Download Second-Order Linear Differential Equations

Second-Order Linear Differential Equations
A second-order linear differential equation has the form
1
Px
d 2y
dy
Rxy Gx
2 Qx
dx
dx
where P, Q, R, and G are continuous functions. We saw in Section 7.1 that equations of
this type arise in the study of the motion of a spring. In Additional Topics: Applications of
Second-Order Differential Equations we will further pursue this application as well as the
application to electric circuits.
In this section we study the case where Gx 0, for all x, in Equation 1. Such equations are called homogeneous linear equations. Thus, the form of a second-order linear
homogeneous differential equation is
2
Px
d 2y
dy
Rxy 0
Qx
dx 2
dx
If Gx 0 for some x, Equation 1 is nonhomogeneous and is discussed in Additional
Topics: Nonhomogeneous Linear Equations.
Two basic facts enable us to solve homogeneous linear equations. The first of these says
that if we know two solutions y1 and y2 of such an equation, then the linear combination
y c1 y1 c2 y2 is also a solution.
3 Theorem If y1x and y2x are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the function
yx c1 y1x c2 y2x
is also a solution of Equation 2.
Proof Since y1 and y2 are solutions of Equation 2, we have
Pxy1 Qxy1 Rxy1 0
and
Pxy2 Qxy2 Rxy2 0
Therefore, using the basic rules for differentiation, we have
Pxy Qxy Rxy
Pxc1 y1 c2 y2 Qxc1 y1 c2 y2 Rxc1 y1 c2 y2
Pxc1 y1 c2 y2 Qxc1 y1 c2 y2 Rxc1 y1 c2 y2
c1Pxy1 Qxy1 Rxy1 c2 Pxy2 Qxy2 Rxy2
c10 c20 0
Thus, y c1 y1 c2 y2 is a solution of Equation 2.
The other fact we need is given by the following theorem, which is proved in more
advanced courses. It says that the general solution is a linear combination of two linearly
independent solutions y1 and y2. This means that neither y1 nor y2 is a constant multiple
of the other. For instance, the functions f x x 2 and tx 5x 2 are linearly dependent,
but f x e x and tx xe x are linearly independent.
1
2 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
4 Theorem If y1 and y2 are linearly independent solutions of Equation 2, and Px
is never 0, then the general solution is given by
yx c1 y1x c2 y2x
where c1 and c2 are arbitrary constants.
Theorem 4 is very useful because it says that if we know two particular linearly independent solutions, then we know every solution.
In general, it is not easy to discover particular solutions to a second-order linear equation. But it is always possible to do so if the coefficient functions P, Q, and R are constant
functions, that is, if the differential equation has the form
ay by cy 0
5
where a, b, and c are constants and a 0.
It’s not hard to think of some likely candidates for particular solutions of Equation 5 if
we state the equation verbally. We are looking for a function y such that a constant times
its second derivative y plus another constant times y plus a third constant times y is equal
to 0. We know that the exponential function y e rx (where r is a constant) has the property that its derivative is a constant multiple of itself: y re rx. Furthermore, y r 2e rx.
If we substitute these expressions into Equation 5, we see that y e rx is a solution if
ar 2e rx bre rx ce rx 0
ar 2 br ce rx 0
or
But e rx is never 0. Thus, y e rx is a solution of Equation 5 if r is a root of the equation
ar 2 br c 0
6
Equation 6 is called the auxiliary equation (or characteristic equation) of the differential equation ay by cy 0. Notice that it is an algebraic equation that is obtained
from the differential equation by replacing y by r 2, y by r, and y by 1.
Sometimes the roots r1 and r 2 of the auxiliary equation can be found by factoring. In
other cases they are found by using the quadratic formula:
r1 7
b sb 2 4ac
2a
r2 b sb 2 4ac
2a
We distinguish three cases according to the sign of the discriminant b 2 4ac.
b2 4ac 0
In this case the roots r1 and r 2 of the auxiliary equation are real and distinct, so y1 e r 1 x
and y2 e r 2 x are two linearly independent solutions of Equation 5. (Note that e r 2 x is not a
constant multiple of e r 1 x.) Therefore, by Theorem 4, we have the following fact.
CASE I
■
If the roots r1 and r 2 of the auxiliary equation ar 2 br c 0 are real and
unequal, then the general solution of ay by cy 0 is
8
y c1 e r 1 x c2 e r 2 x
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 3
In Figure 1 the graphs of the basic solutions
f x e 2 x and tx e3 x of the differential
equation in Example 1 are shown in black and
red, respectively. Some of the other solutions,
linear combinations of f and t , are shown
in blue.
■ ■
8
EXAMPLE 1 Solve the equation y y 6y 0.
SOLUTION The auxiliary equation is
r 2 r 6 r 2r 3 0
whose roots are r 2, 3. Therefore, by (8) the general solution of the given differential equation is
5f+g
y c1 e 2x c2 e3x
f+5g
f+g
f
g
_1
g-f
f-g
_5
1
We could verify that this is indeed a solution by differentiating and substituting into the
differential equation.
EXAMPLE 2 Solve 3
FIGURE 1
d 2y
dy
y 0.
2 dx
dx
SOLUTION To solve the auxiliary equation 3r 2 r 1 0 we use the quadratic
formula:
r
1 s13
6
Since the roots are real and distinct, the general solution is
y c1 e (1s13 ) x6 c2 e (1s13 ) x6
b 2 4 ac 0
In this case r1 r2 ; that is, the roots of the auxiliary equation are real and equal. Let’s
denote by r the common value of r1 and r 2. Then, from Equations 7, we have
CASE II
9
■
r
b
2a
so
2ar b 0
We know that y1 e rx is one solution of Equation 5. We now verify that y2 xe rx is also
a solution:
ay2 by2 cy2 a2re rx r 2xe rx be rx rxe rx cxe rx
2ar be rx ar 2 br cxe rx
0e rx 0xe rx 0
The first term is 0 by Equations 9; the second term is 0 because r is a root of the auxiliary
equation. Since y1 e rx and y2 xe rx are linearly independent solutions, Theorem 4 provides us with the general solution.
If the auxiliary equation ar 2 br c 0 has only one real root r, then the
general solution of ay by cy 0 is
10
y c1 e rx c2 xe rx
EXAMPLE 3 Solve the equation 4y 12y 9y 0.
SOLUTION The auxiliary equation 4r 2 12r 9 0 can be factored as
2r 32 0
4 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
so the only root is r 32 . By (10) the general solution is
Figure 2 shows the basic solutions
f x e3x2 and tx xe3x2 in
Example 3 and some other members of the
family of solutions. Notice that all of them
approach 0 as x l .
■ ■
y c1 e3x2 c2 xe3x2
b 2 4ac 0
In this case the roots r1 and r 2 of the auxiliary equation are complex numbers. (See Appendix I for information about complex numbers.) We can write
CASE III
f-g 8
f
■
5f+g
f+5g
_2
r1 i
2
f+g
g-f
g
r 2 i
where and are real numbers. [In fact, b2a, s4ac b 22a.] Then,
using Euler’s equation
_5
e i cos i sin FIGURE 2
from Appendix I, we write the solution of the differential equation as
y C1 e r 1 x C2 e r 2 x C1 e ix C2 e ix
C1 e xcos x i sin x C2 e xcos x i sin x
e x C1 C2 cos x iC1 C2 sin x
e xc1 cos x c2 sin x
where c1 C1 C2 , c2 iC1 C2. This gives all solutions (real or complex) of the differential equation. The solutions are real when the constants c1 and c2 are real. We summarize the discussion as follows.
If the roots of the auxiliary equation ar 2 br c 0 are the complex numbers r1 i, r 2 i, then the general solution of ay by cy 0
is
11
y e xc1 cos x c2 sin x
■ ■ Figure 3 shows the graphs of the solutions in Example 4, f x e 3 x cos 2x and
tx e 3 x sin 2x, together with some linear
combinations. All solutions approach 0
as x l .
3
f+g
g
f-g
EXAMPLE 4 Solve the equation y 6y 13y 0.
SOLUTION The auxiliary equation is r 2 6r 13 0. By the quadratic formula, the
roots are
r
6 s36 52
6 s16
3 2i
2
2
By (11) the general solution of the differential equation is
f
_3
2
y e 3xc1 cos 2x c2 sin 2x
Initial-Value and Boundary-Value Problems
_3
FIGURE 3
An initial-value problem for the second-order Equation 1 or 2 consists of finding a solution y of the differential equation that also satisfies initial conditions of the form
yx 0 y0
yx 0 y1
where y0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and
Px 0 there, then a theorem found in more advanced books guarantees the existence
and uniqueness of a solution to this initial-value problem. Examples 5 and 6 illustrate the
technique for solving such a problem.
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 5
EXAMPLE 5 Solve the initial-value problem
y y 6y 0
■ ■ Figure 4 shows the graph of the solution of
the initial-value problem in Example 5. Compare
with Figure 1.
y0 1
y0 0
SOLUTION From Example 1 we know that the general solution of the differential equa-
tion is
yx c1 e 2x c2 e3x
20
Differentiating this solution, we get
yx 2c1 e 2x 3c2 e3x
To satisfy the initial conditions we require that
_2
2
0
FIGURE 4
12
y0 c1 c2 1
13
y0 2c1 3c2 0
From (13) we have c2 23 c1 and so (12) gives
c1 23 c1 1
c1 35
c2 25
Thus, the required solution of the initial-value problem is
y 35 e 2x 25 e3x
■ ■ The solution to Example 6 is graphed in
Figure 5. It appears to be a shifted sine curve
and, indeed, you can verify that another way of
writing the solution is
y s13 sinx where tan 23
EXAMPLE 6 Solve the initial-value problem
y y 0
y0 2
y0 3
SOLUTION The auxiliary equation is r 2 1 0, or r 2 1, whose roots are i. Thus
0, 1, and since e 0x 1, the general solution is
5
yx c1 cos x c2 sin x
_2π
2π
yx c1 sin x c2 cos x
Since
the initial conditions become
_5
FIGURE 5
y0 c1 2
y0 c2 3
Therefore, the solution of the initial-value problem is
yx 2 cos x 3 sin x
A boundary-value problem for Equation 1 consists of finding a solution y of the differential equation that also satisfies boundary conditions of the form
yx 0 y0
yx 1 y1
In contrast with the situation for initial-value problems, a boundary-value problem does
not always have a solution.
EXAMPLE 7 Solve the boundary-value problem
y 2y y 0
y0 1
y1 3
SOLUTION The auxiliary equation is
r 2 2r 1 0
or
r 12 0
6 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
whose only root is r 1. Therefore, the general solution is
■ ■ Figure 6 shows the graph of the solution of
the boundary-value problem in Example 7.
yx c1 ex c2 xex
5
The boundary conditions are satisfied if
_1
y0 c1 1
5
y1 c1 e1 c2 e1 3
The first condition gives c1 1, so the second condition becomes
_5
e1 c2 e1 3
FIGURE 6
Solving this equation for c2 by first multiplying through by e, we get
1 c2 3e
c2 3e 1
so
Thus, the solution of the boundary-value problem is
y ex 3e 1xex
Summary: Solutions of ay by c 0
Roots of ar 2 br c 0
General solution
y c1 e r 1 x c2 e r 2 x
y c1 e rx c2 xe rx
y e xc1 cos x c2 sin x
r1, r2 real and distinct
r1 r2 r
r1, r2 complex: i
Exercises
A Click here for answers.
1–13
S
17–24
Click here for solutions.
17. 2y 5y 3y 0,
18. y 3y 0,
Solve the differential equation.
y0 3,
y0 1,
y0 4
y0 3
1. y 6y 8y 0
2. y 4y 8y 0
19. 4 y 4 y y 0,
3. y 8y 41y 0
4. 2y y y 0
20. 2y 5y 3y 0,
5. y 2y y 0
6. 3y 5y
21. y 16y 0,
7. 4y y 0
8. 16y 24y 9y 0
22. y 2y 5y 0,
y 0,
y 2
10. 9y 4y 0
23. y 2y 2y 0,
y0 2,
y0 1
d 2y
dy
6
4y 0
12.
dt 2
dt
24. y 12y 36y 0,
9. 4y y 0
2
d y
dy
2
y0
11.
dt 2
dt
■
d 2y
dy
y0
13.
dt 2
dt
■
■
■
■
■
■
■
■
■
■
■
■
■
2
■
■
■
■
25. 4 y y 0,
■
■
■
■
■
■
y0 4
y4 4
y1 0,
■
y0 1,
■
y1 1
■
■
■
31. y 4y 13y 0,
y 4
y1 2
y0 1,
y0 2,
30. y 6y 9y 0,
■
y0 1,
y0 3,
29. y 6y 25y 0,
d 2y
dy
2
5y 0
16.
dx 2
dx
■
■
y0 1.5
■
Solve the boundary-value problem, if possible.
28. y 100 y 0,
d y
dy
8
16y 0
dx 2
dx
15.
■
y0 1,
y4 3,
27. y 3y 2y 0,
2
d y
dy
2y 0
dx 2
dx
■
26. y 2y 0,
Graph the two basic solutions of the differential equation
and several other solutions. What features do the solutions have in
common?
14. 6
■
25–32
; 14–16
■
Solve the initial-value problem.
y3 0
y 5
y0 1,
y0 1,
y0 2,
y 2
y1 0
y 2 1
■
■
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 7
32. 9y 18y 10 y 0,
■
■
■
■
■
■
y0 0,
■
■
y 1
■
■
■
■
33. Let L be a nonzero real number.
(a) Show that the boundary-value problem y y 0,
y0 0, yL 0 has only the trivial solution y 0 for
the cases 0 and 0.
■
(b) For the case 0, find the values of for which this problem has a nontrivial solution and give the corresponding
solution.
34. If a, b, and c are all positive constants and yx is a solution
of the differential equation ay by cy 0, show that
lim x l yx 0.
8 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
Answers
S
Click here for solutions.
1. y c1 e 4x c2 e 2x
3. y e4x c1 cos 5x c2 sin 5x
x
x
5. y c1 e c2 xe
7. y c1 cosx2 c2 sinx2
9. y c1 c2 ex4
11. y c1 e (1s2 )t c2 e (1s2 )t
13. y et2 [c1 cos(s3t2) c2 sin(s3t2)]
g
15.
40
f
_0.2
1
_40
All solutions approach 0 as x l and approach as x l .
17. y 2e3x2 ex
19. y e x/2 2xe x2
21. y 3 cos 4x sin 4x
23. y ex2 cos x 3 sin x
e 2x
e x3
1
1
25. y 3 cos( 2 x) 4 sin( 2 x)
27. y 3
e 1
1 e3
29. No solution
31. y e2x 2 cos 3x e sin 3x
33. (b) n 2 2L2, n a positive integer; y C sinn xL
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 9
Solutions: Second-Order Linear Differential Equations
1. The auxiliary equation is r 2 − 6r + 8 = 0 ⇒
solution is y = c1 e4x + c2 e2x .
(r − 4)(r − 2) = 0 ⇒ r = 4, r = 2. Then by (8) the general
3. The auxiliary equation is r2 + 8r + 41 = 0 ⇒ r = −4 ± 5i. Then by (11) the general solution is
y = e−4x (c1 cos 5x + c2 sin 5x).
5. The auxiliary equation is r2 − 2r + 1 = (r − 1)2 = 0
y = c1 ex + c2 xex .
7. The auxiliary equation is 4r 2 + 1 = 0
⇒ r = 1. Then by (10), the general solution is
r = ± 12 i, so y = c1 cos
⇒
9. The auxiliary equation is 4r 2 + r = r(4r + 1) = 0 ⇒
¡1 ¢
¡1 ¢
2 x + c2 sin 2 x .
r = 0, r = − 14 , so y = c1 + c2 e−x/4 .
√
√
√
2, so y = c1 e(1+ 2)t + c2 e(1− 2)t .
³√ ´
³ √ ´i
h
√
⇒ r = − 12 ± 23 i, so y = e−t/2 c1 cos 23 t + c2 sin 23 t .
11. The auxiliary equation is r 2 − 2r − 1 = 0 ⇒ r = 1 ±
13. The auxiliary equation is r 2 + r + 1 = 0
15. r 2 − 8r + 16 = (r − 4)2 = 0 so y = c1 e4x + c2 xe4x .
The graphs are all asymptotic to the x-axis as x → −∞,
and as x → ∞ the solutions tend to ±∞.
17. 2r 2 + 5r + 3 = (2r + 3)(r + 1) = 0, so r = − 32 , r = −1 and the general solution is y = c1 e−3x/2 + c2 e−x .
Then y(0) = 3 ⇒ c1 + c2 = 3 and y0 (0) = −4 ⇒ − 32 c1 − c2 = −4, so c1 = 2 and c2 = 1. Thus the
solution to the initial-value problem is y = 2e−3x/2 + e−x .
19. 4r 2 − 4r + 1 = (2r − 1)2 = 0 ⇒ r =
⇒
c1 = 1 and y0 (0) = −1.5 ⇒
y=e
x/2
x/2
− 2xe
1
2 c1
1
2
and the general solution is y = c1 ex/2 + c2 xex/2 . Then y(0) = 1
+ c2 = −1.5, so c2 = −2 and the solution to the initial-value problem is
.
21. r 2 + 16 = 0 ⇒ r = ±4i and the general solution is y = e0x (c1 cos 4x + c2 sin 4x) = c1 cos 4x + c2 sin 4x.
¡ ¢
¡ ¢
Then y π4 = −3 ⇒ −c1 = −3 ⇒ c1 = 3 and y 0 π4 = 4 ⇒ −4c2 = 4 ⇒ c2 = −1, so the
solution to the initial-value problem is y = 3 cos 4x − sin 4x.
23. r2 + 2r + 2 = 0 ⇒
0
r = −1 ± i and the general solution is y = e−x (c1 cos x + c2 sin x). Then 2 = y(0) = c1
c2 = 3 and the solution to the initial-value problem is y = e−x (2 cos x + 3 sin x).
¡ ¢
¡ ¢
25. 4r2 + 1 = 0 ⇒ r = ± 12 i and the general solution is y = c1 cos 12 x + c2 sin 12 x . Then 3 = y(0) = c1 and
¡ ¢
¡ ¢
−4 = y(π) = c2 , so the solution of the boundary-value problem is y = 3 cos 12 x − 4 sin 12 x .
and 1 = y (0) = c2 − c1
⇒
27. r 2 − 3r + 2 = (r − 2)(r − 1) = 0 ⇒ r = 1, r = 2 and the general solution is y = c1 ex + c2 e2x . Then
1 = y(0) = c1 + c2 and 0 = y(3) = c1 e3 + c2 e6 so c2 = 1/(1 − e3 ) and c1 = e3 /(e3 − 1). The solution of the
boundary-value problem is y =
29. r 2 − 6r + 25 = 0 ⇒
and 2 = y(π) = c1 e3π
e2x
ex+3
.
+
e3 − 1
1 − e3
r = 3 ± 4i and the general solution is y = e3x (c1 cos 4x + c2 sin 4x). But 1 = y(0) = c1
⇒
c1 = 2/e3π , so there is no solution.
10 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
r = −2 ± 3i and the general solution is y = e−2x (c1 cos 3x + c2 sin 3x). But
¡ ¢
2 = y(0) = c1 and 1 = y π2 = e−π (−c2 ), so the solution to the boundary-value problem is
31. r 2 + 4r + 13 = 0 ⇒
y = e−2x (2 cos 3x − eπ sin 3x).
33. (a) Case 1 (λ = 0): y 00 + λy = 0 ⇒ y 00 = 0 which has an auxiliary equation r 2 = 0 ⇒ r = 0 ⇒
y = c1 + c2 x where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c1 and 0 = y(L) = c2 L ⇒ c1 = c2 = 0.
Thus, y = 0.
√
Case 2 (λ < 0): y 00 + λy = 0 has auxiliary equation r2 = −λ ⇒ r = ± −λ (distinct and real since
λ < 0)
⇒
y = c1 e
√
0 = y(L) = c1 e
−λL
Multiplying (∗) by e
√
√
√
−λx
+ c 2 e−
√
− −λL
+ c2 e
−λL
−λx
where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c1 + c2 (∗) and
(†).
³ √
´
√
and subtracting (†) gives c2 e −λL − e− −λL = 0 ⇒
c2 = 0 and thus
c1 = 0 from (∗). Thus, y = 0 for the cases λ = 0 and λ < 0.
√
√
⇒ y = c1 cos λ x + c2 sin λ x
√
where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c1 and 0 = y(L) = c2 sin λL since c1 = 0. Since we
√
√
λ L = nπ where n is an integer
cannot have a trivial solution, c2 6= 0 and thus sin λ L = 0 ⇒
(b) y 00 + λy = 0 has an auxiliary equation r 2 + λ = 0 ⇒
⇒
r = ±i
√
λ
λ = n2 π 2 /L2 and y = c2 sin(nπx/L) where n is an integer.