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MATH 4389

Almus

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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
0
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
00 0
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),
39 then, for any two twice diﬀerentiable functions y (x)1and y (x), 2
L[y 1x)+ y (2)] = L[y (x1] + L[y (x2]
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 (I) → C(I)
2
where C (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
00 0 0
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
00 0
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.
40 For reasons which will become clear, almost all of our attention is focused on
homogeneous equations.
Homogeneous Equations
As deﬁned above, a second order, linear, homogeneous diﬀerential equation is an
equation that can be written in the form
00 0
y + p(x)y + q(x)y = 0 (3)
where p and q are continuous functions on some interval I.
The Trivial Solution: The ﬁrst thing to note is that the zero function, y(x)=0
for all x ∈ I, (also denoted by y ≡ 0) is a solution of (1). The zero solution is called
the trivial solution . Obviously our main interest is in ﬁnding nontrivial solutions.
Let S = {y = y(x): y is a solution of (1)}; S is a subset of C (I).
THEOREM 2. Let y = u(x),y = v(x) ∈S , and let C be any real number.
Then
y(x)= u(x)+ v(x) ∈S and
y(x)= Cu(x) ∈S .
2
That is, S is a subspace of C (I). Indeed, S is the null space of the linear
diﬀerential operator L.
Theorem 1 can be restated as: If y = y (x),1 = y (x) ∈2 and C ,1 2 are
real numbers, then
C y + C y ∈S .
1 1 2 2
The expression
C y + C y
1 1 2 2
is called a linear combination of y1 and y .2
Note that the equation
y(x)= C y1 1)+ C y (2 2 (4)
where C 1 and C 2 are arbitrary constants, has the form of the general solution of
equation (1). So the question is: If y1 and y 2 are solutions of (1), is the expression
(2) the general solution of (1)? That is, can every solution of (1) be written as a
linear combination of y 1 and y ?2It turns out that (2) may or not be the general
solution; it depends on the relation between the solutions y 1 and y .2
41 Suppose that y = y (x) an1 y = y (x) are solu2ions of equation (1). Under
what conditions is (2) the general solution of (1)?
Let u = u(x)b e 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 1a)+ C y (2 2 α
0 0
C 1y1(a)+ C y 2a2= β
If we multiply the ﬁrst equation by y (a), the second equation by −y (a), and add,2
2
we get
[y1(a)y (a) − y (2)y (a)]C = αy1(a) − βy (a). 2
2 1 2
0
Similarly, if we multiply the ﬁrst equation by −y 1(a), the second equation by y (a), 1
and add, we get
[y1(a)y (a) − y (2)y (a)]C = −α2 (a)+ βy (a). 1
2 1 1
We are guaranteed that this pair of equations has solutions C ,C 1 2 if and only if
y 1a)y (2) − y (a2y (a)16=0
in which case
αy 2(a) − βy (2) −αy 1(a)+ βy (a1
C 1 = 0 0 and C 2 = 0 0 .
y1(a)y 2a) − y (2)y (a1 y1(a)y 2a) − y (2)y (a1
Since a was chosen to be any point on I, we conclude that (2) is the general
solution of (1) if and only if
y1(x)y 2x) − y (x2y (x)16= 0 for all x ∈ I.
DEFINITION 2. (Wronskian) Let y = y (x) and y = y (x) b1 solutions of 2
(1). The function W deﬁned by
0 0
W[y ,1 ](2)= y (x)y1(x) −2y (x)y (2) 1
is called the Wronskian of y ,y . 1 2
42 We use the notation W[y ,y ]1x)2to emphasize that the Wronskian is a function
of x that is determined by two solutions y 1,y2 of equation (1). When there is no
danger of confusion, we’ll shorten the notation to W(x).
Remark Note that
y 1x) y (2) 0 0
W(x)= y (x) y (x) = y 1x)y (2) − y (2)y (x1.
1 2
THEOREM 3. Let y = y 1(x) and y = y (x2 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 1(x)+ C y 2 2 is the general solution of
(1)
DEFINITION 3. (Fundamental Set) A pair of solutions y = y (x),y = y (x) 1 2
of equation (1) forms a fundamental set of solutions if
W[y ,1 ]2x) 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 ,y 1 ≡ 2,
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 and c ,
1 2
not both zero, such that
c1f(x)+ c g2x) ≡ 0 no 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.
43 The term Wronskian deﬁned above for two solutions of equation (1) can be ex-
tended to any two diﬀerentiable functions f and g. Let f = f(x) and g = g(x)
be diﬀerentiable functions on an interval I. The function W[f,g] deﬁned by
0 0
W[f,g](x)= f(x)g (x) − g(x)f (x)
is called the Wronskian of f, g.
There is a connection between linear dependence/independence and Wronskian.
THEOREM 4. Let f = f(x) and g = g(x) be diﬀerentiable functions on an
interval I.fI f and g are linearly dependent on I, then W(x) = 0 for all
x ∈ I (W ≡ 0o n I).
This theorem can be stated equivalently as: Let f = f(x) and g = g(x)e b
diﬀerentiable functions on an interval I.If W(x) 6= 0 for at least one x ∈ I, then
f and g are linearly independent on I.
Going back to diﬀerential equations, Theorem 4 can be restated as
Theorem 4’ Let y = y (x) a1d y = y (x) be s2lutions of equation (1). Exactly
one of the following holds:
(i) W(x) = 0 for all x ∈ I; y 1 and y 2 are linear dependent.
(ii) W(x) 6= 0 for all x ∈ I; y 1 and y 2 are linearly independent and y =
C 1 1x)+ C y 2 2 is the general solution of (1).
The statements “y (1),y (x)2form a fundamental set of solutions of (1)” and
“y1(x),y (2) are linearly independent solutions of (1)” are synonymous.
The results of this section can be captured in one statement
2
The set S of solutions of (1), a subspace of C (I), has dimension 2, the order of the equation.
Exercises 3.1
In Exercises 1 – 2, verify that the functions y 1 and y 2 are solutions of the given
diﬀerential equation. Do they constitute a fundamental set of solutions of the equa-
tion?
44 1. y − 4y +4 y =0; y1(x)= e ,y (x)2 xe . 2x
2. x y − x(x +2) y +( x +2) y =0; y 1x)= x, y (x2= xe . x
00 0
3. Given the diﬀerential equation y − 3y − 4y =0.
rx
(a) Find two values of r such that y = e 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) =
0
1,y (0) = 0.
2 4
4. Given the diﬀerential equation y 0− y0− y =0.
x x2
r
(a) Find two values of r such that y = x 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) =
0
2,y (1) = −1.
(d) Find the solution of the equation satisfying the initial conditions y(2) =
0
y (2) = 0.
2 00 0
5. Given the diﬀerential equation (x +2 x − 1)y − 2(x +1) y +2 y =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.
6. Let y = y (x) be a solution of (1): y +p(x)y +q(x)y = 0 where p and q are
1
continuous function on an interval I. Let a ∈ I and assume that y (x) 6=0 1
on I. Set Z Rt
x e− ap(u) du
y2(x)= y 1x) dt.
a y1(t)
Show that y 2 is a solution of (1) and that y1 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
7. y00− 2y 0+ 2 y =0; y (x)= x.
x x2 1
45 8. y −0 2x − 1 y 0+ x − 1 y =0; y (x)= e .x
x x 1
9. Let y = y (x1 and y = y (x) be2solutions of equation (1):
00 0
y + p(x)y + q(x)y =0
on an interval I. Let a ∈ I and suppose that
y (a)= α, y (a)= β and y (a)= γ, y (a)= δ.
1 1 2 2
Under what conditions on α ,β,,δ will the functions y 1 and y 2 be linearly
independent on I?
10. Suppose that y = y (x) 1nd y = y (x) are s2lutions of (1). Show that if
y1(x) 6=0 on I and W[y ,y ]1x)2≡ 0n o I, then y (x2= λy (x)n1 o 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
0
y + ay =0 ,
the equation modeling exponential growth and decay. This suggests that equation
rx
(1) may also have an exponential function y = e as a solution.
rx 0 rx 00 2 rx
If y = e , then y = re and y = r e . Substitution into (1) gives
2 rx rx rx rx 2
r e + a(re )+ b(e )= e r + ar + b =0 .
Since e rx 6= 0 for all x, we conclude that y = e rx is a solution of (1) if and only if
2
r + ar + b =0 . (2)
Thus, if r is a root of the quadratic equation (2), then y = e rx is a solution
of equation (1); we can ﬁnd solutions of (1) by ﬁnding the roots of the quadratic
equation (2).
46 DEFINITION 1. Given the diﬀerential equation (1). The corresponding quadratic
equation (2)
2
r + ar + b =0
is called the characteristic equation of (1); the quadratic polynomial r 2+ ar + b
is called the characteristic polynomial. The roots of the characteristic equation are
called the characteristic roots .
The nature of the solutions of the diﬀerential equation (1) depends on the nature
of the roots of its characteristic equation (2). There are three cases to consider:
(1) Equation (2) has two, distinct real roots, r =1α, r = 2.
(2) Equation (2) has only one real root, r = α.
(3) Equation (2) has complex conjugate roots, r = α 1 iβ, r 2= α − iβ, β 6=0.
Case I: The characteristic equation has two, distinct real roots, r 1 =
α, r2 = β. In this case,
αx βx
y 1x)= e and y 2x)= e
are solutions of (1). Since α 6= β, y1 and y 2 are not constant multiples
of each other, the pair y ,1 2 forms a fundamental set of solutions of
equation (1) and
y = C 1 αx + C 2 βx
is the general solution.
Note: We can use the Wronskian to verify the independence of y 1 and
y2:
0 0 αx βx βx αx (α+β)x
W(x)= y y 1y2y =2e 1 βe −e (αe )=( α−β)e 6=0 .
Example 1. Find the general solution of the diﬀerential equation
00 0
y +2 y − 8y =0 .
SOLUTION The characteristic equation is
2
r +2 r − 8=0
(r + 4)(r − 2) = 0
47 The characteristic roots are: r = −4,r = 2. The functions y (x)= e −4x ,y (x)=
1 2 1 2
e2x form a fundamental set of solutions of the diﬀerential equation and
−4x 2x
y = C 1 + C 2
is the general solution of the equation.
Case II: The characteristic equation has only one real root, r = α. 2
Then
αx αx
y1(x)= e and y2(x)= xe
are linearly independent solutions of equation (1) and
αx αx
y = C 1 + C 2e
is the general solution.
Proof: We know that y (x)= e αx is one solution of the diﬀerential
1
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
2 2 2 2
r + ar + b =( r − α) = r − 2αr + α =0
and the diﬀerential equation (1) must have the form
00 0 2
y − 2αy + α y =0 . (*)
Now, z = Ce αx, C any constant, is also a solution of (*), but z is not
independent of y since it is simply a multiple of y . We replace C by
1 1
a function u which is to be determined (if possible) so that y = ue αx is
a solution of (*).3 Calculating the derivatives of y, we have
y = ue αx
y0 = αue αx+ u e αx
00 2 αx 0 αx 00αx
y = α ue +2 αu e + u e
Substitution into (*) gives
α ue αx +2 αu e0 αx + u e0αx − 2α[αue αx + u e ]+ α ue2 αx =0 .
2In this case, α is said to be a double root of the characteristic equation.
3This is an application of a general method called variation of parameteWe will use the
method several times in the work that follows.
48 This reduces to
00 αx 00
u e = 0 which becomes u = 0 since eαx 6=0 .
Now, u 0= 0 is the simplest second order, linear diﬀerential equation with
constant coeﬃcients; the general solution is u = C 1+C x2= C ·11 C ·x 2
, and u 1x)=1 and u 2x)= x form a fundamental set of solutions.
Since y = ue αx, we conclude that
y 1(x)=1 · e αx = eαx and y2(x)= xe αx
are solutions of (*). It’s easy to see that1yand y 2 form a fundamental
set of solutions of (*). This can also be checked by using the Wronskian:
αx αx αx αx 2αx
W(x)= e [e + αxe ] − αxe = e 6=0 .
Finally, the general solution of (*) is
αx αx
y = C 1e + C 2e
Example 2. Find the general solution of the diﬀerential equation
00 0
y − 6y +9 y =0 .
SOLUTION The characteristic equation is
2
r − 6r +9 = 0
(r − 3)2 =0
There is only one characteristic root: r1 = r 2 = 3. The functions y1(x)=
3x 3x
e ,y 2(x)= xe are linearly independent solutions of the diﬀerential equation
and
y = C e 3x+ C xe 3x
1 2
is the general solution.
Case III: The characteristic equation has complex conjugate roots:
r 1 α + iβ, r 2 = α + iβ, β 6=0
In this case
αx αx
y1(x)= e cos βx and y2(x)= e sin βx
49 are linearly independent solutions of equation (1) and
y = C 1e αx cos βx + C 2 αx sin βx = e αx [C1cos βx + C si2 βx]
is the general solution.
(α+iβ)x
Proof: It is true that the functions z (x1= e and z (2)=
e(α−iβ)x are linearly independent solutions of (1), but these are complex-
valued functions and want real-valued solutions of (1). The characteristic
equation in this case is
2 2 2 2
r + ar + b =( r − [α + iβ ])(r − [α − iβ ]) = r − 2αr + α + β =0
and the diﬀerential equation (1) has the form
00 0 2 2
y − 2αy + α + β y =0 . (*)
We’ll proceed in a manner similar to Case II. Set y = ue αx where u is
to be determined (if possible) so that y is a solution of (*). Calculating
the derivatives of y, we have
αx
y = ue
0 αx 0 αx
y = αue + u e
y00 = α ue αx +2 αu e0 αx + u e αx
Substitution into (*) gives
α ue αx +2 αu e0 αx + u e αx − 2α[αue αx + u e ]+ α + β2 2 ue αx =0 .
This reduces to
u00eαx+β ue αx = 0 which becomes u +β u = 0 since e αx 6=0 .
Now,
00 2
u + β u =0
is the equation of simple harmonic motion (for example, it models the
oscillatory motion of a weight suspended on a spring). The functions
u1(x)=cos βx and u (x2 = sin βx form a fundamental set of solutions.
(Verify this.)
Since y = ue αx, we conclude that
y (x)= e αx cos βx and y (x)= e αxsin βx
1 2
50 are solutions of (*). It’s easy to see th1tand 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:
αx αx αx
y = C1e cos βx + C 2 sin βx = e [C1cos βx + C 2in βx].
Example 3. Find the general solution of the diﬀerential equation
y − 4y +13 y =0 .
SOLUTION The characteristic equation is: r − 4r + 13 = 0. By the quadratic
formula, the roots are
p √ √
−(−4) ± (−4) − 4(1)(13) 4 ± 16 − 52 4 ± −36 4 ± 6i
r1,r2 = = = = =2 ±3i.
2 2 2 2
The characteristic roots are the complex numbers: r 12+3 i, r =22−3i. The
2x 2x
functions y 1x)= e cos 3x, y2(x)= e sin 3x are linearly independent solutions
of the diﬀerential equation and
2x 2x 2x
y = C 1 cos 3x + C2e sin 3x = e [C1cos 3x + C 2in 3x]
is the general solution.
Example 4. (Important Special Case) Find the general solution of the diﬀerential
equation
y + β y =0 .
SOLUTION The characteristic equation is: r + β = 0. The characteristic roots
are the complex numbers
r1,r2 =0 ± βi
The functions y 1x)= e 0xcos βx = cos βx, y 2x)= e sin β3x = sin βx are linearly
independent solutions of the diﬀerential equation and
y = C cos βx + C sin βx
1 2
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.
51 Example 5. Find a second order, linear, homogeneous diﬀerential equation with
2x −3x
constant coeﬃcients that has the functions u(x)= e ,v (x)= e as solutions.
2x
SOLUTION Since e 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 −3x 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
2
(r − 2)(r + 3) = 0 which expands to r + r − 6=0 .
Therefore, the diﬀerential equation is
y00+ y − 6y =0 .
Example 6. Find a second order, linear, homogeneous diﬀerential equation with
constant coeﬃcients that has y(x)= e cos 2x as a solution.
x
SOLUTION Since e cos 2x is a solution, the characteristic equation must have the
complex numbers 1+2i and 1−2i as roots. (Although we didn’t state it explicitly,
e sin 2x must also be a solution.) The characteristic equation must be
(r − [1 + 2i])(r − [1 − 2i]) = 0 which expands to r 2− 2r +5=0
and the diﬀerential equation is
00 0
y − 2y +5 y =0 .
Exercises 3.2
Find the general solution of the given diﬀerential equation.
1. y 00+2 y − 8y =0.
00 0
2. y − 13y +42 y =0.
00 0
3. y − 10y +25 y =0.
4. y 00+2 y +5 y =0.
5. y +4 y +13 y =0.
00
6. y =0.
00 0
7. y +2 y =0.
52 8. 2y +5 y − 3y =0.
9. y − 9y =0.
10. y 00+16 y =0.
11. y − 2y +2 y =0.
12. y 00− y − 30y =0.
Find the solution of the initial-value problem.
13. y − 5y +6 y =0; y(0) = 1,y (0) = 1.
14. y 00+4 y +3 y =0; y(0) = y (0) = 0.
15. y +2 y + y =0; y(0) = −3,y (0) = 1.
16. y 00+4 y =0; y(0) = 1,y (0) = −2.
Find a diﬀerential equation y 00+ ay + by = 0 that is satisﬁed by the given
function(s).
2x −5x
17. y 1(x)= e ,y (x)= 2 .
3x
18. y(x)=2 xe .
19. y(x) = cos 2x.
2x −6x
20. y 1x)=3 e ,y (x)= −2e .
−2x
21. y(x)= e sin 4x.
00 0
Find a diﬀerential equation y + ay + by = 0 whose general solution is the
given expression.
22. y = C ex/2 + C e .2x
1 2
23. y = C e3x + C e −4x .
1 2
24. y = C 1e−x cos 3x + C e2 −x sin 3x.
25. y = C e 1 2x + C x2 . 2x
26. y = C 1 cos 4x + C s2n 4x.
27. Find the solution y = y(x) of the initial-value problem y −y −2y =0; y(0) =
α, y (0) = 2. Then ﬁnd α such that y(x) → 0s a x →∞ .
53 00
28. Find the solution y = y(x) of the initial-value problem 4y − y =0 ; y(0) =
2,y (0) = β. Then ﬁnd β such that y(x) → 0s a x →∞ .
Euler Equations: A second order linear homogeneous equation of the form
2
2d y dy
x dx 2 + αx dx + βy = 0 (E)
where α and β are constants, is called an Euler equation .
29. Prove that the Euler equation (E) can be transformed into the second order
equation with constant coeﬃcients
d2y dy
2 + a + by =0
dz dz
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.
2 00 0
30. x y − xy − 8y =0.
2 00 0
31. x y − 3xy +4 y =0.
32. x y − xy +5 y =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
00 0
y + p(x)y + q(x)y =0 . (2)
In this context, equation (2) is called the reduced equation of equation (1).
54 General Results
THEOREM 1. If z = z (x) and1z = z (x) are s2lutions of equation (1), then
y(x)= z 1x) − z 2x)
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 (x) and 1 = y (x) be line2rly independent solutions
of the reduced equation (2) and let z = z(x) be a particular solution of (1). If
u = u(x)si any solution of (1), then there exist constants C 1 and C 2 such that
u(x)= C y1 1)+ C y (2 2 z(x).
According to Theorem 2, if y = y 1(x) and y = y (x2 are linearly independent
solutions of the reduced equation (2) and z = z(x) is a particular solution of (1),
then
y = C y (x)+ C y (x)+ z(x) (3)
1 1 2 2
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):
|{z} = |1 1(x)+{z y2 2) } + |{z}.
general solution of (1) general solution of (2) 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 (x) and1z = z (x) are pa2ticular 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)+ z (x2 is a particular solution of
00 0
y + p(x)y + q(x)y = f(x)+ g(x).
55 This result can be extended to nonhomogeneous equations whose right-hand side
is the sum of an arbitrary number of functions.
COROLLARY If z = z 1(x) is a particular solution of
y + p(x)y + q(x)y = f (x),1
z = z (x) is a particular solution of
2
y + p(x)y + q(x)y = f (x),
2
and so on
z = z nx) is a particular solution of
00 0
y + p(x)y + q(x)y = f (x),n
then z(x)= z (x)1 z (x)+ 2·· + z (x) is n particular solution of
00 0
y + p(x)y + q(x)y = f (x)+1f (x)+ ·2· + f (x). n
The importance of Theorem 7 and its Corollary is that we need only consider
nonhomogeneous equations in which the function on the right-hand side consists of
one term only.
Variation of Parameters
By our work above, to ﬁnd the general solution of (1) we need to ﬁnd:
(i) a linearly independent pair of solutions y ,y 1 2 of the reduced equation (2), and
(ii) a particular solution z of (1).
The method of variation of parameters uses a pair of linearly independent solutions
of the reduced equation to construct a particular solution of (1).
Let y (1) and y (x) 2e linearly independent solutions of the reduced equation
y + p(x)y + q(x)y =0 .
Then
y = C y (x)+ C y (x)
1 1 2 2
56 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 (x) 2
is a particular solution of the nonhomogeneous equation (1). The replacement of the
parameters C and C by the “variables” u and v is the basis for the term
1 2
“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
0 0 0 0 0
z = uy + y1u + v1 + y v . 2 2
For our second condition on u and v, we set
y u + y v =0 .0 (a)
1 2
0
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
0
the equation for z becomes
0 0 0
z = uy + v1 2 (b)
and
z = uy + y u + vy + y v . 00 0 0
1 1 2 2
Now substitute z, z 0 (given by (b)), and z 00 into the left side of equation (1).
This gives
z 00+ pz + qz =( uy + y u + vy + y v )+ p(uy + vy )+ q(uy + vy ) 0 0
1 1 2 2 1 2 1 2
= u(y + py + qy )+ v(y + py + qy )+ y u + y v . 0 0 0 0
1 1 1 2 2 2 1 2
Since y 1 and y 2 are solutions of (2),
00 0 00 0
y1 + py +1qy = 0 1 and y2+ py + 2y =0 2
and so
z + pz + qz = y u + y v . 0 0 0
1 2
The condition that z should satisfy (1) is
y u + y v = f(x). (c)
1 2
57 Equations (a) and (c) constitute a system of two equations in the two unknowns
u and v:
0 0
y1 u + y 2 =0
0 0 0 0
y1u + y v2 = f(x)
0 0 0
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
0 −y 2f 0 y1 f
u = y y − y y 0 and v = y y − y y 0
1 2 2 1 1 2 2 1
We know that the denominators here are non-zero because the expression
0 0
y 1(x)y2(x) − y 2x)y (1)= W(x)
is the Wronskian of y 1 and y ,2and y ,y 1 2 are linearly independent solutions of
the reduced equation.
We can now get u and v by integrating:
Z Z
−y 2(x)f(x) y1(x)f(x)
u = W(x) dx and v = W(x) dx.
Finally
Z Z
−y (x)f(x) y (x)f(x)
z(x)= y 1(x) 2 dx + y2(x) 1 dx (4)
W(x) W(x)
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 − 5y +6 y =4 e . 2x (*)
2x 3x
SOLUTION The functions y 1x)= e ,y 2(x)= e are linearly independent
solutions of the reduced equation. The Wronskian of y ,y is
1 2
W(x)= y y − y y = e . 0 5x
1 2 2 1
58 By the method of variation of parameters, a particular solution of the nonhomoge-
neous equation is
2x 3x
z(x)= u(x)e + v(x)e
where, from (4),
Z 3x 2x Z
−e (4e )
u(x)= e5x dx = −4dx = −4x
and Z Z
e (4e2x) −x −x
v(x)= 5x dx = 4e dx = −4e .
e
(NOTE: Since we are seeking only one function u and one function v we have not
included arbitrary constants in the integration steps.)
Now
z(x)= −4xe 2x− 4e −x e3x= −4xe 2x− 4e 2x
is a particular solution of the nonhomogeneous equation (*) and
2x 2x 2x 3x
y = C 1 + C 23x − 4xe2x − 4e = C 1 + C 2 − 4xe2x
2x 2x
is the general solution (we “absorbed” −4e in the C1e term). As you can check
−4xe 2x is a solution of the nonhomogeneous equation.
Exercises 3.3
Verify that the given functions y 1 and y 2 form a fundamental set of solutions
of the reduced equation of the given nonhomogeneous equation; then ﬁnd a partic-
ular solution of the nonhomogeneous equation and give the general solution of the
equation.
2
1. y − y =3 − x −2; y1(x)= x ,y (x2= x −1.
x2
00 1 0 1 2
2. y − y + 2y = ; y 1(x)= x, y 2x)= x ln x.
x x x
00 0 2 x
3. (x − 1)y − xy + y =( x − 1) ; y1(x)= x, y 2x)= e .
2 00 0
4. x y − xy + y =4 x ln x.
Find the general solution of the given nonhomogeneous diﬀerential equation.
5. y − 4y +4 y = 1 x−1e .
3
59 −2x
00 0 e
6. y +4 y +4 y = 2 .
x
00 0 −x
7. y +2 y + y = e ln x.
2 00 0
8. The function y (x1= x is a solution of x y + xy − y = 0. Find the general
solution of the diﬀerential equation
x y + xy − y =2 x.
9. The functions y (x)1 x + x ln x, y (x)= x + 2 2 and y (x3= x 2 are
solutions of a second order, linear, nonhomogeneous equation. What is the
general solution of the equation?
3.4. Undetermined Coeﬃcients
Solving a linear nonhomgeneous equation depends, in part, on ﬁnding a particular
solution of the equation. We have seen one method for ﬁnding a particular solution,
the method of variation of parameters. In this section we present another method,
the method of undetermined coeﬃcients.
Remark: Limitations of the method. In contrast to variation of parameters,
which can be applied to any nonhomogeneous equation, the method of undetermined
coeﬃcients can be applied only to nonhomogeneous equations of the form
00 0
y + ay + by = f(x) (1)
where a and b are constants and the nonhomogeneous term f is a polynomial,
an exponential function, a sine, a cosine, or a combination of such functions.
To motivate the method of undetermined coeﬃcients, consider the linear operator
on the left side of (1):
00 0
y + ay + by. (2)
rx
If we calculate (2) for an exponential function z = Ae ,A a constant, we have
rx 0 rx 00 2 rx
z = Ae ,z = Are ,z = Ar e
and
00 0 2 rx rx rx 2 rx
y + ay + by = Ar e + a(Are )+ b(Ae = Ar + aAr + bA e
= Ke rx where K = Ar + aAr + bA.
60 That is, the operator (2) “transforms” Ae rx into a constant multiple of e .W rx e
can use this result to determine a particular solution of a nonhomogeneous equation
of the form
00 0 rx
y + ay + by = ce .
Here is a speciﬁc example.
Example 1. Find a particular solution of the nonhomogeneous equation
00 0 3x
y − 2y +5 y =6 e .
00 0 3x
SOLUTION As we saw above, if we “apply” y −2y +5y to z(x)= Ae we will
3x
get an expression of the form Ke . We want to determine A so that K = 6. The
constant A is called an undetermined coeﬃcient. We have
3x 0 3x 00 3x
z = Ae ,z =3 Ae ,z =9 Ae .
Substituting z and its derivatives into the left side of the diﬀerential equation, we
get
9Ae 3x − 2 3Ae 3x +5 Ae 3x =(9 A − 6A +5 A)e 3x=8 Ae .3x
We want
00 0 3x
z − 2z +5 z =6 e ,
so we set
3x 3x
8Ae =6 e which gives 8A = 6 and A = . 4
3 3x 00 0 3x
Thus, z(x)= 4 e is a particular solution of y − 2y +5 y =6 e . (Verify this.)
You can also verify that
x 3 3x
y = e (C c1s 2x + C sin 2x)+ 4 e
is the general solution of the equation.
0 00
If we set z(x)= A cosβx and calculate z and z , we get
0 00 2
z = Acosβx, z = −βA sin βx, z = −β A cos βx.
Therefore, y + ay + by applied to z gives
z + az + bz = −β A cos βx + a(−βA sin βx)+ b(Acosβx )
=( −β 2A + bA)cos βx +( −aβA)sin βx.
61 That is, y + ay + by “transforms” z = Acos βx into an expression of the form
K cos βx + M sin βx
where K and M are constants which depend on a,b,β and A. We will get exactly
the same type of result if we apply y +ay +by to z = B sinβx. Combining these
00 0
two results, it follows that y + ay + by applied to
z = Acos βx + B sin βx
will produce the expression
K cos βx + M sin βx
whe

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