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Second-Order Linear Differential Equations
A second-order linear differential equation has the form
where , , , and 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 , for all , in Equation 1. Such equa-
tions are called homogeneous linear equations. Thus, the form of a second-order linear
homogeneous differential equation is
If for some , 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 and of such an equation, then the linear combination
is also a solution.
Theorem If and are both solutions of the linear homogeneous equa-
tion (2) and and are any constants, then the function
is also a solution of Equation 2.
Proof Since and are solutions of Equation 2, we have
and
Therefore, using the basic rules for differentiation, we have
Thus, 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 and This means that neither nor is a constant multiple
of the other. For instance, the functions and are linearly dependent,
but f x e x and t x xe x are linearly independent.
f x x^2 t x 5 x^2
y 1 y 2. y 1 y 2
y c 1 y 1 c 2 y 2
c 1 0 c 2 0 0
c 1 P x y 1 Q x y 1 R x y 1 c 2 P x y 2 Q x y 2 R x y 2
P x c 1 y 1 c 2 y 2 Q x c 1 y 1 c 2 y 2 R x c 1 y 1 c 2 y 2
P x c 1 y 1 c 2 y 2 Q x c 1 y 1 c 2 y 2 R x c 1 y 1 c 2 y 2
P x y Q x y R x y
P x y 2 Q x y 2 R x y 2 0
P x y 1 Q x y 1 R x y 1 0
y 1 y 2
y x c 1 y 1 x c 2 y 2 x
c 1 c 2
3 y 1 x y 2 x
y c 1 y 1 c 2 y 2
y 1 y 2
G x 0 x
P x
d^2 y
dx^2
Q x
dy
dx
2 R x y 0
G x 0 x
PQ R G
P x
d^2 y
dx^2
Q x
dy
dx
1 R x y G x
1
Theorem If and are linearly independent solutions of Equation 2, and
is never 0, then the general solution is given by
where and are arbitrary constants.
Theorem 4 is very useful because it says that if we know two particular linearly inde-
pendent solutions, then we know every solution.
In general, it is not easy to discover particular solutions to a second-order linear equa-
tion. But it is always possible to do so if the coefficient functions , , and are constant
functions, that is, if the differential equation has the form
where , , and are constants and.
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 such that a constant times
its second derivative plus another constant times plus a third constant times is equal
to 0. We know that the exponential function (where is a constant) has the prop-
erty that its derivative is a constant multiple of itself:. Furthermore,.
If we substitute these expressions into Equation 5, we see that is a solution if
or
But is never 0. Thus, is a solution of Equation 5 if is a root of the equation
Equation 6 is called the auxiliary equation (or characteristic equation ) of the differen-
tial equation. Notice that it is an algebraic equation that is obtained
from the differential equation by replacing by , by , and by.
Sometimes the roots and of the auxiliary equation can be found by factoring. In
other cases they are found by using the quadratic formula:
We distinguish three cases according to the sign of the discriminant.
CASE I ■■
In this case the roots and of the auxiliary equation are real and distinct, so
and are two linearly independent solutions of Equation 5. (Note that is not a
constant multiple of .) Therefore, by Theorem 4, we have the following fact.
If the roots and of the auxiliary equation are real and
unequal, then the general solution of is
y c 1 e r^1 x^ c 2 e r^2 x
ay by cy 0
8 r 1 r 2^ ar^2 ^ br^ ^ c^ ^0
e r^1 x
y 2 e r^2 x e r^2 x
r 1 r 2 y 1 e r^1 x
b^2 4 ac 0
b^2 4 ac
r 2
b s b^2 4 ac
2 a
r 1
b s b^2 4 ac
2 a
r 1 r 2
y r^2 y r y 1
ay by cy 0
6^ ar^2 ^ br^ ^ c^ ^0
e rx y e rx r
ar^2 br c e rx^ 0
ar^2 e rx^ bre rx^ ce rx^ 0
y e rx
y re rx y r^2 e rx
y e rx r
y y y
y
ab c a 0
5 ay by cy 0
PQ R
c 1 c 2
y x c 1 y 1 x c 2 y 2 x
4 y 1 y 2 P x
2 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
4 ■
so the only root is. By (10) the general solution is
CASE III ■■
In this case the roots and of the auxiliary equation are complex numbers. (See Appen-
dix I for information about complex numbers.) We can write
where and are real numbers. [In fact, , .] Then,
using Euler’s equation
from Appendix I, we write the solution of the differential equation as
where ,. This gives all solutions (real or complex) of the dif-
ferential equation. The solutions are real when the constants and are real. We sum-
marize the discussion as follows.
If the roots of the auxiliary equation are the complex num-
bers , , then the general solution of
is
EXAMPLE 4 Solve the equation.
SOLUTION The auxiliary equation is. By the quadratic formula, the
roots are
By (11) the general solution of the differential equation is
Initial-Value and Boundary-Value Problems
An initial-value problem for the second-order Equation 1 or 2 consists of finding a solu-
tion of the differential equation that also satisfies initial conditions of the form
where and are given constants. If , , , and are continuous on an interval and
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.
P x 0
y 0 y 1 PQ R G
y x 0 y 0 y x 0 y 1
y
y e^3 x c 1 cos 2 x c 2 sin 2 x
r
6 s 36 52
6 s 16
3 2 i
r^2 6 r 13 0
y 6 y 13 y 0
y e^ ^ x c 1 cos x c 2 sin x
r 1 i r 2 i ay by cy 0
11 ar^2 br c 0
c 1 c 2
c 1 C 1 C 2 c 2 i C 1 C 2
e^ ^ x c 1 cos x c 2 sin x
e^ ^ x C 1 C 2 cos x i C 1 C 2 sin x
C 1 e^ ^ x cos x i sin x C 2 e^ ^ x cos x i sin x
y C 1 e r^1 x^ C 2 e r^2 x^ C 1 e ^ ^ i^ x^ C 2 e ^ ^ i^ x
e i^ ^ cos i sin
b 2 a s 4 ac b^2 2 a
r 1 i r 2 i
r 1 r 2
b^2 4 ac 0
y c 1 e ^3 x ^2 c 2 xe ^3 x ^2
■ ■ Figure 2 shows the basic solutions r ^32
and in Example 3 and some other members of the family of solutions. Notice that all of them approach 0 as x l.
f x e ^3 x ^2 t x xe ^3 x ^2
FIGURE 2
8
_
_2 2
5f+g f+5g
f
g
f-g
f+g g-f
■ ■ (^) Figure 3 shows the graphs of the solu- tions in Example 4, and , together with some linear combinations. All solutions approach 0 as x l .
t x e^3 x^ sin 2 x
f x e^3 x^ cos 2 x
FIGURE 3
3
_
_3 2
f
g f-g
f+g
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
■ 5
EXAMPLE 5 Solve the initial-value problem
SOLUTION From Example 1 we know that the general solution of the differential equa-
tion is
Differentiating this solution, we get
To satisfy the initial conditions we require that
From (13) we have and so (12) gives
Thus, the required solution of the initial-value problem is
EXAMPLE 6 Solve the initial-value problem
SOLUTION The auxiliary equation is , or , whose roots are. Thus
, , and since , the general solution is
Since
the initial conditions become
Therefore, the solution of the initial-value problem is
A boundary-value problem for Equation 1 consists of finding a solution y of the dif-
ferential equation that also satisfies boundary conditions of the form
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
SOLUTION The auxiliary equation is
r^2 2 r 1 0 or r 1 ^2 0
y 2 y y 0 y 0 1 y 1 3
y x 0 y 0 y x 1 y 1
y x 2 cos x 3 sin x
y 0 c 1 2 y 0 c 2 3
y x c 1 sin x c 2 cos x
y x c 1 cos x c 2 sin x
0 1 e^0 x^ 1
r^2 1 0 r^2 1 i
y y 0 y 0 2 y 0 3
y 35 e^2 x^ 25 e ^3 x
c 1 23 c 1 1 c 1 35 c 2 25
c 2 23 c 1
13 y 0 2 c 1 3 c 2 0
12 y 0 c 1 c 2 1
y x 2 c 1 e^2 x^ 3 c 2 e ^3 x
y x c 1 e^2 x^ c 2 e ^3 x
y y 6 y 0 y 0 1 y 0 0
■ ■ (^) Figure 4 shows the graph of the solution of the initial-value problem in Example 5. Compare with Figure 1.
FIGURE 4
20
_2 0 2
■ ■ (^) 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 sin x where tan (^23)
FIGURE 5
5
_
_2π 2π
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
■ 7
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
33. Let be a nonzero real number. (a) Show that the boundary-value problem , , has only the trivial solution for the cases 0 and 0.
y 0 0 y L 0 y 0
y y 0
L
9 y 18 y 10 y 0 y 0 0 y 1 (b) For the case^ , find the values of^ for which this prob- lem has a nontrivial solution and give the corresponding solution.
34. If , , and are all positive constants and is a solution of the differential equation , show that lim (^) x l y x 0.
ay by cy 0
ab c y x
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
8 ■
Answers
All solutions approach 0 as and approach as.
**17. 19.
29.** No solution 31. 33. (b) n^2 2 L^2 , n a positive integer; y C sin n x L
y e ^2 x 2 cos 3 x e^ ^ sin 3 x
y
e x ^3 e^3 1
e^2 x 1 e^3
y 3 cos( 12 x ) 4 sin( 12 x )
y 3 cos 4 x sin 4 x y e x 2 cos x 3 sin x
y 2 e ^3 x ^2 e x y e x /2^ 2 xe x ^2
x l x l
g f
40
_
_0.2 1
y e t ^2 [ c 1 cos(s 3 t 2) c 2 sin(s 3 t 2)]
y c 1 c 2 e x ^4 y c 1 e (1s2 ) t^ c 2 e (1s2 ) t
y c 1 e x^ c 2 xe x y c 1 cos x 2 c 2 sin x 2
y c 1 e^4 x^ c 2 e^2 x y e ^4 x c 1 cos 5 x c 2 sin 5 x
S Click here for solutions.
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
31. r^2 + 4r + 13 = 0 ⇒ r = − 2 ± 3 i and the general solution is y = e−^2 x(c 1 cos 3x + c 2 sin 3x). But 2 = y(0) = c 1 and 1 = y
¡ (^) π 2
= e−π^ (−c 2 ), so the solution to the boundary-value problem is y = e−^2 x(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 = c 1 + c 2 x where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c 1 and 0 = y(L) = c 2 L ⇒ c 1 = c 2 = 0. Thus, y = 0.
Case 2 (λ < 0 ) : y^00 + λy = 0 has auxiliary equation r^2 = −λ ⇒ r = ±
−λ (distinct and real since λ < 0 ) ⇒ y = c 1 e
√−λx
√−λx where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c 1 + c 2 (∗) and 0 = y(L) = c 1 e
√−λL
√−λL (†).
Multiplying (∗) by e
√−λL and subtracting (†) gives c 2
e
√−λL − e−
√−λL^ ´ = 0 ⇒ c 2 = 0 and thus c 1 = 0 from (∗). Thus, y = 0 for the cases λ = 0 and λ < 0.
(b) y^00 + λy = 0 has an auxiliary equation r^2 + λ = 0 ⇒ r = ±i
λ ⇒ y = c 1 cos
λ x + c 2 sin
λ x where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c 1 and 0 = y(L) = c 2 sin
λL since c 1 = 0. Since we cannot have a trivial solution, c 2 6 = 0 and thus sin
λ L = 0 ⇒
λ L = nπ where n is an integer ⇒ λ = n^2 π^2 /L^2 and y = c 2 sin(nπx/L) where n is an integer.
10 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
