Set 2.9:
2
2 1
1,2
yp1
2Aex
A
yp2
2x
4Be Be2x
B
y
=
=
=
=
=
=
=
=
=
0
1
Axex
2ex
1
Be2x
6e2x
2
(C1 + x)ex + C2 ex + 2e2x
6
32 + 10 + 3 = 0
1,2
yp1
10A + 3Ax + 3B
A
yp2
=
=
=
=
D
=
y
1
3
Ax + B
9x
3, B = 10
C sin x + D cos x
3(C sin x + D cos x) +
10(C cos
Set 1.6:
46 Clairaut.
Regular solutions:
y = Cx +
1
C
Singular solution:
1
1
p =
2
p
x
!
x
1
= + 1
= 2 x
x
x
x =
y
Set 1.8:
12 Original family:
y0 =
1
x
Orthogonal family:
y0
dy
= x
= x dx
x2
y = +C
2
(an upside down parabola, slid up and down).
16 Orig
Set 1.3:
4 Separable.
dy
y2
1
y
= 3x2 dx
= x3 C
y
=
1
C + x3
8 Scale independent.
u + xu0
= 1+u
dx
du =
x
u = ln |x| + C
y = x(ln |x| + C)
10 Suggested substitution.
v0 4
dv
v2 + 4
v
1
arctan
2
2
v
y
= v2
= dx
= x+C
= 2 tan(2x + C)
= 2 tan(2x + C) 4x
16 S
Set 2.1:
4
2xz 0
dz
z
z
y
= 3z
3 dx
=
2x
1 x3/2
= C
= C1 x5/2 + C2
6
ytrial
0
ytrial
00
ytrial
sin x
x
0
sin x
sin x
0
= c
+c
x
x
00
cos x sin x
sin x
00 sin x
0
= c
+c
+ 2c
2
x
x
x
x
= c(x)
Substitute:
sin x
sin x
c00 sin x + 2c0 cos x
+ 2c0
x
x
c00 si
1.
x3 + 2yex
xy 0 2y
dy
y
y
x3 c0
c(x)
y
= xex y 0
= x3 ex
dx
= 2
x
= c(x) x2
= x3 ex
= C ex
= Cx2 x2 ex
Answer:
x = y 2 (C ey )
2.
m(m 1) 3m + 4 = m2 4m + 4
yields m = 2 as a double root. Substituting
yp = (A ln x + B)x
in the original equation yields (a
Set 2.14
12: Characteristic polynomial has clearly = 1 as one root, which leads to
(3 32 + 3 1) ( 1) = 2 2 + 1
telling us that all 3 roots are equal to 1. Thus
y := (C1 + C2 x + C3 x2 )ex
Since
y(0) = C1 = 2
y 0 (0) = C1 + C2 = 2
y 00 (0) = C1 + 2C2 + 2C3
Set 1.6:
32 Bernoulli, a = 1, 1 a = 2
u0 + 2u
du
u
u
c0 e2x
= 2x
= 2dx
= c(x) e2x
= 2x
Z
c = 2 xe2x dx =
( 1 x)e2x + C
2
1
u =
x + Ce2x
2r
1
y =
x + Ce2x
2
33 Bernoulli, a = 4, 1 a = 3
u0 u
du
u
ln u
y
0 x
x
c e + ce cex
c0
c
u
= 2x 1
= dx
=
=
=
=
=
=