Worked Solutions
1
Chapter 20: Method of Educated Guess
20.1 a.
For y p : Let y(x) = y p (x) = Ae2x . Then
52e2x = y ′′ + 9y =
Solving for A , we get A =
Ae2x
52
13
′′
+ 9Ae2x = 4Ae2x + 9Ae2x = 13Ae2x
= 4 . Thus,
y p (x) = 4e2x
.
For yh :
y ′′ + 9y = 0
֒→
֒→
r2 + 9 = 0
֌
√
r = ± −9 = ±3i
yh (x) = c1 cos(3x) + c2 sin(3x) .
The general solution to the nonhomogeneous equation is then
y(x) = y p (x) + yh (x) = 4e2x + c1 cos(3x) + c2 sin(3x)
20.1 c.
For y p : Let y(x) = y p (x) = Ae−4x . Then
30e−4x = y ′′ + 4y ′ − 5y
′′
′
= Ae−4x + 4 Ae−4x − 5Ae−4x
= 16Ae−4x − 16Ae−4x − 5Ae−4x = −5Ae−4x
Solving for A , we get A =
30
−5
.
= −6 . Thus,
y p (x) = −6e−4x
.
For yh :
y ′′ + 4y ′ − 5y = 0
֒→
֒→
֒→
0 = r 2 + 4r − 5 = (r − 1)(r + 5)
r = 1
and
r = −5
yh (x) = c1 e x + c2 e−5x
.
The general solution to the nonhomogeneous equation is then
y(x) = y p (x) + yh (x) = −6e−4x + c1 e x + c2 e−5x
20.2.
.
For y p : Let y(x) = y p (x) = Ae3x . Then
−5e3x = y ′′ − 3y ′ − 10y
′′
′
= Ae3x − 3 Ae3x − 10 Ae3x
.
.
2
Method of Educated Guess
= 9Ae3x − 9Ae3x − 10 Ae3x = −10 Ae3x
−5
−10
Solving for A , we get A =
=
1
2
.
. Thus,
1 3x
e
2
y p (x) =
.
For yh :
y ′′ − 3y ′ − 10y = 0
0 = r 2 − 3r − 10 = (r + 2)(r − 5)
֒→
֒→
r = −2
֒→
yh (x) = c1 e−2x + c2 e5x
and
r = 5
.
The general solution to the nonhomogeneous equation is then
1 3x
e
2
y(x) = y p (x) + yh (x) =
+ c1 e−2x + c2 e5x
,
(⋆)
and the derivative is
3 3x
e
2
y ′ (x) =
− 2c1 e−2x + 5c2 e5x
.
Applying the initial conditions, we have
and
5 = y(0) =
1 3·0
e
2
+ c1 e−2·0 + c2 e5·0 =
1
2
3 = y ′ (0) =
3 3·0
e
2
− 2c1 e−2·0 + 5c2 e5·0 =
+ c1 + c2
3
2
− 2c1 + 5c2
Solving for the constants and plugging them back into formula (⋆) for y :
5 =
֒→
֒→
֒→
֒→
20.3 a.
1
2
c1 =
and
+ c1 + c2
9
2
c1 =
9
2
c1 =
and
− c2
9
2
3
2
and
− c2
−
y(x) =
3
2
= 3
1 3x
e
2
3
2
3 =
= −2
h
c2 =
and
+ 3e−2x +
9
2
− 2c1 + 5c2
− c2
21
7·2
+ 5c2
=
c2 =
3 5x
e
2
i
3
2
3
2
.
For y p : Let y(x) = y p (x) = A cos(2x) + B sin(2x) . Then
10 cos(2x) + 15 sin(2x) = y ′′ + 9y
= [A cos(2x) + B sin(2x)]′′
+ 9 [ A cos(2x) + B sin(2)]
.
Worked Solutions
3
= [−4A cos(2x) − 4B sin(2x)]
+ [9A cos(2x) + 9B sin(2x)]
= 5A cos(2x) + 5B sin(2x) .
Cutting out the middle yields
10 cos(2x) + 15 sin(2x) = 5A cos(2x) + 5B sin(2x) .
Comparing the terms on either side then gives us
cos(2x) terms:
10 = 5A
sin(2x) terms:
15 = 5B
.
= 2 and B =
15
5
and
10
5
Solving for A and B , we get A =
y p (x) = 2 cos(2x) + 3 sin(2x)
= 3 . Thus,
.
For yh :
y ′′ + 9y = 0
0 = r2 + 9
֒→
֒→
√
r = ± −9 = ±3i
֌
yh (x) = c1 cos(3x) + c2 sin(3x) .
The general solution to the nonhomogeneous equation is then
y(x) = y p (x) + yh (x)
20.3 c.
= 2 cos(2x) + 3 sin(2x) + c1 cos(3x) + c2 sin(3x)
x x For y p : Let y(x) = y p (x) = A cos
+ B sin
. Then
3
.
3
x x 26 cos
− 12 sin
= y ′′ + 3y ′
3
3
=
h
x i′′
x + B sin
A cos
3
3
h
x x i′
+ 3 A cos
+ B sin
3
3
x A
B
= − cos
−
sin(2)
9
3
9
x x A
B
+ 3 − sin
+
cos
3
3
3
3
A
x
B
x
= − + B cos
+ −A −
sin
9
3
Comparing terms:
x cos
terms:
3
26 = −
A
+ B
9
9
3
.
4
Method of Educated Guess
and
sin
x terms:
3
− 12 = −A −
B
9
.
Solving for A and B , and plugging the results back into the formula for y p :
26 = −
A
+ B
9
and
− 12 = −A −
B
9
֒→
A = 9[B − 26]
and
− 12 = −9[B − 26] −
֒→
A = 9[B − 26]
and
B =
֒→
B
9
9
[12 + 9 · 26] = 27
82
A = 9[27 − 26] = 9
and
B = = 27
x x y p (x) = 9 cos
+ 27 sin
.
֒→
3
3
For yh :
y ′′ + 3y ′ = 0
֒→
0 = r 2 + 3r = r (r + 3)
֒→
r =0
֒→
and
r = −3
yh (x) = c1 + c2 e−3x
.
The general solution to the nonhomogeneous equation is then
x x y(x) = y p (x) + yh (x) = 9 cos
+ 27 sin
+ c1 + c2 e−3x
3
20.5 a.
3
.
For y p : Since g(x) , the right side of the differential equation, is a constant, we’ll let
y(x) = y p (x) = A . Plugging that into the differential equation, we then have
−200 = y ′′ − 3y ′ − 10y = [ A]′′ − [ A]′ − 10 A = 0 − 0 − 10 A
So, A =
−200
−10
= 20 and
y p (x) = 20
.
For yh :
y ′′ − 3y ′ − 10y = 0
֒→
0 = r 2 − 3r − 10 = (r + 2)(r − 5)
֒→
r = −2
֒→
yh (x) = c1 e−2x + c2 e5x
and
r = 5
.
So the general solution to the nonhomogeneous differential equation is
y(x) = y p (x) + yh (x) = 20 + c1 e−2x + c2 e5x
.
.
Worked Solutions
20.5 c.
5
For y p : Since g(x) , the right side of the differential equation, is a second degree
polynomial, we’ll let y(x) = y p (x) = Ax 2 + Bx + C . Plugging that into the
differential equation, we then have
18x 2 + 3x + 4 = y ′′ − 6y ′ + 9y
′′
′
= Ax 2 + Bx + C − 6 Ax 2 + Bx + C
+ 9 Ax 2 + Bx + C
= 2 A − 6 [2 Ax + B] + 9 Ax 2 + Bx + C
= [9A]x 2 + [−12 A + 9B]x + [2 A − 6B + 9C] .
Cutting out the middle and comparing terms, we get the system
x 2 terms:
18 = 9A
x terms:
3 = −12 A + 9B
constant terms:
.
4 = 2 A − 6B + 9C
Consequently,
18
= 2 ,
9
3 + 12 A
3 + 12 · 2
B =
=
= 3 ,
9
9
4 − 2 A + 6B
4−2·2+6·3
15
C =
=
=
= 2
9
9
−3
A =
,
and the particular solution is
y p (x) = Ax 2 + Bx + C = 2x 2 + 3x + 2
.
For yh :
y ′′ − 6y ′ + 9y = 0
֒→
֒→
0 = r 2 − 6r + 9 = (r − 3)2
֌
yh (x) = c1 e3x + c2 xe3x
r = 3
.
So the general solution to the nonhomogeneous differential equation is
y(x) = y p (x) + yh (x) = 2x 2 + 3x + 2 + c1 e3x + c2 xe3x
20.7 a.
.
For y p : Since g(x) = 25x cos(2x) , we’ll let
y(x) = y p (x) = [ Ax + B] cos(2x) + [C x + D] sin(2x) .
Computing the derivatives, we get
y ′ (x) = [ A] cos(2x) + [ Ax + B](−2 sin(2x))
+ [C] sin(2x) + [C x + D](2 cos(2x))
(⋆)
6
Method of Educated Guess
= [2C x + A + 2D] cos(2x) + [−2 Ax − 2B + C] sin(2x) ,
and
y ′′ (x) = [2C] cos(2x) + [2C x + A + 2D](−2 sin(2x))
+ [−2 A] sin(2x) + [−2 Ax − 2B + C](2 cos(2x))
= [−4Ax − 4B + 4C] cos(2x) + [−4C x − 4A − 4D] sin(2x) .
Plugging these into the differential equation, we then have
25x cos(2x) = y ′′ + 9y
= [−4Ax − 4B + 4C] cos(2x) + [−4C x − 4A − 4D] sin(2x)
+ 9 [ Ax + B] cos(2x) + [C x + D] sin(2x)
= [5Ax + 5B + 4C] cos(2x) + [5C x − 4A + 5D] sin(2x) .
Cutting out the middle and comparing terms, we get the system
x cos(2x) terms:
25 = 5A
cos(2x) terms:
0 = 5B + 4C
x sin(2x) terms:
0 = 5C
sin(2x) terms:
0 = −4A + 5D
Hence,
A =
25
5
= 5
, C = 0
,
4
5
B = − C = 0
and
D =
4
A
5
= 4
,
and particular solution (⋆) ends up being
y p (x) = [5x + 0] cos(2x) + [0x + 4] sin(x) = 5x cos(2x) + 4 sin(2x) .
For yh : From several of the previous problems, we already know that the general solution
to the corresponding homogeneous equation y ′ + 9y = 0 is
yh (x) = c1 cos(3x) + c2 sin(3x)
Consequently, the general solution to the given nonhomogeneous differential equation is
y(x) = y p (x) + yh (x)
= 5x cos(2x) + 4 sin(2x) + c1 cos(3x) + c2 sin(3x) .
20.7 c.
For y p : Since g(x) = 54x 2 e3x , we’ll let
y(x) = y p (x) = Ax 2 + Bx + C e3x
.
Computing the derivatives, we get
y ′ (x) = [2 Ax + B] e3x + Ax 2 + Bx + C 3e3x
= 3Ax 2 + (2 A + 3B)x + (B + 3C) e3x
(⋆)
Worked Solutions
7
and
y ′′ (x) = [6Ax + (2 A + 3B)] e3x + 3Ax 2 + (2 A + 3B)x + (B + 3C) 3e3x
= 9Ax 2 + (12 A + 9B)x + (2 A + 6B + 9C) e3x .
Plugging these into the differential equation, we then have
54x 2 e3x = y ′′ + 9y
= 9Ax 2 + (12 A + 9B)x + (2 A + 6B + 9C) e3x
+ 9 Ax 2 + Bx + C e3x
= 18Ax 2 + (12 A + 18B)x + (2 A + 6B + 18C) e3x
.
Cutting out the middle and comparing terms, we get:
x 2 e3x terms:
xe
3x
54 = 18A
terms:
0 = 12 A + 18B
e3x terms:
0 = 2 A + 6B + 18C
Hence,
A =
54
18
C = −
= 3 ,
12
2
A = − ·3
18
3
1
− [2 · 3 − 6 · 2]
18
B = −
1
[2 A + 6B]
18
=
and particular solution (⋆) ends up being
h
i
1
y p (x) = 3x 2 − 2x + e3x
3
,
= −2
=
1
3
,
.
For yh : From several of the previous problems, we already know that the general solution
to the corresponding homogeneous equation y ′ + 9y = 0 is
yh (x) = c1 cos(3x) + c2 sin(3x)
Consequently, the general solution to the given nonhomogeneous differential equation is
h
i
1
y(x) = y p (x) + yh (x) = 3x 2 − 2x + e3x + c1 cos(3x) + c2 sin(3x) .
3
20.7 e.
For y p : We’ll let
y(x) = y p (x) = [ Ax + B] cos(2x) + [C x + D] sin(2x) .
Computing the derivatives, we get
y ′ (x) = [ A] cos(2x) + [ Ax + B](−2 sin(2x))
+ [C] sin(2x) + [C x + D](2 cos(2x))
= [2C x + A + 2D] cos(2x) + [−2 Ax − 2B + C] sin(2x)
(⋆)
8
Method of Educated Guess
and
y ′′ (x) = [2C] cos(2x) + [2C x + A + 2D](−2 sin(2x))
+ [−2 A] sin(2x) + [−2 Ax − 2B + C](2 cos(2x))
= [−4Ax − 4B + 4C] cos(2x) + [−4C x − 4A − 4D] sin(2x)
Plugging that into the differential equation, we then have
[−6x − 8] cos(2x) + [8x − 11] sin(2x)
= y ′′ − 2y ′ + y
= [−4Ax − 4B + 4C] cos(2x) + [−4C x − 4A − 4D] sin(2x)
− 2 [2C x + A + 2D] cos(2x) + [−2 Ax − 2B + C] sin(2x)
+ [ Ax + B] cos(2x) + [C x + D] sin(x)
= [(−3A + 4C)x + (−2 A − 3B + 4C − 4D)] cos(2x)
+ [(4A − 3C)x + (−4A + 4B − 2C − 3D)] sin(2x) .
Cutting out the middle and comparing terms, we get the system
x cos(2x) terms:
−6 = −3A + 4C
cos(2x) terms:
−8 = −2 A − 3B + 4C − 4D
x sin(2x) terms:
sin(2x) terms:
8 = 4A − 3C
−11 = −4A + 4B − 2C − 3D
Since there are two equations just involving A and C , we will solve that pair first:
−6 = −3A + 4C
֒→
A =
6 + 4C
3
and
and
8 = 4A − 3C
6 + 4C
7C
8 = 4
− 3C = 8 +
3
֒→
A =
6 + 4C
3
֒→
A =
6+4·0
= 2
3
and
3
C =
and
3(8 − 8)
= 0
7
C = = 0
.
With these values for A and C , the other equations in the system for the coefficients
become
−8 = −2 · 2 − 3B + 4 · 0 − 4D
and
−11 = −4 · 2 + 4B − 2 · 0 − 3D
֌
4 = 3B + 4D
− 3 = 4B − 3D
֌
.
Solving for B and D :
4 = 3B + 4D
֒→
B =
4 − 4D
3
and
and
− 3 = 4B − 3D
4 − 4D
16
25D
−3 = 4
− 3D =
−
3
3
3
Worked Solutions
֒→
֒→
9
B =
4 − 4D
3
B =
and
4−4·1
= 0
3
D =
3
16
3+
25
3
and
D = 1
= 1
.
So particular solution (⋆) becomes
y p (x) = [2x + 0] cos(2x) + [0x + 1] sin(2x) = 2x cos(2x) + sin(2x) .
For yh :
y ′′ − 2y ′ + y = 0
֒→
֒→
0 = r 2 − 2r + 1 = (r − 1)2
֌
yh (x) = c1 e x + c2 xe x
r = 1
.
Hence, the general solution to the given nonhomogeneous differential equation is
y(x) = y p (x) + yh (x) = 2x cos(2x) + sin(2x) + c1 e x + c2 xe x
20.9 a.
.
For these problems, the “first guess” will not work. For this reason we start by finding
the general solution yh to the corresponding homogeneous differential equation:
y ′′ − 3y ′ − 10y = 0
֒→
0 = r 2 − 3r − 10 = (r + 2)(r − 5)
֒→
r = −2
֒→
yh (x) = c1 e−2x + c2 e5x
and
r = 5
.
For y p : Since g(x) = −3e−2x ,
the “first guess” for y p (x) is Ae−2x
.
But that is a term in yh . So we must consider the “second guess”,
“second guess” for y p (x) = x × “first guess” = x Ae−2x = Axe−2x
.
Since this is not a solution to the corresponding homogeneous equation, it is an appropriate
“guess”. Accordingly, we will let
y(x) = y p (x) = Axe−2x
֒→
֌
y ′ (x) = Ae−2x − 2 Axe−2x
y ′′ (x) = −4Ae−2x + 4x Ae−2x
.
Using this with the given differential equation yields
−3e−2x = y ′′ − 3y ′ − 10y
= −4Ae−2x + 4x Ae−2x − 3 Ae−2x − 2 Axe−2x − 10 Axe−2x
10
Method of Educated Guess
= [4 + 6 − 10] Axe−2x + [−4 − 3] Ae−2x
= [0] Axe−2x + [−7] Ae−2x = −7Ae−2x
Thus, A =
−3
−7
3
7
=
.
, and
y p (x) = Axe−2x =
3 −2x
xe
7
,
with the general solution being
3 −2x
xe
7
y(x) = y p (x) + yh (x) =
20.9 c.
+ c1 e−2x + c2 e5x
.
For yh :
y ′′ + 4y ′ = 0
֒→
0 = r 2 + 4r = r (r + 4)
֒→
r = 0
֒→
and
r = −4
yh (x) = c1 + c2 e−4x
.
For y p :
“first guess” = Ax 2 + Bx + C
,
which we must reject since one term, the constant C , is also a solution to the corresponding homogeneous differential equation. So we consider
“second guess” = x Ax 2 + Bx + C = Ax 3 + Bx 2 + C x .
No terms in this can be viewed as a term in yh . So, in finding y p we will let
y(x) = y p (x) = Ax 3 + Bx 2 + C x
֒→
y ′ (x) = 3Ax 2 + 2Bx + C
֌
y ′′ (x) = 6Ax + 2B
.
Plugging into the differential equation,
x 2 = y ′′ + 4y ′
= [6Ax + 2B] + 4 3Ax 2 + 2Bx + C
= [12 A]x 2 + [6A + 8B]x + [2B + 4C]
,
leads to the system
x 2 terms:
1 = 12 A
x terms:
0 = 6A + 8B
constant terms:
0 = 2B + 4C
.
Hence,
A =
1
12
,
6
8
B = − A = −
1
16
2
4
, C = − B =
1
32
,
Worked Solutions
11
y p (x) = Ax 3 + Bx 2 + C x =
1 3
x
12
−
1 2
x
16
1
x
32
+
,
and the general solution is
y(x) =
20.9 e.
1 3
x
12
−
1 2
x
16
+
1
x
32
+ c1 + c2 e−4x
.
For yh :
y ′′ − 6y ′ + 9y = 0
0 = r 2 − 6r + 9 = (r − 3)2
֒→
yh (x) = c1 e3x + c2 xe3x
֒→
r = 3
֌
.
For y p :
“First guess” = Ae3x
,
which we reject since it is a term in yh (with A = c1 ).
“Second guess” = x Ae3x = Axe3x
,
which we also reject since it is also a term in yh . But
“Third guess” = x Axe3x = Ax 2 e3x
is not a solution to the corresponding homogeneous equation. Thus, to find y p we let
y(x) = y p (x) = Ax 2 e3x
֌
y ′ (x) = 2 Axe3x + 3Ax 2 e3x
y ′′ (x) = 2 Ae3x + 12 Axe3x + 9Ax 2 e3x
֒→
.
Plugging into the differential equation, we get
10e3x = y ′′ − 6y ′ + 9
= 2 Ae3x + 12 Axe3x + 9Ax 2 e3x − 6 2 Axe3x + 3Ax 2 e3x
+ 9 Ax 2 e3x
= 2 Ae3x + [12 − 12] Axe3x + [9 − 18 + 9] Ax 2 e3x = 2 Ae3x
So A =
10
2
= 5,
y p (x) = Ax 2 e3x = 5x 2 e3x
,
and the general solution is
y(x) = 5x 2 e3x + c1 e3x + c2 xe3x
20.10 a.
.
For yh :
y ′′ − 3y ′ − 10y = 0
֒→
0 = r 2 − 3r − 10 = (r + 2)(r − 5)
.
12
Method of Educated Guess
yh (x) = c1 e−2x + c2 e5x
֒→
.
For y p :
“First guess” =
Ax 2 + Bx + C e2x = Ax 2 e2x + Bxe2x + Ce2x
.
Since no term in this is also a term in yh , this guess is what we will use,
y(x) = y p (x) =
Ax 2 + Bx + C e2x
.
Computing the derivatives:
y ′ (x) = [2 Ax + B] e2x + Ax 2 + Bx + C 2e2x
= 2 Ax 2 + (2 A + 2B)x + (B + 2C) e2x
and
y ′′ (x) = [4Ax + (2 A + 2B)] e2x + 2 Ax 2 + (2 A + 2B)x + (B + 2C) 2e2x
= 4Ax 2 + (8A + 4B)x + (2 A + 4B + 4C) e2x .
Plugging into the differential equation:
72x 2 − 1 e2x = y ′′ − 3y ′ − 10y
= 4Ax 2 + (8A + 4B)x + (2 A + 4B + 4C) e2x
− 3 2 Ax 2 + (2 A + 2B)x + (B + 2C) e2x
− 10 Ax 2 + Bx + C e2x
= −12 Ax 2 + (2 A − 12B)x + (2 A + B − 12C) e2x
.
Cutting out the middle and comparing terms, we get the system
x 2 e2x terms:
72 = −12 A
xe2x terms:
0 = 2 A − 12B
e2x terms:
−1 = 2 A + B − 12C
Hence,
A =
72
−12
= −6
,
B =
2
A
12
= −1 , C =
1
[1 + 2 A +
12
particular solution (⋆) is
y p (x) = −6x 2 − x − 1 e3x
,
and the general solution is
y(x) = −6x 2 − x − 1 e3x + c1 e−2x + c2 e5x
.
B] = −1 ,
Worked Solutions
20.10 c.
13
For yh :
y ′′ − 10y ′ + 25y = 0
0 = r 2 − 10r + 25 = (r − 5)2
֒→
֌
yh (x) = c1 e5x + c2 xe5x
֒→
r = 5
.
For y p ,
“First guess” = Ae−5x
.
Since no term in this is also a term in yh , this is what we will use. Plugging y(x) =
y p (x) = Ae−5x into the differential equation:
6e−5x = y ′′ − 10y ′ + 25y
′′
′
= Ae−5x − 10 e−5x + 25e−5x
= 25Ae−5x + 50 Ae−5x + 25e−5x = 100 Ae−5x
Hence, A =
6
100
=
3
50
.
,
y p (x) =
3 −5x
e
50
,
and the general solution is
y(x) =
20.10 d.
3 −5x
e
50
+ c1 e5x + c2 xe5x
.
For yh :
y ′′ − 10y ′ + 25y = 0
֒→
֒→
0 = r 2 − 10r + 25 = (r − 5)2
֌
yh (x) = c1 e5x + c2 xe5x
r = 5
.
For y p :
“First guess” = Ae5x
,
which we reject since it is a term in yh (with A = c1 ).
“Second guess” = x Ae5x = Axe3x
,
which we also reject since it is also a term in yh . But
“third guess” = x Axe5x = Ax 2 e5x
is not a solution to the corresponding homogeneous equation. Thus, to find y p we let
y(x) = y p (x) = Ax 2 e5x
֒→
֌
y ′ (x) = 2 Axe5x + 5Ax 2 e5x
y ′′ (x) = 2 Ae5x + 20 Axe5x + 25Ax 2 e5x
Plugging into the differential equation,
6e5x = y ′′ − 10y ′ + 25y
.
14
Method of Educated Guess
= 2 Ae5x + 20 Axe5x + 25Ax 2 e5x − 10 2 Axe5x + 5Ax 2 e5x
+ 25 Ax 2 e5x
= 2 Ae5x + [20 − 20] Axe5x + [25 − 50 − 25] Ax 2 e5x = 2 Ae5x
So A =
6
2
= 3,
and the general solution is
y p (x) = Ax 2 e5x = 3x 2 e5x
,
y(x) = 3x 2 e5x + c1 e5x + c2 xe5x
20.10 e.
.
.
For yh :
y ′′ + 4y ′ + 5y = 0
֒→
2
r + 4r + 5 = 0
֒→
r =
֌
−4 ±
√
42 − 4 · 5
= −2 ± 1i
2
yh (x) = c1 e−2x cos(x) + c2 e−2x sin(x) .
For y p ,
“First guess” = A cos(3x) + B sin(3x)
.
Since no term in this is also a term in yh , this is what we will use for y = y p :
24 sin(3x) = y ′′ + 4y ′ + 5y
= [ A cos(3x) + B sin(3x)]′′ + 4 [ A cos(3x) + B sin(3x)]′
+ 5 [A cos(3x) + B sin(3x)]
= [−9A cos(3x) − 9B sin(3x)] + 4 [−3A sin(3x) + 3B cos(3x)]
+ 5 [A cos(3x) + B sin(3x)]
= [−4A + 12B] cos(3x) + [−12 A − 4B] sin(3x) .
Comparing terms, we get
cos(3x) terms:
0 = −4A + 12B
sin(3x) terms:
24 = −12 A − 4B
At this point, you should be able to figure out that the solution to this system is A = −
3
5
and B = − . Hence,
9
5
y p (x) = − cos(3x) −
3
5
sin(3x)
and the general solution is
9
5
y(x) = − cos(3x) −
20.10 g.
3
5
sin(3x) + c1 e−2x cos(x) + c2 e−2x sin(x) .
For yh :
y ′′ − 4y ′ + 5y = 0
9
5
Worked Solutions
֒→
15
2
r + 4r + 5 = 0
֌
r =
4±
√
42 − 4 · 5
= −2 ± 1i
2
yh (x) = c1 e2x cos(x) + c2 e2x sin(x)
֒→
.
For y p ,
“First guess” = Ae2x cos(x) + Be2x sin(x) ,
which we reject since both terms are also terms in yh . Continuing:
“Second guess” = x Ae2x cos(x) + Be2x sin(x)
= Axe2x cos(x) + Bxe2x sin(x)
.
Neither term in this guess is a term in yh , so this is what we will use for y = y p . For
convenience in computing the derivatives, it may be best to write this as
y(x) = y p (x) = xe2x [ A cos(x) + B sin(x)] .
Computing the derivatives:
y ′ (x) = e2x + 2xe2x [ A cos(x) + B sin(x)] + xe2x [−A sin(x) + B cos(x)]
= e2x A + [2 A + B]x cos(x) + B + [2B − A]x sin(x)
and
y ′′ (x) = 2e2x
A + [2 A + B]x cos(x) + B + [2B − A]x sin(x)
+ e2x 0 + [2 A + B] cos(x) + 0 + [2B − A] sin(x)
+ e2x A + [2 A + B]x (− sin(x)) + B + [2B − A]x cos(x)
= e2x [4A + 2B] + [3A + 4B]x cos(x)
+ [4B − 2 A] + [3B − 4A]x sin(x) .
Plugging into the differential equation:
e2x sin(x) = y ′′ + 4y ′ + 5y
= e2x [4A + 2B] + [3A + 4B]x cos(x)
+ [4B − 2 A] + [3B − 4A]x sin(x)
− 4e2x A + [2 A + B]x cos(x) + B + [2B − A]x sin(x)
+ 5e2x [ Ax cos(x) + Bx sin(x)]
= e2x 2B + 0x cos(x) + − 2 A + 0x sin(x) .
Comparing terms, we get
e2x cos(x) terms:
e2x sin(x) terms:
0 = 2B
1 = −2 A
1
2
Hence, A = − , B = 0 ,
1
2
1
2
y p (x) = − xe2x cos(x) + 0xe2x sin(x) = − xe2x cos(x)
16
Method of Educated Guess
and the general solution is
1
2
y(x) = − xe2x cos(x) + c1 e2x cos(x) + c2 e2x sin(x) .
20.10 j.
From several problems above, we know
yh (x) = c1 e2x cos(x) + c2 e2x sin(x) .
For y p :
“First guess” = A
.
Since no term in this is also a term in yh , this is what we will use for y = y p :
100 = y ′′ − 4y ′ + 5y = [ A]′′ − 4[ A]′ + 5A = 5A
So A =
100
5
= 20 ,
.
y p (x) = 20
and the general solution is
y(x) = 20 + c1 e2x cos(x) + c2 e2x sin(x)
20.10 l.
.
From several problems above, we know
yh (x) = c1 e2x cos(x) + c2 e2x sin(x) .
For y p :
“First guess” = Ax 2 + Bx + C
.
Since no term in this is also a term in yh , this is what we will use for y = y p :
10x 2 + 4x + 8 = y ′′ − 4y ′ + 5y
′′
′
= Ax 2 + Bx + C − 4 Ax 2 + Bx + C
+ 5 Ax 2 + Bx + C
= [2 A] − 4 [2 Ax + B] + 5 Ax 2 + Bx + C
= 5Ax 2 + [−8A + 5B]x + [2 A − 4B + 5C] .
Comparing terms:
x 2 terms:
10 = 5A
x terms:
4 = −8A + 5B
constant terms:
8 = 2 A − 4B + 5C
So,
A =
10
= 2
5
,
B =
4 + 8A
= 4
5
8 − 2 A + 4B
= 4
5
, C =
y p (x) = 2x 2 + 4x + 4
,
and the general solution is
y(x) = 2x 2 + 4x + 4 + c1 e2x cos(x) + c2 e2x sin(x) .
,
Worked Solutions
20.10 n.
17
For yh :
y ′′ + y = 0
r2 + 1 = 0
֒→
֒→
֌
√
r = ± −1 = ±1i
yh (x) = c1 cos(x) + c2 sin(x) .
For y p ,
“First guess” = A cos(x) + B sin(x)
,
which we reject since both terms are also terms in yh . Continuing:
“Second guess” = x [A cos(x) + B sin(x)] = Ax cos(x) + Bxe sin(x) .
Neither term in this guess is a term in yh , so this is what we will use for y p . Computing
the derivatives:
y(x) = y p (x) = x[ A cos(x) + B sin(x)]
֒→
y ′ (x) = [ A cos(x) + B sin(x)] + x[−A sin(x) + B cos(x)]
֒→
y ′′ (x) = 2[−A sin(x) + B cos(x)] − x[ A cos(x) + B sin(x)] .
Plugging into the differential equation:
6 cos(x) − 3 sin(x) = y ′′ + y
= 2[−A sin(x) + B cos(x)] − x[ A cos(x) + B sin(x)]
+ x[ A cos(x) + B sin(x)]
= 2B cos(x) − 2 A sin(x) .
Comparing terms, we get
Hence, A =
cos(x) terms:
6 = 2B
sin(x) terms:
−3 = −2 A
−3
−2
=
3
2
, B=
6
2
.
= 3,
y p (x) =
3
x
2
cos(x) + 3x sin(x) ,
and the general solution is
y(x) =
20.11 a.
3
x
2
cos(x) + 3x sin(x) + c1 cos(x) + c2 sin(x) .
From several problems above, we know
yh (x) = c1 e2x cos(x) + c2 e2x sin(x) .
For y p :
“First guess” =
A0 x 3 + A1 x 2 + A2 x + A3 e−x sin(x)
+ B0 x 3 + B1 x 2 + B2 x + B3 e−x cos(x)
Since no term in this is also a term in yh , this guess is what we use for y p .
.
18
Method of Educated Guess
20.11 c.
For yh :
y ′′ − 5y ′ + 6y = 0
0 = r 2 − 5r + 6 = (r − 2)(r − 3)
֒→
yh (x) = c1 e2x + c2 e3x
֒→
.
For y p :
“First guess” =
Ax 2 + Bx + C e−7x = Ax 2 e−7x + Bxe−7x + Ce−7x
.
Since no term in this is also a term in yh , this guess is what we use for y p .
20.11 e.
From previous problems, we know
yh (x) = c1 e2x + c2 e3x
.
“First guess” = Ae−8x
.
For y p :
Since no term in this is also a term in yh , this guess is what we use for y p .
20.11 g.
From previous problems, we know
yh (x) = c1 e2x + c2 e3x
.
For y p :
“First guess” =
Ax 2 + Bx + C e3x = Ax 2 e3x + Bxe3x + Ce3x
.
which we reject because the last term is also a term in yh . Continuing,
“Second guess” = x Ax 2 + Bx + C e3x = Ax 3 e3x + Bx 2 e3x + C xe3x
.
Since no term in this is also a term in yh , this guess is what we use for y p .
20.11 i.
From previous problems, we know
yh (x) = c1 e2x + c2 e3x
.
For y p :
“First guess” = Ax 2 + Bx + C e3x cos(2x) + Dx 2 + E x + F e3x sin(2x) .
Since no term in this is also a term in yh , this guess is what we use for y p .
20.11 k.
For yh :
y ′′ − 4y ′ + 20y = 0
֒→
֒→
2
r − 4r + 20 = 0
֌
r =
4±
p
(−4)2 − 4 · 20
= 2 ± 4i
2
yh (x) = c1 e2x cos(4x) + c2 e2x sin(4x)
.
Worked Solutions
19
For y p :
“First guess” = Ae2x cos(4x) + Be2x sin(4x) ,
which we reject because each term is also a term in yh . Continuing,
“Second guess” = x Ae2x cos(4x) + Be2x sin(4x)
= Axe2x cos(4x) + Bxe2x sin(4x)
.
Since no term in this is also a term in yh , this guess is what we use for y p .
20.12 a.
From the exercises in chapter 17, we know
yh (x) = c1 + c2 x + c3 x 2 + c4 e4x
.
For y p :
“First guess” = Ae−2x
.
Since no term in this is also a term in yh , this guess is what we will use for y = y p .
Plugging it into the differential equation, we obtain
(4)
(3)
12e−2x = y (4) − 4y (3) = Ae−2x
− 4 Ae−2x
= (−2)4 Ae−2x − 4(−2)3 Ae−2x = 48Ae−2x
So A =
20.12 b.
12
48
=
1
4
, and y p (x) =
1 −2x
e
4
.
.
From the exercises in chapter 17, we know
yh (x) = c1 + c2 x + c3 x 2 + c4 e4x
.
For y p :
“First guess” = Ae4x
.
which we reject because it is a term in yh . Continuing,
“Second guess” = x Ae4x = Axe4x
.
Since no term in this is also a term in yh , this guess is what we will use for y = y p .
Doing so and computing the necessary derivatives:
y(x) = y p (x) = Axe4x
֌
y ′′ (x) = 8Ae4x + 16Axe4x
֒→
֌
y ′ (x) = Ae4x + 4Axe4x
y (3) (x) = 48Ae4x + 64Axe4x
y (4) (x) = 256Ae4x + 256Axe4x
֒→
.
Plugging these into the differential equation, we obtain
32e4x = y (4) − 4y (3)
= 256Ae4x + 256Axe4x − 4 48Ae4x + 64Axe4x
= 64Ae4x + 0 Axe4x = 64Ae4x
So A =
32
64
=
1
2
, and y p (x) =
1 4x
xe
2
.
.
20
Method of Educated Guess
20.12 c.
From the exercises in chapter 17, we know
yh (x) = c1 + c2 x + c3 x 2 + c4 e4x
.
For y p :
“First guess” = A cos(2x) + B sin(2x)
.
Since no term in this is also a term in yh , this guess is what we will use for y = y p .
Plugging it into the differential equation, we obtain
10 sin(2x) = y (4) − 4y (3)
= [ A cos(2x) + B sin(2x)](4) − 4 [ A cos(2x) + B sin(2x)](3)
= [16A cos(2x) + 16B sin(2x)] − 4 [8A sin(2x) − 8B cos(2x)]
= [16A + 32B] cos(2x) + [−32 A + 16B] sin(2x) .
Comparing terms, we get
cos(x) terms:
0 = 16A + 32B
sin(x) terms:
10 = −32 A + 16B
You can easily solve this system, obtaining A = −
1
4
y p (x) = − cos(2x) +
20.12 e.
1
4
and B =
1
8
sin(2x) .
1
8
. Thus,
From the exercises in chapter 17, we know
yh (x) = c1 e x + c2 cos(x) + c3 sin(x) .
For y p :
“First guess” = Ax 2 + Bx + C
.
Since no term in this is also a term in yh , this guess is what we will use for y = y p .
Plugging it into the differential equation, we obtain
x 2 = y (3) − y ′′ + y ′ − y
(3)
′′
Ax 2 + Bx + C
− Ax 2 + Bx + C
′
+ Ax 2 + Bx + C − Ax 2 + Bx + C
= [0] − [2 A] + [2 Ax + B] + −Ax 2 − Bx − C
=
= −Ax 2 + [2 A − B]x + [2 A + B − C]
.
Comparing terms, we get
x 2 terms:
x terms:
constant terms:
1 = −A
0 = 2A − B
.
0 = 2A + B − C
Hence, A = −1 , B = −2 A = 2 , C = 2 A + B = 0 , and y p (x) = −x 2 − 2x
.
Worked Solutions
20.12 g.
21
From the exercises in chapter 17, we know
yh (x) = c1 e x + c2 cos(x) + c3 sin(x) .
For y p :
“First guess” = Ae x
,
which we reject because it is a term in yh . Continuing,
“Second guess” = x Ae x = Axe x
,
Since no term in this is also a term in yh , this guess is what we will use for y = y p .
Doing so and computing the necessary derivatives:
y(x) = y p (x) = Axe x
֒→
y ′′ (x) = 2 Ae x + Axe x
֌
y ′ (x) = Ae x + Axe x
y (3) (x) = 3Ae x + Axe x
֌
.
Plugging these into the differential equation, we obtain
6e x = y (3) − y ′′ + y ′ − y
= 3Ae x + Axe x − 2 Ae x + Axe x + Ae x + Axe x − Axe x
= 2 Ae x + 0 Axe x = 2 Ae x
So A =
20.13 a.
6
2
= 3 , and y p (x) = 3xe x
.
.
From the exercises in chapter 17, we know
yh (x) = c1 + [c2 + c3 x] cos(3x) + [c4 + c5 x] sin(3x)
.
For y p :
“First guess” =
Ax 2 + Bx + C e3x = Ax 2 e3x + Bxe3x + Ce3x
Since no term in this is also a term in yh , this guess is what we use for y p .
20.13 c.
From the exercises in chapter 17, we know
yh (x) = c1 + [c2 + c3 x] cos(3x) + [c4 + c5 x] sin(3x)
.
For y p :
“First guess” =
Ax 2 + Bx + C e3x cos(3x)
+ Dx 2 + E x + F e3x sin(3x)
.
Since no term in this is also a term in yh , this guess is what we use for y p .
20.13 e.
From the exercises in chapter 17, we know
yh (x) = c1 e x + c2 cos(x) + c3 sin(x) .
For y p :
“First guess” = [ Ax + B] cos(x) + [C x + D] sin(x)
.
22
Method of Educated Guess
= Ax cos(x) + B cos(x) + C x sin(x) + D sin(x) ,
which we reject because it has two terms that are also in yh . Continuing,
“Second guess” = x [Ax + B] cos(x) + [C x + D] sin(x)
= Ax 2 + Bx cos(x) + [C x 2 + Dx] sin(x)
= Ax 2 cos(x) + Bx cos(x) + C x 2 sin(x) + Dx sin(x) .
Since no term in this is also a term in yh , this guess is what we use for y p .
20.13 g.
From the exercises in chapter 17, we know
yh (x) = c1 e x + c2 cos(x) + c3 sin(x) .
For y p :
“First guess” =
Ax 5 + Bx 4 + C x 3 + Dx 2 + E x + F e2x
= Ax 5 e2x + Bx 4 e2x + C x 3 e2x + Dx 2 e2x + E xe2x + Fe2x
Since no term in this is also a term in yh , this guess is what we use for y p .
20.14 a.
From the answer to exercise 20.1 b, we know that one solution to
y ′′ − 6y ′ + 9y = 27e6x = g1 (x)
is
y p1 = 3e6x
,
and from the answer to exercise 20.3 b, we know that one solution to
y ′′ − 6y ′ + 9y = 25 sin(6x) = g2 (x)
is
y p2 (x) =
4
9
1
3
cos(6x) −
sin(6x) .
So, by superposition, it follows that one solution to
y ′′ − 6y ′ + 9y = 27e6x + 25 sin(6x) = g1 (x) + g2 (x)
is
y p (x) = y p1 (x) + y p2 (x) = 3e6x +
20.14 c.
4
9
cos(6x) −
1
3
sin(6x) .
Using the identity
sin2 (x) =
1
2
−
1
2
cos(2x)
we can rewrite the differential equation as
y ′′ − 4y ′ + 5y = g1 (x) + g2 (x)
where
g1 (x) =
1
2
and
1
2
g2 (x) = − cos(2x) .
.
Worked Solutions
23
Our solution is y p = y p1 + y p2 where y p1 and y p2 are, respectively, particular solutions
to
y ′′ − 4y ′ + 5y = g1 (x)
and
y ′′ − 4y ′ + 5y = g2 (x) ,
which we will find by the method of guess.
First, we must find (again) yh :
y ′′ − 4y ′ + 5y = 0
2
֒→
r − 4r + 5 = 0
֌
4±
r =
p
(−4)2 − 4 · 5
= 2 ± 1i
2
yh (x) = c1 e2x cos(x) + c2 e2x sin(x) .
֒→
For y p1 : Since g1 is a constant,
“First guess” = A
.
Since this is not a term in yh , this guess is what we will use for y = y p1 . Plugging it
into the differential equation:
5
2
So A =
= g1 (x) = y ′′ − 4y ′ + 5y = [ A]′′ − [ A]′ + 5A = 5A
5
5·2
=
1
2
.
, and
y p1 (x) =
1
2
.
1
2
For y p2 : Since g2 (x) = − cos(2x) ,
“First guess” = A cos(2x) + B sin(2x)
.
Since no term is also a term in yh , this guess is what we will use for y = y p2 . Plugging
it into the differential equation:
1
2
− cos(2x) = g2 (x)
= y ′′ − 4y ′ + 5y
= [ A cos(2x) + B sin(2x)]′′ − 4[ A cos(2x) + B sin(2x)A]′
+ 5[ A cos(2x) + B sin(2x)]
= [−4A cos(2x) − 4B sin(2x)] − 4[−2 A sin(2x) + 2B cos(2x)A]
+ 5[ A cos(2x) + B sin(2x)]
= [ A − 8B] cos(2x) + [8A + B] sin(2x) . .
Comparing terms, we get
cos(x) terms:
sin(x) terms:
−
1
2
= A − 8B
0 = 8A + B
24
Method of Educated Guess
Solving this yields A = −
1
26
and B =
y p2 (x) = −
1
26
4
13
. Hence,
cos(2x) +
4
13
sin(2x) .
Finally, combining the above yields
y p (x) = y p1 + y p2 =
1
2
−
1
26
cos(2x) +
4
13
sin(2x) .