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Chapter 3
Second Order Linear Differential
Equations
3.1. Introduction; Basic Terminology
Recall that a first order linear differential equation is an equation which can be written in the form y′^ + p(x)y = q(x)
where p and q are continuous functions on some interval I. A second order linear differential equation has an analogous form.
SECOND ORDER LINEAR DIFFERENTIAL EQUATION: A second or- der, linear differential equation is an equation which can be written in the form
y′′^ + p(x)y′^ + q(x)y = f(x) (1)
where p, q, and f are continuous functions on some interval I.
The functions p and q are called the coefficients of the equation; the function f on the right-hand side is called the forcing function or the nonhomogeneous term
. The term “forcing function” comes from the applications of second-order equations; an explanation of the alternative term “ nonhomogeneous” is given below.
A second order equation which is not linear is said to be nonlinear.
Remarks on “Linear.” Set L[y] = y′′^ + p(x)y′^ + q(x)y. If we view L as an “operator” that transforms a twice differentiable function y = y(x) into the continuous function
L[y(x)] = y′′(x) + p(x)y′(x) + q(x)y(x),
then, for any two twice differentiable functions y 1 (x) and y 2 (x),
L[y 1 (x) + y 2 (x)] = L[y 1 (x)] + L[y 2 (x)]
and, for any constant c, L[cy(x)] = cL[y(x)].
As introduced in Section 2.1, L is a linear transformation, specifically, a linear differential operator: L : C^2 (I) → C(I)
where C^2 (I) is the vector space of twice continuously differentiable functions on I and C(I) is the vector space of continuous functions on I.
The first thing we need to know is that an initial-value problem has a solution, and that it is unique.
THEOREM 1. (Existence and Uniqueness Theorem:) Given the second order linear equation (1). Let a be any point on the interval I, and let α and β be any two real numbers. Then the initial-value problem
y′′^ + p(x) y′^ + q(x) y = f(x), y(a) = α, y′(a) = β
has a unique solution.
A proof of this theorem is beyond the scope of this course.
Remark: We can solve any first order linear differential equation; Chapter 2 gives a method for finding the general solution of any first order linear equation. In contrast, there is no general method for solving second (or higher) order linear differential equations. There are, however, methods for solving certain special types of second order linear equations and we’ll consider these in this chapter.
DEFINITION 1. (Homogeneous/Nonhomogeneous Equations) The linear differential equation (1) is homogeneous 1 if the function f on the right side is 0 for all x ∈ I. In this case, equation (1) becomes
y′′^ + p(x) y′^ + q(x) y = 0. (2)
Equation (1) is nonhomogeneous if f is not the zero function on I, i.e., (1) is nonhomogeneous if f(x) 6 = 0 for some x ∈ I. (^1) This use of the term “homogeneous” is completely different from its use to categorize the first order equation y′^ = f(x, y) in Exercises 2.2.
Suppose that y = y 1 (x) and y = y 2 (x) are solutions of equation (1). Under what conditions is (2) the general solution of (1)?
Let u = u(x) be any solution of (1) and choose any point a ∈ I. Suppose that
α = u(a), β = u′(a).
Then u is a member of the two-parameter family (2) if and only if there are values for C 1 and C 2 such that
C 1 y 1 (a) + C 2 y 2 (a) = α
C 1 y′ 1 (a) + C 2 y′ 2 (a) = β
If we multiply the first equation by y 2 ′(a), the second equation by −y 2 (a), and add, we get [y 1 (a)y 2 ′(a) − y 2 (a)y 1 ′(a)]C 1 = αy′ 2 (a) − βy 2 (a).
Similarly, if we multiply the first equation by −y′ 1 (a), the second equation by y 1 (a), and add, we get
[y 1 (a)y′ 2 (a) − y 2 (a)y′ 1 (a)]C 2 = −αy 1 ′(a) + βy 1 (a).
We are guaranteed that this pair of equations has solutions C 1 , C 2 if and only if
y 1 (a)y 2 ′(a) − y 2 (a)y 1 ′(a) 6 = 0
in which case
C 1 = αy 2 ′(a)^ −^ βy^2 (a) y 1 (a)y 2 ′(a) − y 2 (a)y 1 ′(a) and^ C^2 =^
−αy′ 1 (a) + βy 1 (a) y 1 (a)y 2 ′(a) − y 2 (a)y 1 ′(a).
Since a was chosen to be any point on I, we conclude that (2) is the general solution of (1) if and only if
y 1 (x)y 2 ′(x) − y 2 (x)y 1 ′(x) 6 = 0 for all x ∈ I.
DEFINITION 2. (Wronskian) Let y = y 1 (x) and y = y 2 (x) be solutions of (1). The function W defined by
W y 1 , y 2 = y 1 (x)y′ 2 (x) − y 2 (x)y′ 1 (x)
is called the Wronskian of y 1 , y 2.
We use the notation W y 1 , y 2 to emphasize that the Wronskian is a function of x that is determined by two solutions y 1 , y 2 of equation (1). When there is no danger of confusion, we’ll shorten the notation to W (x).
Remark Note that
W (x) =
y 1 (x) y 2 (x) y 1 ′(x) y′ 2 (x)
∣ =^ y^1 (x)y
′ 2 (x) − y 2 (x)y′ 1 (x).
THEOREM 3. Let y = y 1 (x) and y = y 2 (x) be solutions of equation (1), and let W (x) be their Wronskian. Exactly one of the following holds:
(i) W (x) = 0 for all x ∈ I and y 1 is a constant multiple of y 2.
(ii) W (x) 6 = 0 for all x ∈ I and y = C 1 y 1 (x) + C 2 y 2 (x) is the general solution of (1)
DEFINITION 3. (Fundamental Set) A pair of solutions y = y 1 (x), y = y 2 (x) of equation (1) forms a fundamental set of solutions if
W y 1 , y 2 6 = 0 for all x ∈ I.
Linear Dependence; Linear Independence
By Theorem 3, if y 1 and y 2 are solutions of equation (1) such that W [y 1 , y 2 ] ≡ 0, then y 1 is a constant multiple of y 2. The question as to whether or not one function is a multiple of another function and the consequences of this are of fundamental importance in differential equations and in linear algebra.
In this sub-section we are dealing with functions in general, not just solutions of the differential equation (1)
DEFINITION 4. (Linear Dependence; Linear Independence) Given two functions f = f(x), g = g(x) defined on an interval I. The functions f and g are linearly dependent on I if and only if there exist two real numbers c 1 and c 2 , not both zero, such that
c 1 f(x) + c 2 g(x) ≡ 0 on I.
The functions f and g are linearly independent on I if they are not linearly dependent.
Linear dependence can be stated equivalently as: f and g are linearly dependent on I if and only if one of the functions is a constant multiple of the other.
- y′′^ − 4 y′^ + 4y = 0; y 1 (x) = e^2 x, y 2 (x) = xe^2 x.
- x^2 y′′^ − x(x + 2)y′^ + (x + 2)y = 0; y 1 (x) = x, y 2 (x) = xex.
- Given the differential equation y′′^ − 3 y′^ − 4 y = 0.
(a) Find two values of r such that y = erx^ is a solution of the equation. (b) Determine a fundamental set of solutions and give the general solution of the equation. (c) Find the solution of the equation satisfying the initial conditions y(0) = 1 , y′(0) = 0.
- Given the differential equation y′′^ −
x
y′^ −
x^2
y = 0.
(a) Find two values of r such that y = xr^ is a solution of the equation. (b) Determine a fundamental set of solutions and give the general solution of the equation. (c) Find the solution of the equation satisfying the initial conditions y(1) = 2 , y′(1) = −1. (d) Find the solution of the equation satisfying the initial conditions y(2) = y′(2) = 0.
- Given the differential equation (x^2 + 2x − 1)y′′^ − 2(x + 1)y′^ + 2y = 0.
(a) Show that the equation has a linear polynomial and a quadratic polyno- mial as solutions. b Find two linearly independent solutions of the equation and give the gen- eral solution.
- Let y = y 1 (x) be a solution of (1): y′′^ +p(x)y′^ +q(x)y = 0 where p and q are continuous function on an interval I. Let a ∈ I and assume that y 1 (x) 6 = 0 on I. Set y 2 (x) = y 1 (x)
∫ (^) x
a
e−^
∫ (^) t a p(u)^ du y 12 (t) dt. Show that y 2 is a solution of (1) and that y 1 and y 2 are linearly independent. Use Exercise 6 to find a fundamental set of solutions of the given equation starting from the given solution y 1.
- y′′^ − (^) x^2 y′^ + x^22 y = 0; y 1 (x) = x.
- y′′^ − 2 x^ x− 1 y′^ + x^ − x 1 y = 0; y 1 (x) = ex.
- Let y = y 1 (x) and y = y 2 (x) be solutions of equation (1): y′′^ + p(x)y′^ + q(x)y = 0 on an interval I. Let a ∈ I and suppose that y 1 (a) = α, y′ 1 (a) = β and y 2 (a) = γ, y 2 ′(a) = δ. Under what conditions on α, β, γ, δ will the functions y 1 and y 2 be linearly independent on I?
- Suppose that y = y 1 (x) and y = y 2 (x) are solutions of (1). Show that if y 1 (x) 6 = 0 on I and W y 1 , y 2 ≡ 0 on I, then y 2 (x) = λy 1 (x) on I.
3.2. Homogenous Equations with Constant Coefficients
We have emphasized that there are no general methods for solving second (or higher) order linear differential equations. However, there are some special cases for which solution methods do exist. In this and the following sections we consider such a case, linear equations with constant coefficients.
A second order, linear, homogeneous differential equation with constant coefficients is an equation which can be written in the form
y′′^ + ay′^ + by = 0 (1)
where a and b are real numbers.
You have seen that the function y = e−ax^ is a solution of the first-order linear equation y′^ + ay = 0,
the equation modeling exponential growth and decay. This suggests that equation (1) may also have an exponential function y = erx^ as a solution.
If y = erx, then y′^ = r erx^ and y′′^ = r^2 erx. Substitution into (1) gives r^2 erx^ + a (r erx) + b (erx) = erx^
r^2 + ar + b
Since erx^6 = 0 for all x, we conclude that y = erx^ is a solution of (1) if and only if
r^2 + ar + b = 0. (2)
Thus, if r is a root of the quadratic equation (2), then y = erx^ is a solution of equation (1); we can find solutions of (1) by finding the roots of the quadratic equation (2).
The characteristic roots are: r 1 = − 4 , r 2 = 2. The functions y 1 (x) = e−^4 x, y 2 (x) = e^2 x^ form a fundamental set of solutions of the differential equation and
y = C 1 e−^4 x^ + C 2 e^2 x
is the general solution of the equation.
Case II: The characteristic equation has only one real root, r = α.^2 Then y 1 (x) = eαx^ and y 2 (x) = x eαx are linearly independent solutions of equation (1) and
y = C 1 eαx^ + C 2 x eαx
is the general solution. Proof: We know that y 1 (x) = eαx^ is one solution of the differential equation; we need to find another solution which is independent of y 1. Since the characteristic equation has only one real root, α, the equation must be r^2 + ar + b = (r − α)^2 = r^2 − 2 αr + α^2 = 0 and the differential equation (1) must have the form
y′′^ − 2 α y′^ + α^2 y = 0. (*)
Now, z = C eαx, C any constant, is also a solution of (), but z is not independent of y 1 since it is simply a multiple of y 1. We replace C by a function u which is to be determined (if possible) so that y = ueαx^ is a solution of ().^3 Calculating the derivatives of y, we have
y = u eαx
y′^ = α u eαx^ + u′^ eαx y′′^ = α^2 u eαx^ + 2α u′^ eαx^ + u′′^ eαx
Substitution into (*) gives
α^2 u eαx^ + 2α u′^ eαx^ + u′′^ eαx^ − 2 α [α u eαx^ + u′^ eαx] + α^2 u eαx^ = 0. (^2) In this case, α is said to be a double root of the characteristic equation. (^3) This is an application of a general method called variation of parameters. We will use the method several times in the work that follows.
This reduces to
u′′^ eαx^ = 0 which becomes u′′^ = 0 since eαx^6 = 0.
Now, u′′^ = 0 is the simplest second order, linear differential equation with constant coefficients; the general solution is u = C 1 + C 2 x = C 1 · 1 + C 2 · x , and u 1 (x) = 1 and u 2 (x) = x form a fundamental set of solutions. Since y = u eαx, we conclude that
y 1 (x) = 1 · eαx^ = eαx^ and y 2 (x) = x eαx
are solutions of (). It’s easy to see that y 1 and y 2 form a fundamental set of solutions of (). This can also be checked by using the Wronskian:
W (x) = eαx^ [eαx^ + α x eαx] − α x eαx^ = e^2 αx^6 = 0.
Finally, the general solution of (*) is
y = C 1 eαx^ + C 2 x eαx^
Example 2. Find the general solution of the differential equation
y′′^ − 6 y′^ + 9y = 0.
SOLUTION The characteristic equation is
r^2 − 6 r + 9 = 0 (r − 3)^2 = 0
There is only one characteristic root: r 1 = r 2 = 3. The functions y 1 (x) = e^3 x, y 2 (x) = x e^3 x^ are linearly independent solutions of the differential equation and y = C 1 e^3 x^ + C 2 x e^3 x
is the general solution.
Case III: The characteristic equation has complex conjugate roots:
r 1 = α + i β, r 2 = α + i β, β 6 = 0
In this case
y 1 (x) = eαx^ cos βx and y 2 (x) = eαx^ sin βx
are solutions of (*). It’s easy to see that y 1 and y 2 form a fundamental set of solutions. This can also be checked by using the Wronskian Finally, we conclude that the general solution of equation (1) is: y = C 1 eαx^ cos βx + C 2 eαx^ sin βx = eαx^ [C 1 cos βx + C 2 sin βx].
Example 3. Find the general solution of the differential equation
y′′^ − 4 y′^ + 13y = 0.
SOLUTION The characteristic equation is: r^2 − 4 r + 13 = 0. By the quadratic formula, the roots are
r 1 , r 2 =
(−4)^2 − 4(1)(13)
4 ± 6 i 2 = 2±^3 i.
The characteristic roots are the complex numbers: r 1 = 2 + 3 i, r 2 = 2 − 3 i. The functions y 1 (x) = e^2 x^ cos 3x, y 2 (x) = e^2 x^ sin 3x are linearly independent solutions of the differential equation and
y = C 1 e^2 x^ cos 3x + C 2 e^2 x^ sin 3x = e^2 x^ [C 1 cos 3x + C 2 sin 3x]
is the general solution.
Example 4. (Important Special Case) Find the general solution of the differential equation y′′^ + β^2 y = 0.
SOLUTION The characteristic equation is: r^2 + β^2 = 0. The characteristic roots are the complex numbers r 1 , r 2 = 0 ± β i
The functions y 1 (x) = e^0 x^ cos βx = cos βx, y 2 (x) = e^0 sin β 3 x = sin βx are linearly independent solutions of the differential equation and
y = C 1 cos βx + C 2 sin βx
is the general solution.
Recovering a Differential Equation from Solutions
You can also work backwards using the results above. That is, we can determine a second order, linear, homogeneous differential equation with constant coefficients that has given functions u and v as solutions. Here are some examples.
Example 5. Find a second order, linear, homogeneous differential equation with constant coefficients that has the functions u(x) = e^2 x, v(x) = e−^3 x^ as solutions.
SOLUTION Since e^2 x^ is a solution, 2 must be a root of the characteristic equation and r−2 must be a factor of the characteristic polynomial. Similarly, e−^3 x^ a solution means that − 3 is a root and r − (−3) = r + 3 is a factor of the characteristic polynomial. Thus the characteristic equation must be
(r − 2)(r + 3) = 0 which expands to r^2 + r − 6 = 0.
Therefore, the differential equation is
y′′^ + y′^ − 6 y = 0.
Example 6. Find a second order, linear, homogeneous differential equation with constant coefficients that has y(x) = ex^ cos 2x as a solution.
SOLUTION Since ex^ cos 2x is a solution, the characteristic equation must have the complex numbers 1 + 2i and 1 − 2 i as roots. (Although we didn’t state it explicitly, ex^ sin 2x must also be a solution.) The characteristic equation must be
(r − [1 + 2i])(r − [1 − 2 i]) = 0 which expands to r^2 − 2 r + 5 = 0
and the differential equation is
y′′^ − 2 y′^ + 5y = 0.
Exercises 3.
Find the general solution of the given differential equation.
- y′′^ + 2y′^ − 8 y = 0.
- y′′^ − 13 y′^ + 42y = 0.
- y′′^ − 10 y′^ + 25y = 0.
- y′′^ + 2y′^ + 5y = 0.
- y′′^ + 4y′^ + 13y = 0.
- y′′^ = 0.
- y′′^ + 2y′^ = 0.
- Find the solution y = y(x) of the initial-value problem 4 y′′^ − y = 0; y(0) = 2 , y′(0) = β. Then find β such that y(x) → 0 as x → ∞.
Euler Equations: A second order linear homogeneous equation of the form
x^2 d
(^2) y dx^2 +^ αx
dy dx +^ βy^ = 0^ (E) where α and β are constants, is called an Euler equation.
- Prove that the Euler equation (E) can be transformed into the second order equation with constant coefficients d^2 y dz^2
where a and b are constants, by means of the change of independent variable z = ln x.
Find the general solution of the Euler equations.
- x^2 y′′^ − xy′^ − 8 y = 0.
- x^2 y′′^ − 3 xy′^ + 4y = 0.
- x^2 y′′^ − xy′^ + 5y = 0.
3.3. Nonhomogeneous Equations
In this section we consider the general second order, linear, nonhomogeneous equation
y′′^ + p(x)y′^ + q(x)y = f(x) (1)
where p, q, f are continuous functions on an interval I.
The objectives of this section are to determine the “structure” of the set of solu- tions of (1).
As we shall see, there is a close connection between equation (1) and
y′′^ + p(x)y′^ + q(x)y = 0. (2)
In this context, equation (2) is called the reduced equation of equation (1).
General Results
THEOREM 1. If z = z 1 (x) and z = z 2 (x) are solutions of equation (1), then
y(x) = z 1 (x) − z 2 (x)
is a solution of equation (2).
Thus the difference of any two solutions of the nonhomogeneous equation (1) is a solution of its reduced equation (2).
Our next theorem gives the “structure” of the set of solutions of (1).
THEOREM 2. Let y = y 1 (x) and y = y 2 (x) be linearly independent solutions of the reduced equation (2) and let z = z(x) be a particular solution of (1). If u = u(x) is any solution of (1), then there exist constants C 1 and C 2 such that
u(x) = C 1 y 1 (x) + C 2 y 2 (x) + z(x).
According to Theorem 2, if y = y 1 (x) and y = y 2 (x) are linearly independent solutions of the reduced equation (2) and z = z(x) is a particular solution of (1), then
y = C 1 y 1 (x) + C 2 y 2 (x) + z(x) (3)
represents the set of all solutions of (1). That is, (3) is the general solution of (1). Another way to look at (3) is: The general solution of (1) consists of the general solution of the reduced equation (2) plus a particular solution of (1):
︸︷︷︸^ y general solution of (1)
= C ︸ 1 y 1 (x) +︷︷ C 2 y 2 (x︸) general solution of (2)
- z ︸ (︷︷x) ︸. particular solution of (1)
The next result is sometimes useful in finding particular solutions of nonhomoge- neous equations. It is known as the superposition principle.
THEOREM 3. If z = z 1 (x) and z = z 2 (x) are particular solutions of
y′′^ + p(x)y′^ + q(x)y = f(x) and y′′^ + p(x)y′^ + q(x)y = g(x),
respectively, then z(x) = z 1 (x) + z 2 (x) is a particular solution of
y′′^ + p(x)y′^ + q(x)y = f(x) + g(x).
is the general solution. We replace the arbitrary constants C 1 and C 2 by functions u = u(x) and v = v(x), which are to be determined so that
z(x) = u(x)y 1 (x) + v(x)y 2 (x)
is a particular solution of the nonhomogeneous equation (1). The replacement of the parameters C 1 and C 2 by the “variables” u and v is the basis for the term “variation of parameters.” Since there are two unknowns u and v to be determined we shall impose two conditions on these unknowns. One condition is that z should solve the differential equation (1). The second condition is at our disposal and we shall choose it in a manner that will simplify our calculations.
Differentiating z we get
z′^ = u y 1 ′ + y 1 u′^ + v y 2 ′ + y 2 v′.
For our second condition on u and v, we set
y 1 u′^ + y 2 v′^ = 0. (a)
This condition is chosen because it simplifies the first derivative z′^ and because it will lead to a simple pair of equations in the unknowns u and v. With this condition the equation for z′^ becomes z′^ = u y′ 1 + v y′ 2 (b)
and z′′^ = u y 1 ′′ + y′ 1 u′^ + v y 2 ′′ + y 2 ′ v′.
Now substitute z, z′^ (given by (b)), and z′′^ into the left side of equation (1). This gives
z′′^ + pz′^ + qz = (u y′′ 1 + y′ 1 u′^ + v y′′ 2 + y 2 ′ v′) + p(u y′ 1 + v y′ 2 ) + q(u y 1 + v y 2 )
= u(y′′ 1 + py 1 ′ + qy 1 ) + v(y′′ 2 + py 2 ′ + qy 2 ) + y 1 ′ u′^ + y′ 2 v′.
Since y 1 and y 2 are solutions of (2),
y′′ 1 + py′ 1 + qy 1 = 0 and y′′ 2 + py 2 ′ + qy 2 = 0
and so z′′^ + pz′^ + qz = y 1 ′ u′^ + y′ 2 v′.
The condition that z should satisfy (1) is
y 1 ′ u′^ + y 2 ′ v′^ = f(x). (c)
Equations (a) and (c) constitute a system of two equations in the two unknowns u and v:
y 1 u′^ + y 2 v′^ = 0
y 1 ′ u′^ + y 2 ′ v′^ = f(x)
Obviously this system involves u′^ and v′^ not u and v, but if we can solve for u′ and v′, then we can integrate to find u and v. Solving for u′^ and v′, we find that
u′^ = (^) y −y^2 f 1 y′ 2 −^ y 2 y 1 ′
and v′^ = (^) y y^1 f 1 y′ 2 −^ y 2 y′ 1
We know that the denominators here are non-zero because the expression
y 1 (x)y 2 ′(x) − y 2 (x)y 1 ′(x) = W (x)
is the Wronskian of y 1 and y 2 , and y 1 , y 2 are linearly independent solutions of the reduced equation.
We can now get u and v by integrating:
u =
−y 2 (x)f(x) W (x) dx^ and^ v^ =
y 1 (x)f(x) W (x) dx.
Finally
z(x) = y 1 (x)
∫ (^) −y 2 (x)f(x) W (x) dx^ +^ y^2 (x)
∫ (^) y 1 (x)f(x) W (x) dx^ (4)
is a particular solution of the nonhomogeneous equation (1).
Remark This result illustrates why the emphasis is on linear homogeneous equa- tions. To find the general solution of the nonhomogeneous equation (1) we need a fundamental set of solutions of the reduced equation (2) and one particular solution of (1). But, as we have just shown, if we have a fundamental set of solutions of (2), then we can use them to construct a particular solution of (1). Thus, all we really need to solve (1) is a fundamental set of solutions of its reduced equation (2).
Example 1. Find a particular solution of the nonhomogeneous equation
y′′^ − 5 y′^ + 6 y = 4e^2 x. (*)
SOLUTION The functions y 1 (x) = e^2 x, y 2 (x) = e^3 x^ are linearly independent solutions of the reduced equation. The Wronskian of y 1 , y 2 is
W (x) = y 1 y 2 ′ − y 2 y′ 1 = e^5 x.
