of the characteristic equation, α is a root of the characteristic equation, and α + iβ is a root of the characteristic equation, respectively.
Chapter 3. Second Order Linear Equations
For some problems this entire procedure is easy to carry out by hand, but in many cases it requires considerable algebra. Once you understand clearly how the method works, a computer algebra system can be of great assistance in executing the details.
The method of undetermined coefficients is self-correcting in the sense that if one assumes too little for Y (t), then a contradiction is soon reached that usually points the way to the modification that is needed in the assumed form. On the other hand, if one assumes too many terms, then some unnecessary work is done and some coefficients turn out to be zero, but at least the correct answer is obtained.
Proof of the Method of Undetermined Coefficients.
In the preceding discussion we have described the method of undetermined coefficients on the basis of several examples. To prove that the procedure always works as stated, we now give a general argument, in which we consider several cases corresponding to different forms for the nonhomogeneous term g(t).
g(t) = P (t) = a tn + a tn−1 + · · · + a . In this case Eq. (23) becomes
n
0
1
n
ay + by + cy = a tn + a tn−1 + · · · + a .
(24)
0
1
n
To obtain a particular solution we assume that Y (t) = A tn + A tn−1 + · · · + At2 + At + A .
(25)
0
1
n−2
n−1
n
Substituting in Eq. (24), we obtain a[n(n − 1)A tn−2 + · · · + 2A ] + b(n A tn−1 + · · · + A)
0
n−2
0
n−1
+ c(A tn + A tn−1 + · · · + A ) = a tn + · · · + a .
(26)
0
1
n
0
n
Equating the coefficients of like powers of t gives c A = a ,
0
0
c A + nb A = a ,
1
0
1
...
c A + b A + 2a A = a .
n
n−1
n−2
n
Provided that c = 0, the solution of the first equation is A = a /c, and the remaining
0
0
equations determine A , . . . , A successively. If c = 0, but b = 0, then the polynomial
1
n
on the left side of Eq. (26) is of degree n − 1, and we cannot satisfy Eq. (26). To be sure that aY (t) + bY (t) is a polynomial of degree n, we must choose Y (t) to be a polynomial of degree n + 1. Hence we assume that Y (t) = t (A tn + · · · + A ).
0
n
There is no constant term in this expression for Y (t), but there is no need to include such a term since a constant is a solution of the homogeneous equation when c = 0.
Since b = 0, we have A = a /b(n + 1), and the other coefficients A , . . . , A can be
0
0
1
n
determined similarly. If both c and b are zero, we assume that Y (t) = t2(A tn + · · · + A ).
0
n
The term aY (t) gives rise to a term of degree n, and we can proceed as before. Again the constant and linear terms in Y (t) are omitted, since in this case they are both solutions of the homogeneous equation.
3.6
Nonhomogeneous Equations; Method of Undetermined Coefficientsg(t) = eαt P (t). The problem of determining a particular solution of n
ay + by + cy = eαt P (t)
(27)
n
can be reduced to the preceding case by a substitution. Let Y (t) = eαt u(t); then Y (t) = eαt [u(t) + αu(t)] and Y (t) = eαt [u(t) + 2αu(t) + α2u(t)].
Substituting for y, y, and y in Eq. (27), canceling the factor eαt , and collecting terms, we obtain au(t) + (2aα + b)u(t) + (aα2 + bα + c)u(t) = P (t).
(28)
n
The determination of a particular solution of Eq. (28) is precisely the same problem, except for the names of the constants, as solving Eq. (24). Therefore, if aα2 + bα + c
is not zero, we assume that u(t) = A tn + · · · + A ; hence a particular solution of
0
n
Eq. (27) is of the form Y (t) = eαt (A tn + A tn−1 + · · · + A ).
(29)
0
1
n
On the other hand, if aα2 + bα + c is zero, but 2aα + b is not, we must take u(t) to be of the form t (A tn + · · · + A ). The corresponding form for Y (t) is t times the
0
n
expression on the right side of Eq. (29). Note that if aα2 + bα + c is zero, then eαt is a solution of the homogeneous equation. If both aα2 + bα + c and 2aα + b are zero (and this implies that both eαt and teαt are solutions of the homogeneous equation), then the correct form for u(t) is t2(A tn + · · · + A ). Hence Y (t) is t2 times the expression on
0
n the right side of Eq. (29).
g(t) = eαt P (t) cos β t or eαt P (t) sin β t. These two cases are similar, so we con n
n
sider only the latter. We can reduce this problem to the preceding one by noting that, as a consequence of Euler’s formula, sin βt = (eiβt − e−iβt )/2i. Hence g(t) is of the form g(t) = P (t) e(α+iβ)t − e(α−iβ)t
n
2i
and we should choose Y (t) = e(α+iβ)t (A tn + · · · + A ) + e(α−iβ)t (B tn + · · · + B ), 0
n
0
n
or equivalently, Y (t) = eαt (A tn + · · · + A ) cos βt + eαt (B tn + · · · + B ) sin βt.
0
n
0
n
Usually, the latter form is preferred. If α ± iβ satisfy the characteristic equation corresponding to the homogeneous equation, we must, of course, multiply each of the polynomials by t to increase their degrees by one.
If the nonhomogeneous function involves both cos βt and sin βt, it is usually con venient to treat these terms together, since each one individually may give rise to the Chapter 3. Second Order Linear Equations same form for a particular solution. For example, if g(t) = t sin t + 2 cos t, the form for Y (t) would be Y (t) = (A t + A ) sin t + (B t + B ) cos t, 0
1
0
1
provided that sin t and cos t are not solutions of the homogeneous equation.
PROBLEMS
In each of Problems 1 through 12 find the general solution of the given differential equation.
0
0
0
0
In each of Problems 13 through 18 find the solution of the given initial value problem.
In each of Problems 19 through 26:
䉴
䉴 䉴 䉴 䉴 䉴 䉴
27. Determine the general solution of
N
m
䉴 28. In many physical problems the nonhomogeneous term may be specified by different for-
t > π,
䉴 29. Follow the instructions in Problem 28 to solve the differential equation
(i)
1
2
1
2
32. In this problem we indicate an alternate procedure7 for solving the differential equation (i)
1
2
(a) Verify that Eq. (i) can be written in the factored form
1
2
1
2
1 2
2
following two first order equations:
1
2
(see Example 1) (see Problem 7) (see Problem 6) (see Problem 4)
3.7 Variation of Parameters
In this section we describe another method of finding a particular solution of a nonhomogeneous equation. This method, known as is due to Lagrange and complements the method of undetermined coefficients rather well. The main advantage of variation of parameters is that it is a general method; in principle at least, it can be applied to any equation, and it requires no detailed assumptions about 7R. S. Luthar, “Another Approach to a Standard Differential Equation,” Two Year College Mathematics Journal 10
(1979), pp. 200 –201; also see D. C. Sandell and F. M. Stein, “Factorization of Operators of Second Order Linear Homogeneous Ordinary Differential Equations,” Two Year College Mathematics Journal 8 (1977), pp. 132–141, for a more general discussion of factoring operators.
Chapter 3. Second Order Linear Equations the form of the solution. In fact, later in this section we use this method to derive a formula for a particular solution of an arbitrary second order linear nonhomogeneous differential equation. On the other hand, the method of variation of parameters eventually requires that we evaluate certain integrals involving the nonhomogeneous term in the differential equation, and this may present difficulties. Before looking at this method in the general case, we illustrate its use in an example.
Find a particular solution of E X A M P L E
1
y + 4y = 3 csc t.
(1)
Observe that this problem does not fall within the scope of the method of un determined coefficients because the nonhomogeneous term g(t) = 3 csc t involves a quotient (rather than a sum or a product) of sin t or cos t. Therefore, we need a different approach. Observe also that the homogeneous equation corresponding to Eq. (1) is y + 4y = 0, (2)
and that the general solution of Eq. (2) is y (t) = c cos 2t + c sin 2t.
(3)
c
1
2
The basic idea in the method of variation of parameters is to replace the constants c1 and c in Eq. (3) by functions u (t) and u (t), respectively, and then to determine these
2
1
2
functions so that the resulting expression y = u (t) cos 2t + u (t) sin 2t (4)
1
2
is a solution of the nonhomogeneous equation (1).
To determine u and u we need to substitute for y from Eq. (4) in Eq. (1). However,
1
2
even without carrying out this substitution, we can anticipate that the result will be a single equation involving some combination of u , u , and their first two derivatives.
1
2
Since there is only one equation and two unknown functions, we can expect that there are many possible choices of u and u that will meet our needs. Alternatively, we
1
2
may be able to impose a second condition of our own choosing, thereby obtaining two equations for the two unknown functions u and u . We will soon show (following
1
2
Lagrange) that it is possible to choose this second condition in a way that makes the computation markedly more efficient.
Returning now to Eq. (4), we differentiate it and rearrange the terms, thereby obtaining y = −2u (t) sin 2t + 2u (t) cos 2t + u (t) cos 2t + u (t) sin 2t.
(5)
1
2
1
2
Keeping in mind the possibility of choosing a second condition on u and u , let us
1
2
require the last two terms on the right side of Eq. (5) to be zero; that is, we require that u (t) cos 2t + u (t) sin 2t = 0.
(6)
1
2
It then follows from Eq. (5) that y = −2u (t) sin 2t + 2u (t) cos 2t.
(7)
1
2
3.7
Variation of Parameters
Although the ultimate effect of the condition (6) is not yet clear, at the very least it has simplified the expression for y. Further, by differentiating Eq. (7), we obtain y = −4u (t) cos 2t − 4u (t) sin 2t − 2u (t) sin 2t + 2u (t) cos 2t.
(8)
1
2
1
2
Then, substituting for y and y in Eq. (1) from Eqs. (4) and (8), respectively, we find that u and u must satisfy
1
2
−2u (t) sin 2t + 2u (t) cos 2t = 3 csc t.
(9)
1
2
Summarizing our results to this point, we want to choose u and u so as to satisfy
1
2
Eqs. (6) and (9). These equations can be viewed as a pair of linear algebraic equations for the two unknown quantities u (t) and u (t). Equations (6) and (9) can be solved in
1
2
various ways. For example, solving Eq. (6) for u (t), we have
2
u (t) = −u (t) cos 2t .
(10)
2
1
sin 2t
Then, substituting for u (t) in Eq. (9) and simplifying, we obtain
2
u (t) = − 3 csc t sin 2t = −3 cos t.
(11)
1
2
Further, putting this expression for u (t) back in Eq. (10) and using the double angle
1
formulas, we find that u (t) = 3 cos t cos 2t = 3(1 − 2 sin2 t) = 3 csc t − 3 sin t.
(12)
2
sin 2t 2 sin t
2
Having obtained u (t) and u (t), the next step is to integrate so as to obtain u (t)
1
2
1
and u (t). The result is
2
u (t) = −3 sin t + c
(13)
1
1
and u (t) = 3 ln | csc t − cot t| + 3 cos t + c .
(14)
2
2
2
Finally, on substituting these expressions in Eq. (4), we have y = −3 sin t cos 2t + 3 ln | csc t − cot t| sin 2t + 3 cos t sin 2t
2
+ c cos 2t + c sin 2t,
1
2
or y = 3 sin t + 3 ln | csc t − cot t| sin 2t + c cos 2t + c sin 2t.
(15)
2
1
2
The terms in Eq. (15) involving the arbitrary constants c and c are the general solution
1
2
of the corresponding homogeneous equation, while the remaining terms are a particular solution of the nonhomogeneous equation (1). Therefore Eq. (15) is the general solution of Eq. (1).
In the preceding example the method of variation of parameters worked well in determining a particular solution, and hence the general solution, of Eq. (1). The next
Chapter 3. Second Order Linear Equations question is whether this method can be applied effectively to an arbitrary equation.
Therefore we consider y + p(t)y + q(t)y = g(t), (16)
where p, q, and g are given continuous functions. As a starting point, we assume that we know the general solution y (t) = c y (t) + c y (t)
(17)
c 1 1
2 2 of the corresponding homogeneous equation y + p(t)y + q(t)y = 0.
(18)
This is a major assumption because so far we have shown how to solve Eq. (18) only if it has constant coefficients. If Eq. (18) has coefficients that depend on t, then usually the methods described in Chapter 5 must be used to obtain y (t).
c
The crucial idea, as illustrated in Example 1, is to replace the constants c and c in
1
2
Eq. (17) by functions u (t) and u (t), respectively; this gives
1
2
y = u (t)y (t) + u (t)y (t).
(19)
1
1
2
2
Then we try to determine u (t) and u (t) so that the expression in Eq. (19) is a solution
1
2
of the nonhomogeneous equation (16) rather than the homogeneous equation (18).
Thus we differentiate Eq. (19), obtaining y = u (t)y (t) + u (t)y (t) + u (t)y (t) + u (t)y (t).
(20)
1
1
1
1
2
2
2
2
As in Example 1, we now set the terms involving u (t) and u (t) in Eq. (20) equal to
1
2
zero; that is, we require that u (t)y (t) + u (t)y (t) = 0.
(21)
1
1
2
2
Then, from Eq. (20), we have y = u (t)y (t) + u (t)y (t).
(22)
1
1
2
2
Further, by differentiating again, we obtain y = u (t)y (t) + u (t)y(t) + u (t)y (t) + u (t)y(t).
(23)
1
1
1
1
2
2
2
2
Now we substitute for y, y, and y in Eq. (16) from Eqs. (19), (22), and (23), respectively. After rearranging the terms in the resulting equation we find that u (t)[y(t) + p(t)y (t) + q(t)y (t)]
1
1
1
1
+ u (t)[y(t) + p(t)y (t) + q(t)y (t)]
2
2
2
2
+ u (t)y (t) + u (t)y (t) = g(t).
(24)
1
1
2
2
Each of the expressions in square brackets in Eq. (24) is zero because both y and y
1
2
are solutions of the homogeneous equation (18). Therefore Eq. (24) reduces to u (t)y (t) + u (t)y (t) = g(t).
(25)
1
1
2
2
Equations (21) and (25) form a system of two linear algebraic equations for the derivatives u (t) and u (t) of the unknown functions. They correspond exactly to
1
2
Eqs. (6) and (9) in Example 1.
3.7
Variation of Parameters
By solving the system (21), (25) we obtain y (t)g(t)y (t)g(t)u (t) = −
2
,u (t) =
1
,
(26)
1
W (y ,y )(t)
2
W (y ,y )(t)
1
2
1
2
where W (y ,y ) is the Wronskian of y and y . Note that division by W is permissible
1
2
1
2
since y and y are a fundamental set of solutions, and therefore their Wronskian is
1
2
nonzero. By integrating Eqs. (26) we find the desired functions u (t) and u (t), namely,
1
2
y (t)g(t)y (t)g(t)u (t) = −
2
,u (t) =
1 .
(27)
1
W (y ,y )(t) dt + c1
2
W (y ,y )(t) dt + c2
1
2
1
2
Finally, substituting from Eq. (27) in Eq. (19) gives the general solution of Eq. (16).
We state the result as a theorem.
Theorem 3.7.1
If the functions p, q, and g are continuous on an open interval I , and if the functions y and y are linearly independent solutions of the homogeneous equation (18)
1
2
corresponding to the nonhomogeneous equation (16), y + p(t)y + q(t)y = g(t), then a particular solution of Eq. (16) is
y (t)g(t)y (t)g(t)Y (t) = −y (t)
2
(t)
1
1
W (y ,y )(t) dt + y2 W (y ,y )(t) dt, (28)
1
2
1
2
and the general solution is y = c y (t) + c y (t) + Y (t), (29)
1 1
2 2 as prescribed by Theorem 3.6.2.
By examining the expression (28) and reviewing the process by which we derived it, we can see that there may be two major difficulties in using the method of variation of parameters. As we have mentioned earlier, one is the determination of y (t) and y (t), 1
2
a fundamental set of solutions of the homogeneous equation (18), when the coefficients in that equation are not constants. The other possible difficulty is in the evaluation of the integrals appearing in Eq. (28). This depends entirely on the nature of the functions y , y , and g. In using Eq. (28), be sure that the differential equation is exactly in the
1
2
form (16); otherwise, the nonhomogeneous term g(t) will not be correctly identified.
A major advantage of the method of variation of parameters is that Eq. (28) provides an expression for the particular solution Y (t) in terms of an arbitrary forcing function g(t). This expression is a good starting point if you wish to investigate the effect of variations in the forcing function, or if you wish to analyze the response of a system to a number of different forcing functions.
PROBLEMS
In each of Problems 1 through 4 use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients.
Chapter 3. Second Order Linear Equations
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
21. Show that the solution of the initial value problem
(i)
0
0
0
0
(ii)
0
0
0
0
(iii)
0
0
0
1
2
1
2
(
(
(
(
t
0
1
2
1
2
0
0
23. (a) Use the result of Problem 22 to show that the solution of the initial value problem
(i)
0
0
is
(ii)
(b) Find the solution of the initial value problem
0
0
3.7
Variation of Parameters
24. Use the result of Problem 22 to find the solution of the initial value problem
0
0
25. Use the result of Problem 22 to find the solution of the initial value problem
0
0
26. Use the result of Problem 22 to find the solution of the initial value problem
0
0
27. By combining the results of Problems 24 through 26, show that the solution of the initial value problem
0
0
(i)
1
2
0
28. The method of reduction of order (Section 3.5) can also be used for the nonhomogeneous equation
(i)
1
1
(ii)
1
1
1
1
1
1
(see Problem 15)
1
(see Problem 16)
1
Complex Roots of the Characteristic Equation
1
2
set of solutions on that interval.
1
2
a fundamental set of solutions on that interval.
1
2
0
0
3.4 Complex Roots of the Characteristic Equation
We continue our discussion of the equation ay + by + cy = 0,
(1)
where a, b, and c are given real numbers. In Section 3.1 we found that if we seek solutions of the form y = ert , then r must be a root of the characteristic equation ar 2 + br + c = 0.
(2)
If the roots r and r are real and different, which occurs whenever the discriminant
1
2
b2 − 4ac is positive, then the general solution of Eq. (1) is y = c er t
t
1
+ c er2 .
(3)
1
2
Suppose now that b2 − 4ac is negative. Then the roots of Eq. (2) are conjugate complex numbers; we denote them by r = λ + iµ,r = λ − iµ,
(4)
1
2
where λ and µ are real. The corresponding expressions for y are y (t) = exp[(λ + iµ)t],y (t) = exp[(λ − iµ)t].
(5)
1
2
Our first task is to explore what is meant by these expressions, which involve evaluating the exponential function for a complex exponent. For example, if λ = −1, µ = 2, and t = 3, then from Eq. (5) y (3) = e−3+6i .
(6)
1
What does it mean to raise the number e to a complex power? The answer is provided by an important relation known as Euler’s formula.
Chapter 3. Second Order Linear EquationsEuler’s Formula.
To assign a meaning to the expressions in Eqs. (5) we need to give a definition of the complex exponential function. Of course, we want the definition to reduce to the familiar real exponential function when the exponent is real. There are several ways to accomplish this extension of the exponential function. Here we use a method based on infinite series; an alternative is outlined in Problem 28.
Recall from calculus that the Taylor series for et about t = 0 is
∞
tn
et = , −∞ < t < ∞.
(7) n!
n=0
If we now assume that we can substitute i t for t in Eq. (7), then we have
∞ (it)neit = n!
n=0
∞
(−
∞
(−
=
1)nt2n + 1)n−1t2n−1 ,
(
i
(8) 2n)!
(2n − 1)! n=0
n=1
where we have separated the sum into its real and imaginary parts, making use of the fact that i 2 = −1, i3 = −i, i4 = 1, and so forth. The first series in Eq. (8) is precisely the Taylor series for cos t about t = 0, and the second is the Taylor series for sin t
about t = 0. Thus we have eit = cos t + i sin t.
(9)
Equation (9) is known as Euler’s formula and is an extremely important mathematical relationship. While our derivation of Eq. (9) is based on the unverified assumption that the series (7) can be used for complex as well as real values of the independent variable, our intention is to use this derivation only to make Eq. (9) seem plausible. We now put matters on a firm foundation by adopting Eq. (9) as the definition of eit . In other words, whenever we write eit , we mean the expression on the right side of Eq. (9).
There are some variations of Euler’s formula that are also worth noting. If we replace t by −t in Eq. (9) and recall that cos(−t) = cos t and sin(−t) = − sin t, then we have e−it = cos t − i sin t.
(10)
Further, if t is replaced by µt in Eq. (9), then we obtain a generalized version of Euler’s formula, namely, eiµt = cos µt + i sin µt.
(11)
Next, we want to extend the definition of the exponential function to arbitrary complex exponents of the form (λ + iµ)t. Since we want the usual properties of the exponential function to hold for complex exponents, we certainly want exp[(λ + iµ)t] to satisfy e(λ+iµ)t = eλt eiµt .
(12)
Then, substituting for eiµt from Eq. (11), we obtain e(λ+iµ)t = eλt (cos µt + i sin µt) = eλt cos µt + ieλt sin µt.
(13)
We now take Eq. (13) as the definition of exp[(λ + iµ)t]. The value of the exponential function with a complex exponent is a complex number whose real and imaginary parts are given by the terms on the right side of Eq. (13). Observe that the real and imaginary parts of exp[(λ + iµ)t] are expressed entirely in terms of elementary realvalued functions. For example, the quantity in Eq. (6) has the value e−3+6i = e−3 cos 6 + ie−3 sin 6 ∼ = 0.0478041 − 0.0139113i.
With the definitions (9) and (13) it is straightforward to show that the usual laws of exponents are valid for the complex exponential function. It is also easy to verify that the differentiation formula d (ert) = rert
(14)
dt
also holds for complex values of r .
Real-Valued Solutions.
The functions y (t) and y (t), given by Eqs. (5) and with
1
2
the meaning expressed by Eq. (13), are solutions of Eq. (1) when the roots of the characteristic equation (2) are complex numbers λ ± iµ. Unfortunately, the solutions y1 and y are complex-valued functions, whereas in general we would prefer to have real 2
valued solutions, if possible, because the differential equation itself has real coefficients.
Such solutions can be found as a consequence of Theorem 3.2.2, which states that if y and y are solutions of Eq. (1), then any linear combination of y and y is also a
1
2
1
2
solution. In particular, let us form the sum and then the difference of y and y . We
1
2
have y (t) + y (t) = eλt (cos µt + i sin µt) + eλt (cos µt − i sin µt)
1
2
= 2eλt cos µt
and y (t) − y (t) = eλt (cos µt + i sin µt) − eλt (cos µt − i sin µt)
1
2
= 2ieλt sin µt.
Hence, neglecting the constant multipliers 2 and 2i , respectively, we have obtained a pair of real-valued solutions u(t) = eλt cos µt,v(t) = eλt sin µt.
(15)
Observe that u and v are simply the real and imaginary parts, respectively, of y .
1
By direct computation you can show that the Wronskian of u and v is W (u, v)(t) = µe2λt .
(16)
Thus, as long as µ = 0, the Wronskian W is not zero, so u and v form a fundamental set of solutions. (Of course, if µ = 0, then the roots are real and the discussion in this section is not applicable.) Consequently, if the roots of the characteristic equation are complex numbers λ ± iµ, with µ = 0, then the general solution of Eq. (1) is y = c eλt cos µt + c eλt sin µt,
(17)
1
2
where c and c are arbitrary constants. Note that the solution (17) can be written down
1
2
as soon as the values of λ and µ are known.
Chapter 3. Second Order Linear Equations
Find the general solution of E X A M P L E
1
y + y + y = 0.
(18)
The characteristic equation is r 2 + r + 1 = 0, and its roots are
√
−1 ± (1 − 4)1/2
3
r = = −1 ± i.
2
2
2
√ Thus λ = −1/2 and µ = 3/2, so the general solution of Eq. (18) is
√
√
y = c e−t/2 cos( 3t/2) + c e−t/2 sin( 3t/2) .
(19)
1
2
Find the general solution of E X A M P L E
2
y + 9y = 0.
(20)
The characteristic equation is r 2 + 9 = 0 with the roots r = ±3i; thus λ = 0 and µ = 3. The general solution is y = c cos 3t + c sin 3t; (21)
1
2
note that if the real part of the roots is zero, as in this example, then there is no exponential factor in the solution.
Find the solution of the initial value problem E X A M P L E
3
16y − 8y + 145y = 0,y(0) = −2,y(0) = 1.
(22)
The characteristic equation is 16r 2 − 8r + 145 = 0 and its roots are r = 1/4 ± 3i.
Thus the general solution of the differential equation is y = c et/4 cos 3t + c et/4 sin 3t.
(23)
1
2
To apply the first initial condition we set t = 0 in Eq. (23); this gives y(0) = c = −2.
1
For the second initial condition we must differentiate Eq. (23) and then set t = 0. In this way we find that y(0) = 1 c + 3c = 1, 4 1
2
from which c = 1/2. Using these values of c and c in Eq. (23), we obtain
2
1
2
y = −2et/4 cos 3t + 1 et/4 sin 3t (24)
2
as the solution of the initial value problem (22).
We will discuss the geometrical properties of solutions such as these more fully in Section 3.8, so we will be very brief here. Each of the solutions u and v in represents an oscillation, because of the trigonometric factors, and also either grows or decays exponentially, depending on the sign of λ (unless λ = 0). In Example 1 we have λ = −1/2 < 0, so solutions are decaying oscillations. The graph of a typical solution of Eq. (18) is shown in Figure 3.4.1. On the other hand, λ = 1/4 > 0 in Example 3, so solutions of the differential equation (22) are growing oscillations. The graph of the solution (24) of the given initial value problem is shown in Figure 3.4.2. The intermediate case is illustrated in Example 2 in which λ = 0. In this case the solution neither grows nor decays exponentially, but oscillates steadily; a typical solution of Eq. (20) is shown in Figure 3.4.3.
y
2
1
4
6
2
8
t
FIGURE 3.4.1
A typical solution of y + y + y = 0.
y
10
y = –2et/4 cos 3t + 1 et/4 sin 3t
2
5
2
4
6
8 t
–5
–10
FIGURE 3.4.2
Solution of 16y − 8y + 145y = 0, y(0) = −2, y(0) = 1.
Chapter 3. Second Order Linear Equations
y
1
2
4
6
8
10
t
–1
FIGURE 3.4.3
A typical solution of y + 9y = 0.
PROBLEMS
In each of Problems 1 through 6 use Euler’s formula to write the given expression in the form a + ib.
In each of Problems 7 through 16 find the general solution of the given differential equation.
䉴 23. Consider the initial value problem
䉴 24. Consider the initial value problem
䉴 25. Consider the initial value problem
3.4Complex Roots of the Characteristic Equation
䉴 26. Consider the initial value problem
1
2
(i)
1
2
1
2
1
2
1
2
29. Using Euler’s formula, show that
t
t
1
2
1
2
dt
(i)
1
2
1
2
1
2
Change of Variables.
Often a differential equation with variable coefficients, (i)
into an equation with constant coefficients by a change of the independent variable. Let
Chapter 3. Second Order Linear Equations later.
(a) Show that
d y
2
dt
dt d x
dt
(b) Show that the differential equation (i) becomes
d y p(t) d x
(ii)
dt
dt
d x
be 1; hence
(iii)
(iv)
An equation of the form
equations are discussed in detail in Section 5.5.
3.5 Repeated Roots; Reduction of Order
In earlier sections we showed how to solve the equation ay + by + cy = 0
(1)
are either real and different, or are complex conjugates. Now we consider the third possibility, namely, that the two roots r and r are equal. This case occurs when the
1
2
discriminant b2 − 4ac is zero, and it follows from the quadratic formula that r = r = −b/2a.
(3)
1
2
The difficulty is immediately apparent; both roots yield the same solution y (t) = e−bt/2a
(4)
1
of the differential equation (1), and it is not obvious how to find a second solution.
Solve the differential equation E X A M P L E
1
y + 4y + 4y = 0.
(5)
The characteristic equation is r 2 + 4r + 4 = (r + 2)2 = 0, so r = r = −2. Therefore one solution of Eq. (5) is y (t) = e−2t . To find the general
1
2
1
solution of Eq. (5) we need a second solution that is not a multiple of y . This second
1
solution can be found in several ways (see Problems 20 through 22); here we use a method originated by D’Alembert 5 in the eighteenth century. Recall that since y (t) is
1
a solution of Eq. (1), so is cy (t) for any constant c. The basic idea is to generalize this
1
observation by replacing c by a function v(t) and then trying to determine v(t) so that the product v(t)y (t) is a solution of Eq. (1).
1
To carry out this program we substitute y = v(t)y (t) in Eq. (1) and use the resulting
1
equation to find v(t). Starting with y = v(t)y (t) = v(t)e−2t , (6)
1
we have y = v(t)e−2t − 2v(t)e−2t (7) and y = v(t)e−2t − 4v(t)e−2t + 4v(t)e−2t .
(8)
By substituting the expressions in Eqs. (6), (7), and (8) in Eq. (5) and collecting terms, we obtain [v(t) − 4v(t) + 4v(t) + 4v(t) − 8v(t) + 4v(t)]e−2t = 0, which simplifies to v(t) = 0.
(9)
5Jean d’Alembert (1717–1783), a French mathematician, was a contemporary of Euler and Daniel Bernoulli, and is known primarily for his work in mechanics and differential equations. D’Alembert’s principle in mechanics and d’Alembert’s paradox in hydrodynamics are named for him, and the wave equation first appeared in his paper on vibrating strings in 1747. In his later years he devoted himself primarily to philosophy and to his duties as science editor of Diderot’s Encyclope´die.
Chapter 3. Second Order Linear Equations
Therefore v(t) = c1 and v(t) = c t + c ,
(10)
1
2
where c and c are arbitrary constants. Finally, substituting for v(t) in Eq. (6), we
1
2
obtain y = c te−2t + c e−2t .
(11)
1
2
The second term on the right side of Eq. (11) corresponds to the original solution y (t) = exp(−2t), but the first term arises from a second solution, namely y (t) =
1
2
t exp(−2t). These two solutions are obviously not proportional, but we can verify that they are linearly independent by calculating their Wronskian:
e−2t
te−2t
W (y , y )(t) =
1
2
−2e−2t
(1 − 2t)e−2t = e−4t − 2te−4t + 2te−4t = e−4t = 0.
Therefore y (t) = e−2t ,y (t) = te−2t
(12)
1
2
form a fundamental set of solutions of Eq. (5), and the general solution of that equation is given by Eq. (11). Note that both y (t) and y (t) tend to zero as t → ∞; consequently,
1
2
all solutions of Eq. (5) behave in this way. The graph of a typical solution is shown in Figure 3.5.1.
y
2
1
0.5
1
1.5
2
t
FIGURE 3.5.1
A typical solution of y + 4y + 4y = 0.
The procedure used in Example 1 can be extended to a general equation whose characteristic equation has repeated roots. That is, we assume that the coefficients in Eq. (1) satisfy b2 − 4ac = 0, in which case y (t) = e−bt/2a
1
is a solution. Then we assume that y = v(t)y (t) = v(t)e−bt/2a
(13)
1
and substitute in Eq. (1) to determine v(t). We have y = v(t)e−bt/2a − b v(t)e−bt/2a
(14)
2a
and y = v(t)e−bt/2a − b v(t)e−bt/2a + b2 v(t)e−bt/2a.
(15)
a
4a2
Then, by substituting in Eq. (1), we obtain
a v(t) − b v(t) + b2 v(t) + b v(t) − b v(t) + cv(t) e−bt/2a = 0.
(16)
a
4a2
2a
Canceling the factor exp(−bt/2a), which is nonzero, and rearranging the remaining terms, we find that
b2 av(t) + (−b + b)v(t) + − b2 + c v(t) = 0.
(17)
4a
2a
The term involving v(t)is obviously zero. Further, the coefficient of v(t) is c − (b2/4a), which is also zero because b2 − 4ac = 0 in the problem that we are considering. Thus, just as in Example 1, Eq. (17) reduces to v(t) = 0; therefore, v(t) = c t + c .
1
2
Hence, from Eq. (13), we have y = c te−bt/2a + c e−bt/2a.
(18)
1
2
Thus y is a linear combination of the two solutions y (t) = e−bt/2a,y (t) = te−bt/2a.
(19)
1
2
The Wronskian of these two solutions is
e−bt/2a
te−bt/2a
W (y , y )(t) = = e−bt/a.
(20)
1
2
− b e−bt/2a 1 − bt
e−bt/2a
2a
2a
Since W (y , y )(t) is never zero, the solutions y and y given by Eq. (19) are a
1
2
1
2
fundamental set of solutions. Further, Eq. (18) is the general solution of Eq. (1) when the roots of the characteristic equation are equal. In other words, in this case, there is one exponential solution corresponding to the repeated root, while a second solution is obtained by multiplying the exponential solution by t.
Find the solution of the initial value problem E X A M P L E
2
y − y + 0.25y = 0,y(0) = 2,y(0) = 1 .
(21)
3
The characteristic equation is r 2 − r + 0.25 = 0,Chapter 3. Second Order Linear Equations so the roots are r = r = 1/2. Thus the general solution of the differential equation is
1
2
y = c et/2 + c tet/2.
(22)
1
2
The first initial condition requires that y(0) = c = 2.
1
To satisfy the second initial condition, we first differentiate Eq. (22) and then set t = 0.
This gives y(0) = 1 c + c = 1 , 2 1
2
3
so c = −2/3. Thus, the solution of the initial value problem is
2
y = 2et/2 − 2 tet/2.
(23)
3
The graph of this solution is shown in Figure 3.5.2.
y
4
y'(0) = 2: y = 2et/2 + tet/2
3
y'(0) = :
1
y = 2et/2 – 2 tet/2
3
3
2
1
1
2
3
t
–1
FIGURE 3.5.2
Solutions of y − y + 0.25y = 0, y(0) = 2, with y(0) = 1/3 and with y(0) = 2, respectively.
Let us now modify the initial value problem (21) by changing the initial slope; to be specific, let the second initial condition be y(0) = 2. The solution of this modified problem is y = 2et/2 + tet/2 and its graph is also shown in Figure 3.5.2. The graphs shown in this figure suggest that there is a critical initial slope, with a value between 1 and 2, that separates solutions
3
that grow positively from those that ultimately grow negatively. In Problem 16 you are asked to determine this critical initial slope.
3.5
Repeated Roots; Reduction of Order
The geometrical behavior of solutions is similar in this case to that when the roots are real and different. If the exponents are either positive or negative, then the magnitude of the solution grows or decays accordingly; the linear factor t has little influence.
A decaying solution is shown in Figure 3.5.1 and growing solutions in However, if the repeated root is zero, then the differential equation is y = 0 and the general solution is a linear function of t.
Summary.
We can now summarize the results that we have obtained for second order linear homogeneous equations with constant coefficients, ay + by + cy = 0.
(1)
Let r and r be the roots of the corresponding characteristic polynomial
1
2
ar 2 + br + c = 0.
(2)
If r and r are real but not equal, then the general solution of the differential
1
2
equation (1) is y = c er t
t
1
+ c er2 .
(24)
1
2
If r and r are complex conjugates λ ± iµ, then the general solution is
1
2
y = c eλt cos µt + c eλt sin µt.
(25)
1
2
If r = r , then the general solution is
1
2
y = c er t
t
1
+ c ter1 .
(26)
1
2
Reduction of Order.
It is worth noting that the procedure used earlier in this section for equations with constant coefficients is more generally applicable. Suppose we know one solution y (t), not everywhere zero, of
1
y + p(t)y + q(t)y = 0.
(27)
To find a second solution, let y = v(t)y (t); (28)
1
then y = v(t)y (t) + v(t)y (t)
1
1
and y = v(t)y (t) + 2v(t)y (t) + v(t)y(t).
1
1
1
Substituting for y, y, and y in Eq. (27) and collecting terms, we find that y v + (2y + py )v + (y + py + qy )v = 0.
(29)
1
1
1
1
1
1
Since y is a solution of Eq. (27), the coefficient of v in Eq. (29) is zero, so that Eq. (29)
1
becomes y v + (2y + py )v = 0.
(30)
1
1
1
Despite its appearance, Eq. (30) is actually a first order equation for the function v and can be solved either as a first order linear equation or as a separable equation. Once v has been found, then v is obtained by an integration. Finally, y is determined from Chapter 3. Second Order Linear Equations Eq. (28). This procedure is called the method of reduction of order because the crucial step is the solution of a first order differential equation for v rather than the original second order equation for y. Although it is possible to write down a formula for v(t), we will instead illustrate how this method works by an example.
Given that y (t) = t−1 is a solution of
1
E X A M P L E
3
2t2 y + 3t y − y = 0,t > 0, (31)
find a second linearly independent solution.
We set y = v(t)t−1; then y = vt−1 − vt−2,y = vt−1 − 2vt−2 + 2vt−3.
Substituting for y, y, and y in Eq. (31) and collecting terms, we obtain 2t2(vt−1 − 2vt−2 + 2vt−3) + 3t (vt−1 − vt−2) − vt−1 = 2tv + (−4 + 3)v + (4t−1 − 3t−1 − t−1)v
= 2tv − v = 0.
(32)
Note that the coefficient of v is zero, as it should be; this provides a useful check on our algebra.
Separating the variables in Eq. (32) and solving for v(t), we find that v(t) = ct1/2 ; then v(t) = 2ct3/2 + k.
3
It follows that y = 2 ct1/2 + kt−1,
(33)
3
where c and k are arbitrary constants. The second term on the right side of Eq. (33) is a multiple of y (t) and can be dropped, but the first term provides a new independent
1
solution. Neglecting the arbitrary multiplicative constant, we have y (t) = t1/2.
2
PROBLEMS
In each of Problems 1 through 10 find the general solution of the given differential equation.
3.5
Repeated Roots; Reduction of Order 䉴 15. Consider the initial value problem
(b) Determine where the solution has the value zero.
0
0
16. Consider the following modification of the initial value problem in Example 2:
䉴 17. Consider the initial value problem (a) Solve the initial value problem and plot the solution.
0
0
M
M
M
M
18. Consider the initial value problem (a) Solve the initial value problem.
0 can take on the value zero at most once.
1
2
1
2
1
2
1
1
1
2
1
2
1
2
1
2
1
2
2
1
2
1
2
1
1
1
2
2
1
1
(i)
1
Chapter 3. Second Order Linear Equations
1
1
∂
∂
=
).
ert
(ii)
r
∂r
1
1
1
1
1
1
1
1
1
1
√
1
1
31. The differential equation
1
2
2
2
1!
2!
2
1
32. The differential equation
1
an integral.
33. The method of Problem 20 can be extended to second order equations with variable
1
2
2
1
1
2
1
1
2
1
2
2
1
1
6 T. A. Newton, “On Using a Differential Equation to Generate Polynomials,” American Mathematical Monthly
81 (1974), pp. 592 – 601. Also see the references given there.
3.6
Nonhomogeneous Equations; Method of Undetermined Coefficients
1
1
0
0
Euler Equations.
Use the substitution introduced in Problem 38 in Section 3.4 to solve each of the equations in Problems 41 and 42.
3.6 Nonhomogeneous Equations; Method of Undetermined Coefficients
We now return to the nonhomogeneous equation L[y] = y + p(t)y + q(t)y = g(t), (1)
where p, q, and g are given (continuous) functions on the open interval I . The equation L[y] = y + p(t)y + q(t)y = 0, (2)
in which g(t) = 0 and p and q are the same as in Eq. (1), is called the homogeneous equation corresponding to Eq. (1). The following two results describe the structure of solutions of the nonhomogeneous equation (1) and provide a basis for constructing its general solution.
Chapter 3. Second Order Linear Equations
Theorem 3.6.1 If Y and Y are two solutions of the nonhomogeneous equation (1), then their
1
2
difference Y − Y is a solution of the corresponding homogeneous equation (2). If,
1
2
in addition, y and y are a fundamental set of solutions of Eq. (2), then
1
2
Y (t) − Y (t) = c y (t) + c y (t), (3)
1
2
1 1
2 2 where c and c are certain constants.
1
2
To prove this result, note that Y and Y satisfy the equations
1
2
L[Y ](t) = g(t),L[Y ](t) = g(t).
(4)
1
2
Subtracting the second of these equations from the first, we have L[Y ](t) − L[Y ](t) = g(t) − g(t) = 0.
(5)
1
2
However, L[Y ] − L[Y ] = L[Y − Y ], 1
2
1
2
so Eq. (5) becomes L[Y − Y ](t) = 0.
(6)
1
2
Equation (6) states that Y − Y is a solution of Eq. (2). Finally, since all solutions of
1
2
Eq. (2) can be expressed as linear combinations of a fundamental set of solutions by Theorem 3.2.4, it follows that the solution Y − Y can be so written. Hence Eq. (3)
1
2
holds and the proof is complete.
Theorem 3.6.2
The general solution of the nonhomogeneous equation (1) can be written in the form y = φ(t) = c y (t) + c y (t) + Y (t), (7)
1 1
2 2 where y and y are a fundamental set of solutions of the corresponding homogeneous
1
2
equation (2), c and c are arbitrary constants, and Y is some specific solution of the
1
2
nonhomogeneous equation (1).
The proof of Theorem 3.6.2 follows quickly from the preceding theorem. Note that Eq. (3) holds if we identify Y with an arbitrary solution φ of Eq. (1) and Y with the
1
2
specific solution Y . From Eq. (3) we thereby obtain φ(t) − Y (t) = c y (t) + c y (t), (8)
1 1
2 2 which is equivalent to Eq. (7). Since φ is an arbitrary solution of Eq. (1), the expression on the right side of Eq. (7) includes all solutions of Eq. (1); thus it is natural to call it the general solution of Eq. (1).
In somewhat different words, Theorem 3.6.2 states that to solve the nonhomogeneous equation (1), we must do three things: 1.
Find the general solution c y (t) + c y (t) of the corresponding homogeneous 1 1
2 2 equation. This solution is frequently called the complementary solution and may be denoted by y (t).
c
2.
Find some single solution Y (t) of the nonhomogeneous equation. Often this solution is referred to as a particular solution.
3.
Add together the functions found in the two preceding steps.
3.6
Nonhomogeneous Equations; Method of Undetermined Coefficients
We have already discussed how to find y (t), at least when the homogeneous equa c tion (2) has constant coefficients. Therefore, in the remainder of this section and in the next, we will focus on finding a particular solution Y (t) of the nonhomogeneous equation (1). There are two methods that we wish to discuss. They are known as the method of undetermined coefficients and the method of variation of parameters, respectively.
Each has some advantages and some possible shortcomings.
Method of Undetermined Coefficients.
The method of undetermined coefficients re quires that we make an initial assumption about the form of the particular solution Y (t), but with the coefficients left unspecified. We then substitute the assumed expression into Eq. (1) and attempt to determine the coefficients so as to satisfy that equation. If we are successful, then we have found a solution of the differential equation (1) and can use it for the particular solution Y (t). If we cannot determine the coefficients, then this means that there is no solution of the form that we assumed. In this case we may modify the initial assumption and try again.
The main advantage of the method of undetermined coefficients is that it is straight forward to execute once the assumption is made as to the form of Y (t). Its major limitation is that it is useful primarily for equations for which we can easily write down the correct form of the particular solution in advance. For this reason, this method is usually used only for problems in which the homogeneous equation has constant coefficients and the nonhomogeneous term is restricted to a relatively small class of functions. In particular, we consider only nonhomogeneous terms that consist of polynomials, exponential functions, sines, and cosines. Despite this limitation, the method of undetermined coefficients is useful for solving many problems that have important applications. However, the algebraic details may become tedious and a computer algebra system can be very helpful in practical applications. We will illustrate the method of undetermined coefficients by several simple examples and then summarize some rules for using it.
Find a particular solution of E X A M P L E
1
y − 3y − 4y = 3e2t .
(9)
We seek a function Y such that the combination Y (t) − 3Y (t) − 4Y (t) is equal to 3e2t . Since the exponential function reproduces itself through differentiation, the most plausible way to achieve the desired result is to assume that Y (t) is some multiple of e2t , that is, Y (t) = Ae2t , where the coefficient A is yet to be determined. To find A we calculate Y (t) = 2Ae2t ,Y (t) = 4Ae2t , and substitute for y, y, and y in Eq. (9). We obtain (4A − 6A − 4A)e2t = 3e2t.
Hence −6Ae2t must equal 3e2t , so A = −1/2. Thus a particular solution is Y (t) = − 1 e2t .
(10)
2
Chapter 3. Second Order Linear Equations
Find a particular solution of E X A M P L E
2
y − 3y − 4y = 2 sin t.
(11)
By analogy with Example 1, let us first assume that Y (t) = A sin t, where A is a constant to be determined. On substituting in Eq. (11) and rearranging the terms, we obtain −5A sin t − 3A cos t = 2 sin t, or (2 + 5A) sin t + 3A cos t = 0.
(12)
The functions sin t and cos t are linearly independent, so Eq. (12) can hold on an interval only if the coefficients 2 + 5A and 3A are both zero. These contradictory requirements mean that there is no choice of the constant A that makes Eq. (12) true for all t. Thus we conclude that our assumption concerning Y (t) is inadequate. The appearance of the cosine term in Eq. (12) suggests that we modify our original assumption to include a cosine term in Y (t), that is, Y (t) = A sin t + B cos t, where A and B are to be determined. Then Y (t) = A cos t − B sin t,Y (t) = −A sin t − B cos t.
By substituting these expressions for y, y, and y in Eq. (11) and collecting terms, we obtain (−A + 3B − 4A) sin t + (−B − 3A − 4B) cos t = 2 sin t.
(13) To satisfy Eq. (13) we must match the coefficients of sin t and cos t on each side of the equation; thus A and B must satisfy the equations −5A + 3B = 2, −3A − 5B = 0.
Hence A = −5/17 and B = 3/17, so a particular solution of Eq. (11) is Y (t) = − 5 sin t + 3 cos t.
17
17
The method illustrated in the preceding examples can also be used when the right side of the equation is a polynomial. Thus, to find a particular solution of y − 3y − 4y = 4t2 − 1, (14)
we initially assume that Y (t) is a polynomial of the same degree as the nonhomogeneous term, that is, Y (t) = At2 + Bt + C.
To summarize our conclusions up to this point: if the nonhomogeneous term g(t) in Eq. (1) is an exponential function eαt , then assume that Y (t) is proportional to the same exponential function; if g(t) is sin βt or cos βt, then assume that Y (t) is a linear combination of sin βt and cos βt; if g(t) is a polynomial, then assume that Y (t) is a polynomial of like degree. The same principle extends to the case where g(t) is a product of any two, or all three, of these types of functions, as the next example illustrates.
3.6
Nonhomogeneous Equations; Method of Undetermined Coefficients
Find a particular solution of E X A M P L E
3
y − 3y − 4y = −8et cos 2t.
(15)
In this case we assume that Y (t) is the product of et and a linear combination of cos 2t and sin 2t, that is, Y (t) = Aet cos 2t + Bet sin 2t.
The algebra is more tedious in this example, but it follows that Y (t) = (A + 2B)et cos 2t + (−2A + B)et sin 2t and Y (t) = (−3A + 4B)et cos 2t + (−4A − 3B)et sin 2t.
By substituting these expressions in Eq. (15), we find that A and B must satisfy 10 A + 2B = 8, 2 A − 10B = 0.
Hence A = 10/13 and B = 2/13; therefore a particular solution of Eq. (15) is Y (t) = 10 et cos 2t + 2 et sin 2t.
13
13
Now suppose that g(t) is the sum of two terms, g(t) = g (t) + g (t), and suppose
1
2
that Y and Y are solutions of the equations
1
2
ay + by + cy = g (t)
(16)
1
and ay + by + cy = g (t), (17)
2
respectively. Then Y + Y is a solution of the equation
1
2
ay + by + cy = g(t).
(18)
To prove this statement, substitute Y (t) + Y (t) for y in Eq. (18) and make use
1
2
of Eqs. (16) and (17). A similar conclusion holds if g(t) is the sum of any finite number of terms. The practical significance of this result is that for an equation whose nonhomogeneous function g(t) can be expressed as a sum, one can consider instead several simpler equations and then add the results together. The following example is an illustration of this procedure.
Find a particular solution of E X A M P L E
4
y − 3y − 4y = 3e2t + 2 sin t − 8et cos 2t.
(19)
By splitting up the right side of Eq. (19), we obtain the three equations y − 3y − 4y = 3e2t ,
y − 3y − 4y = 2 sin t, and y − 3y − 4y = −8et cos 2t.
Chapter 3. Second Order Linear Equations
Solutions of these three equations have been found in Examples 1, 2, and 3, respectively.
Therefore a particular solution of Eq. (19) is their sum, namely, Y (t) = − 1 e2t + 3 cos t − 5 sin t + 10 et cos 2t + 2 et sin 2t.
2
17
17
13
13
The procedure illustrated in these examples enables us to solve a fairly large class of problems in a reasonably efficient manner. However, there is one difficulty that sometimes occurs. The next example illustrates how it arises.
Find a particular solution of E X A M P L E
5
y + 4y = 3 cos 2t.
(20)
Proceeding as in Example 2, we assume that Y (t) = A cos 2t + B sin 2t. By substi tuting in Eq. (20), we then obtain (4A − 4A) cos 2t + (4B − 4B) sin 2t = 3 cos 2t.
(21)
Since the left side of Eq. (21) is zero, there is no choice of A and B that satisfies this equation. Therefore, there is no particular solution of Eq. (20) of the assumed form. The reason for this possibly unexpected result becomes clear if we solve the homogeneous equation y + 4y = 0 (22)
that corresponds to Eq. (20). A fundamental set of solutions of Eq. (22) is y (t) =
1
cos 2t and y (t) = sin 2t. Thus our assumed particular solution of Eq. (20) is actually
2
a solution of the homogeneous equation (22); consequently, it cannot possibly be a solution of the nonhomogeneous equation (20).
To find a solution of Eq. (20) we must therefore consider functions of a somewhat different form. The simplest functions, other than cos 2t and sin 2t themselves, that when differentiated lead to cos 2t and sin 2t are t cos 2t and t sin 2t. Hence we assume that Y (t) = At cos 2t + Bt sin 2t. Then, upon calculating Y (t) and Y (t), substituting them into Eq. (20), and collecting terms, we find that −4A sin 2t + 4B cos 2t = 3 cos 2t.
Therefore A = 0 and B = 3/4, so a particular solution of Eq. (20) is Y (t) = 3 t sin 2t.
4
The fact that in some circumstances a purely oscillatory forcing term leads to a solution that includes a linear factor t as well as an oscillatory factor is important in some applications; see Section 3.9 for a further discussion.
The outcome of Example 5 suggests a modification of the principle stated previously:
If the assumed form of the particular solution duplicates a solution of the corresponding homogeneous equation, then modify the assumed particular solution by multiplying it by t. Occasionally, this modification will be insufficient to remove all duplication with the solutions of the homogeneous equation, in which case it is necessary to multiply by t a second time. For a second order equation, it will never be necessary to carry the process further than this.
Summary.
We now summarize the steps involved in finding the solution of an initial value problem consisting of a nonhomogeneous equation of the form ay + by + cy = g(t), (23)
where the coefficients a, b, and c are constants, together with a given set of initial conditions:
1.
Find the general solution of the corresponding homogeneous equation.
2.
Make sure that the function g(t) in Eq. (23) belongs to the class of functions discussed in this section, that is, it involves nothing more than exponential functions, sines, cosines, polynomials, or sums or products of such functions. If this is not the case, use the method of variation of parameters (discussed in the next section).
3.
If g(t) = g (t) + · · · + g (t), that is, if g(t) is a sum of n terms, then form n
1
n
subproblems, each of which contains only one of the terms g (t), . . . , g (t). The
1
ni th subproblem consists of the equation ay + by + cy = g (t),
i where i runs from 1 to n.
4.
For the i th subproblem assume a particular solution Y (t) consisting of the ap i
propriate exponential function, sine, cosine, polynomial, or combination thereof.
If there is any duplication in the assumed form of Y (t) with the solutions of the i homogeneous equation (found in step 1), then multiply Y (t) by t, or (if necessary)
i by t2, so as to remove the duplication. See Table 3.6.1.
5.
Find a particular solution Y (t) for each of the subproblems. Then the sum Y (t) +
i
1
· · · + Y (t) is a particular solution of the full nonhomogeneous equation (23).
n
6.
Form the sum of the general solution of the homogeneous equation (step 1) and the particular solution of the nonhomogeneous equation (step 5). This is the general solution of the nonhomogeneous equation.
7.
Use the initial conditions to determine the values of the arbitrary constants remaining in the general solution.
TABLE 3.6.1 The Particular Solution of ay + by + cy = g (t)ig (t)
Y (t)
i
i
P (t) = a tn + a tn−1 + · · · + ats (A tn + A tn−1 + · · · + A )
n
0
1
n
0
1
n
P (t)eαtts (A tn + A tn−1 + · · · + A )eαt
n
0
1
n
sinβt
P (t)eαtts [(A tn + A tn−1 + · · · + A )eαt cos βt
n
cos βt
0
1
n
+(B tn + B tn−1 + · · · + B )eαt sin βt]
0
1
n
Notes. Here s is the smallest nonnegative integer (s = 0, 1, or 2) that will ensure that no term in Y (t) is a solution of the corresponding homogeneous equation. Equivalently, for the three cases,
i
s is the number of times 0 is a root of the characteristic equation, α is a root of the characteristic equation, and α + iβ is a root of the characteristic equation, respectively.
Chapter 3. Second Order Linear Equations
For some problems this entire procedure is easy to carry out by hand, but in many cases it requires considerable algebra. Once you understand clearly how the method works, a computer algebra system can be of great assistance in executing the details.
The method of undetermined coefficients is self-correcting in the sense that if one assumes too little for Y (t), then a contradiction is soon reached that usually points the way to the modification that is needed in the assumed form. On the other hand, if one assumes too many terms, then some unnecessary work is done and some coefficients turn out to be zero, but at least the correct answer is obtained.
Proof of the Method of Undetermined Coefficients.
In the preceding discussion we have described the method of undetermined coefficients on the basis of several examples. To prove that the procedure always works as stated, we now give a general argument, in which we consider several cases corresponding to different forms for the nonhomogeneous term g(t).
g(t) = P (t) = a tn + a tn−1 + · · · + a . In this case Eq. (23) becomes
n
0
1
n
ay + by + cy = a tn + a tn−1 + · · · + a .
(24)
0
1
n
To obtain a particular solution we assume that Y (t) = A tn + A tn−1 + · · · + At2 + At + A .
(25)
0
1
n−2
n−1
n
Substituting in Eq. (24), we obtain a[n(n − 1)A tn−2 + · · · + 2A ] + b(n A tn−1 + · · · + A)
0
n−2
0
n−1
+ c(A tn + A tn−1 + · · · + A ) = a tn + · · · + a .
(26)
0
1
n
0
n
Equating the coefficients of like powers of t gives c A = a ,
0
0
c A + nb A = a ,
1
0
1
...
c A + b A + 2a A = a .
n
n−1
n−2
n
Provided that c = 0, the solution of the first equation is A = a /c, and the remaining
0
0
equations determine A , . . . , A successively. If c = 0, but b = 0, then the polynomial
1
n
on the left side of Eq. (26) is of degree n − 1, and we cannot satisfy Eq. (26). To be sure that aY (t) + bY (t) is a polynomial of degree n, we must choose Y (t) to be a polynomial of degree n + 1. Hence we assume that Y (t) = t (A tn + · · · + A ).
0
n
There is no constant term in this expression for Y (t), but there is no need to include such a term since a constant is a solution of the homogeneous equation when c = 0.
Since b = 0, we have A = a /b(n + 1), and the other coefficients A , . . . , A can be
0
0
1
n
determined similarly. If both c and b are zero, we assume that Y (t) = t2(A tn + · · · + A ).
0
n
The term aY (t) gives rise to a term of degree n, and we can proceed as before. Again the constant and linear terms in Y (t) are omitted, since in this case they are both solutions of the homogeneous equation.
3.6
Nonhomogeneous Equations; Method of Undetermined Coefficientsg(t) = eαt P (t). The problem of determining a particular solution of n
ay + by + cy = eαt P (t)
(27)
n
can be reduced to the preceding case by a substitution. Let Y (t) = eαt u(t); then Y (t) = eαt [u(t) + αu(t)] and Y (t) = eαt [u(t) + 2αu(t) + α2u(t)].
Substituting for y, y, and y in Eq. (27), canceling the factor eαt , and collecting terms, we obtain au(t) + (2aα + b)u(t) + (aα2 + bα + c)u(t) = P (t).
(28)
n
The determination of a particular solution of Eq. (28) is precisely the same problem, except for the names of the constants, as solving Eq. (24). Therefore, if aα2 + bα + c
is not zero, we assume that u(t) = A tn + · · · + A ; hence a particular solution of
0
n
Eq. (27) is of the form Y (t) = eαt (A tn + A tn−1 + · · · + A ).
(29)
0
1
n
On the other hand, if aα2 + bα + c is zero, but 2aα + b is not, we must take u(t) to be of the form t (A tn + · · · + A ). The corresponding form for Y (t) is t times the
0
n
expression on the right side of Eq. (29). Note that if aα2 + bα + c is zero, then eαt is a solution of the homogeneous equation. If both aα2 + bα + c and 2aα + b are zero (and this implies that both eαt and teαt are solutions of the homogeneous equation), then the correct form for u(t) is t2(A tn + · · · + A ). Hence Y (t) is t2 times the expression on
0
n the right side of Eq. (29).
g(t) = eαt P (t) cos β t or eαt P (t) sin β t. These two cases are similar, so we con n
n
sider only the latter. We can reduce this problem to the preceding one by noting that, as a consequence of Euler’s formula, sin βt = (eiβt − e−iβt )/2i. Hence g(t) is of the form g(t) = P (t) e(α+iβ)t − e(α−iβ)t
n
2i
and we should choose Y (t) = e(α+iβ)t (A tn + · · · + A ) + e(α−iβ)t (B tn + · · · + B ), 0
n
0
n
or equivalently, Y (t) = eαt (A tn + · · · + A ) cos βt + eαt (B tn + · · · + B ) sin βt.
0
n
0
n
Usually, the latter form is preferred. If α ± iβ satisfy the characteristic equation corresponding to the homogeneous equation, we must, of course, multiply each of the polynomials by t to increase their degrees by one.
If the nonhomogeneous function involves both cos βt and sin βt, it is usually con venient to treat these terms together, since each one individually may give rise to the Chapter 3. Second Order Linear Equations same form for a particular solution. For example, if g(t) = t sin t + 2 cos t, the form for Y (t) would be Y (t) = (A t + A ) sin t + (B t + B ) cos t, 0
1
0
1
provided that sin t and cos t are not solutions of the homogeneous equation.
PROBLEMS
In each of Problems 1 through 12 find the general solution of the given differential equation.
0
0
0
0
In each of Problems 13 through 18 find the solution of the given initial value problem.
In each of Problems 19 through 26:
䉴
䉴 䉴 䉴 䉴 䉴 䉴
27. Determine the general solution of
N
m
䉴 28. In many physical problems the nonhomogeneous term may be specified by different for-
t > π,
䉴 29. Follow the instructions in Problem 28 to solve the differential equation
(i)
1
2
1
2
32. In this problem we indicate an alternate procedure7 for solving the differential equation (i)
1
2
(a) Verify that Eq. (i) can be written in the factored form
1
2
1
2
1 2
2
following two first order equations:
1
2
(see Example 1) (see Problem 7) (see Problem 6) (see Problem 4)
3.7 Variation of Parameters
In this section we describe another method of finding a particular solution of a nonhomogeneous equation. This method, known as is due to Lagrange and complements the method of undetermined coefficients rather well. The main advantage of variation of parameters is that it is a general method; in principle at least, it can be applied to any equation, and it requires no detailed assumptions about 7R. S. Luthar, “Another Approach to a Standard Differential Equation,” Two Year College Mathematics Journal 10
(1979), pp. 200 –201; also see D. C. Sandell and F. M. Stein, “Factorization of Operators of Second Order Linear Homogeneous Ordinary Differential Equations,” Two Year College Mathematics Journal 8 (1977), pp. 132–141, for a more general discussion of factoring operators.
Chapter 3. Second Order Linear Equations the form of the solution. In fact, later in this section we use this method to derive a formula for a particular solution of an arbitrary second order linear nonhomogeneous differential equation. On the other hand, the method of variation of parameters eventually requires that we evaluate certain integrals involving the nonhomogeneous term in the differential equation, and this may present difficulties. Before looking at this method in the general case, we illustrate its use in an example.
Find a particular solution of E X A M P L E
1
y + 4y = 3 csc t.
(1)
Observe that this problem does not fall within the scope of the method of un determined coefficients because the nonhomogeneous term g(t) = 3 csc t involves a quotient (rather than a sum or a product) of sin t or cos t. Therefore, we need a different approach. Observe also that the homogeneous equation corresponding to Eq. (1) is y + 4y = 0, (2)
and that the general solution of Eq. (2) is y (t) = c cos 2t + c sin 2t.
(3)
c
1
2
The basic idea in the method of variation of parameters is to replace the constants c1 and c in Eq. (3) by functions u (t) and u (t), respectively, and then to determine these
2
1
2
functions so that the resulting expression y = u (t) cos 2t + u (t) sin 2t (4)
1
2
is a solution of the nonhomogeneous equation (1).
To determine u and u we need to substitute for y from Eq. (4) in Eq. (1). However,
1
2
even without carrying out this substitution, we can anticipate that the result will be a single equation involving some combination of u , u , and their first two derivatives.
1
2
Since there is only one equation and two unknown functions, we can expect that there are many possible choices of u and u that will meet our needs. Alternatively, we
1
2
may be able to impose a second condition of our own choosing, thereby obtaining two equations for the two unknown functions u and u . We will soon show (following
1
2
Lagrange) that it is possible to choose this second condition in a way that makes the computation markedly more efficient.
Returning now to Eq. (4), we differentiate it and rearrange the terms, thereby obtaining y = −2u (t) sin 2t + 2u (t) cos 2t + u (t) cos 2t + u (t) sin 2t.
(5)
1
2
1
2
Keeping in mind the possibility of choosing a second condition on u and u , let us
1
2
require the last two terms on the right side of Eq. (5) to be zero; that is, we require that u (t) cos 2t + u (t) sin 2t = 0.
(6)
1
2
It then follows from Eq. (5) that y = −2u (t) sin 2t + 2u (t) cos 2t.
(7)
1
2
3.7
Variation of Parameters
Although the ultimate effect of the condition (6) is not yet clear, at the very least it has simplified the expression for y. Further, by differentiating Eq. (7), we obtain y = −4u (t) cos 2t − 4u (t) sin 2t − 2u (t) sin 2t + 2u (t) cos 2t.
(8)
1
2
1
2
Then, substituting for y and y in Eq. (1) from Eqs. (4) and (8), respectively, we find that u and u must satisfy
1
2
−2u (t) sin 2t + 2u (t) cos 2t = 3 csc t.
(9)
1
2
Summarizing our results to this point, we want to choose u and u so as to satisfy
1
2
Eqs. (6) and (9). These equations can be viewed as a pair of linear algebraic equations for the two unknown quantities u (t) and u (t). Equations (6) and (9) can be solved in
1
2
various ways. For example, solving Eq. (6) for u (t), we have
2
u (t) = −u (t) cos 2t .
(10)
2
1
sin 2t
Then, substituting for u (t) in Eq. (9) and simplifying, we obtain
2
u (t) = − 3 csc t sin 2t = −3 cos t.
(11)
1
2
Further, putting this expression for u (t) back in Eq. (10) and using the double angle
1
formulas, we find that u (t) = 3 cos t cos 2t = 3(1 − 2 sin2 t) = 3 csc t − 3 sin t.
(12)
2
sin 2t 2 sin t
2
Having obtained u (t) and u (t), the next step is to integrate so as to obtain u (t)
1
2
1
and u (t). The result is
2
u (t) = −3 sin t + c
(13)
1
1
and u (t) = 3 ln | csc t − cot t| + 3 cos t + c .
(14)
2
2
2
Finally, on substituting these expressions in Eq. (4), we have y = −3 sin t cos 2t + 3 ln | csc t − cot t| sin 2t + 3 cos t sin 2t
2
+ c cos 2t + c sin 2t,
1
2
or y = 3 sin t + 3 ln | csc t − cot t| sin 2t + c cos 2t + c sin 2t.
(15)
2
1
2
The terms in Eq. (15) involving the arbitrary constants c and c are the general solution
1
2
of the corresponding homogeneous equation, while the remaining terms are a particular solution of the nonhomogeneous equation (1). Therefore Eq. (15) is the general solution of Eq. (1).
In the preceding example the method of variation of parameters worked well in determining a particular solution, and hence the general solution, of Eq. (1). The next
Chapter 3. Second Order Linear Equations question is whether this method can be applied effectively to an arbitrary equation.
Therefore we consider y + p(t)y + q(t)y = g(t), (16)
where p, q, and g are given continuous functions. As a starting point, we assume that we know the general solution y (t) = c y (t) + c y (t)
(17)
c 1 1
2 2 of the corresponding homogeneous equation y + p(t)y + q(t)y = 0.
(18)
This is a major assumption because so far we have shown how to solve Eq. (18) only if it has constant coefficients. If Eq. (18) has coefficients that depend on t, then usually the methods described in Chapter 5 must be used to obtain y (t).
c
The crucial idea, as illustrated in Example 1, is to replace the constants c and c in
1
2
Eq. (17) by functions u (t) and u (t), respectively; this gives
1
2
y = u (t)y (t) + u (t)y (t).
(19)
1
1
2
2
Then we try to determine u (t) and u (t) so that the expression in Eq. (19) is a solution
1
2
of the nonhomogeneous equation (16) rather than the homogeneous equation (18).
Thus we differentiate Eq. (19), obtaining y = u (t)y (t) + u (t)y (t) + u (t)y (t) + u (t)y (t).
(20)
1
1
1
1
2
2
2
2
As in Example 1, we now set the terms involving u (t) and u (t) in Eq. (20) equal to
1
2
zero; that is, we require that u (t)y (t) + u (t)y (t) = 0.
(21)
1
1
2
2
Then, from Eq. (20), we have y = u (t)y (t) + u (t)y (t).
(22)
1
1
2
2
Further, by differentiating again, we obtain y = u (t)y (t) + u (t)y(t) + u (t)y (t) + u (t)y(t).
(23)
1
1
1
1
2
2
2
2
Now we substitute for y, y, and y in Eq. (16) from Eqs. (19), (22), and (23), respectively. After rearranging the terms in the resulting equation we find that u (t)[y(t) + p(t)y (t) + q(t)y (t)]
1
1
1
1
+ u (t)[y(t) + p(t)y (t) + q(t)y (t)]
2
2
2
2
+ u (t)y (t) + u (t)y (t) = g(t).
(24)
1
1
2
2
Each of the expressions in square brackets in Eq. (24) is zero because both y and y
1
2
are solutions of the homogeneous equation (18). Therefore Eq. (24) reduces to u (t)y (t) + u (t)y (t) = g(t).
(25)
1
1
2
2
Equations (21) and (25) form a system of two linear algebraic equations for the derivatives u (t) and u (t) of the unknown functions. They correspond exactly to
1
2
Eqs. (6) and (9) in Example 1.
3.7
Variation of Parameters
By solving the system (21), (25) we obtain y (t)g(t)y (t)g(t)u (t) = −
2
,u (t) =
1
,
(26)
1
W (y ,y )(t)
2
W (y ,y )(t)
1
2
1
2
where W (y ,y ) is the Wronskian of y and y . Note that division by W is permissible
1
2
1
2
since y and y are a fundamental set of solutions, and therefore their Wronskian is
1
2
nonzero. By integrating Eqs. (26) we find the desired functions u (t) and u (t), namely,
1
2
y (t)g(t)y (t)g(t)u (t) = −
2
,u (t) =
1 .
(27)
1
W (y ,y )(t) dt + c1
2
W (y ,y )(t) dt + c2
1
2
1
2
Finally, substituting from Eq. (27) in Eq. (19) gives the general solution of Eq. (16).
We state the result as a theorem.
Theorem 3.7.1
If the functions p, q, and g are continuous on an open interval I , and if the functions y and y are linearly independent solutions of the homogeneous equation (18)
1
2
corresponding to the nonhomogeneous equation (16), y + p(t)y + q(t)y = g(t), then a particular solution of Eq. (16) is
y (t)g(t)y (t)g(t)Y (t) = −y (t)
2
(t)
1
1
W (y ,y )(t) dt + y2 W (y ,y )(t) dt, (28)
1
2
1
2
and the general solution is y = c y (t) + c y (t) + Y (t), (29)
1 1
2 2 as prescribed by Theorem 3.6.2.
By examining the expression (28) and reviewing the process by which we derived it, we can see that there may be two major difficulties in using the method of variation of parameters. As we have mentioned earlier, one is the determination of y (t) and y (t), 1
2
a fundamental set of solutions of the homogeneous equation (18), when the coefficients in that equation are not constants. The other possible difficulty is in the evaluation of the integrals appearing in Eq. (28). This depends entirely on the nature of the functions y , y , and g. In using Eq. (28), be sure that the differential equation is exactly in the
1
2
form (16); otherwise, the nonhomogeneous term g(t) will not be correctly identified.
A major advantage of the method of variation of parameters is that Eq. (28) provides an expression for the particular solution Y (t) in terms of an arbitrary forcing function g(t). This expression is a good starting point if you wish to investigate the effect of variations in the forcing function, or if you wish to analyze the response of a system to a number of different forcing functions.
PROBLEMS
In each of Problems 1 through 4 use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients.
Chapter 3. Second Order Linear Equations
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
21. Show that the solution of the initial value problem
(i)
0
0
0
0
(ii)
0
0
0
0
(iii)
0
0
0
1
2
1
2
(
(
(
(
t
0
1
2
1
2
0
0
23. (a) Use the result of Problem 22 to show that the solution of the initial value problem
(i)
0
0
is
(ii)
(b) Find the solution of the initial value problem
0
0
3.7
Variation of Parameters
24. Use the result of Problem 22 to find the solution of the initial value problem
0
0
25. Use the result of Problem 22 to find the solution of the initial value problem
0
0
26. Use the result of Problem 22 to find the solution of the initial value problem
0
0
27. By combining the results of Problems 24 through 26, show that the solution of the initial value problem
0
0
(i)
1
2
0
28. The method of reduction of order (Section 3.5) can also be used for the nonhomogeneous equation
(i)
1
1
(ii)
1
1
1
1
1
1
(see Problem 15)
1
(see Problem 16)
1