e9 R
1
−−−→
§3.1: 1, 5, 9, 15, 20, 21, 23.
§3.2: 13, 14.
§3.3: 1, 5, 7, 11, 18, 22, 23, 25.
§3.1
00
1
−9e−9
1
9e−9 R1 +R1
−−−−−−
−−→
0
1
R2
1 e10
−10e
−−−→
0 1
1
−e10 R2 +R1
−−−−−−−−→
0
Homework #6, Math 2174
0
1. y + 2y − 3y = 0.
e10
e
e9
0
e10
e + 9e
e9
9 −1
10 e
⇒ c1 =
⇒ r = −3, 1,
⇒ y=
⇒ y = c1 e−3t + c2 et .
1 9
10 e ,
+
e10
10e
9 −1 t
e
10 e
r2 + 5r = 0
2r2 − 3r + 1 = 0
⇒ r(r + 5) = 0
⇒ (2r − 1)(r − 1) = 0
⇒ r = 0, −5,
⇒ r = 21 , 1,
⇒ y = c1 + c2 e−5t .
⇒ y = c1 e 2 t + c2 e t
=
1 −9(t−1)
10 e
1
1
⇒ y 0 = 12 c1 e 2 t + c2 et
(
2 = y(0) = c1 + c2
⇒ 1
1
0
2 = y (0) = 2 c1 + c2
9. y 00 + y 0 − 2y = 0, y(0) = 1, y 0 (0) = 1.
r2 + r − 2 = 0
⇒ (r + 2)(r − 1) = 0
⇒ r = −2, 1,
1 2 −1
1 1 2
2 R1 +R2
−
−
−
−
−
−
−
→
1
0 12 −1
1 21
2
2
1 1 2 −R2 +R1 1 0 3
2R2
−−→
−−−−−−→
0 1 −1
0 1 −1
⇒ y = c1 e−2t + c2 et
⇒ y 0 = −2c1 e−2t + c2 et
(
1 = y(0) = c1 + c2
⇒
1 = y 0 (0) = −2c1 + c2
1
⇒ c1 = 3, c2 = −1,
1 1 2R1 +R1 1 1 1
−−−−−→
1 1
0 3 3
(1/3)R2
1 1 1 −R2 +R1 1 0
−−−−−→
−−−−−−→
0 1 1
0 1
1
−2
⇒ y = 3et/2 − et .
0
1
y attains its maximum value at t
⇒ c1 = 0, c2 = 1,
⇒ y 0 (t) = 0
⇒ y = et
⇒
00
0
⇒
0
15. y + 8y − 9y = 0, y(1) = 1, y (1) = 0.
⇒
3 t/2
− et = 0
2e
2
3
2 u − u = 0 (u
u( 23 − u) = 0
= et/2 )
r2 + 8r − 9 = 0
⇒ u = 0, 3/2
⇒ (r + 9)(r − 1) = 0
⇒ et/2 = 0, 3/2
⇒ r = −9, 1,
⇒ t/2 = ln(3/2)
⇒ y = c1 e−9t + c2 et
⇒ t = 2 ln(3/2) = ln(9/4) .
⇒ y 0 = −9c1 e−2t + c2 et
(
1 = y(1) = c1 e−9 + c2 e
⇒
0 = y 0 (1) = −9c1 e−9 + c2 e
e−9
−9e−9
e 1
e 0
y(t) = 0 for some t
⇒ 3et/2 − et = 0
⇒ 3u − u2 = 0
⇒ u(3 − u) = 0
1
20. 2y 00 − 3y 0 + y = 0, y(0) = 2, y 0 (0) = 21 .
5. y 00 + 5y 0 = 0.
e9
9
1 9
10 e
9 −1
10 e
9 −1
,
10 e
c2 =
1 9 −9t
10 e e
e9
1
=
9
0
9 9
e9 − 10
1 0
e
=
9 −1
e
0 1
10
0
1
r2 + 2r − 3 = 0
⇒ (r + 3)(r − 1) = 0
(u = et/2 )
+
9 t−1
10 e
.
§3.2
⇒ u = 0, 3
⇒ e
t/2
= 0, 3
13. Consider t2 y 00 − 2y = 0 for t > 0.
⇒ t/2 = ln 3
y1 (t) = t2
⇒ t = 2 ln 3 = ln 9 .
⇒ y10 = 2t, y100 = 2
⇒ t2 y100 − 2y1 = t2 · 2 − 2 · t2
21. y 00 − y 0 − 2y = 0, y(0) = α, y 0 (0) = 2.
= (2t2 ) − (2t2 ) = 0
r2 − r − 2 = 0
⇒ y1 is a solution.
⇒ (r − 2)(r + 1) = 0
⇒ r = 2, −1,
y2 (t) = t−1
⇒ y = c1 e2t + c2 e−t
0
⇒ y20 = −t−2 , y200 = 2t−3
−t
2t
⇒ y = 2c1 e − c2 e
(
α = y(0) = c1 + c2
⇒
2 = y 0 (0) = 2c1 − c2
1
2
α (−2)R1 +R1 1 1
−−−−−−−→
2
0 −3
(−1/3)R2
1 1
α
−−−−−−→
0 1 (2α − 2)/3
1 0 (2 + α)/3
−R2 +R1
−−−−
−−→
0 1 (2α − 2)/3
1
−1
⇒ t2 y200 − 2y2 = t2 · (2t−3 ) − 2 · t−1
= (2t−1 ) − (2t−1 ) = 0
⇒ y2 is a solution.
α
2 − 2α
y = c1 t2 + c2 t−1
⇒ y 0 = −2c1 t + c2 t−2 , y 00 = 2c1 + 2c2 t−3
⇒ t2 y 00 − 2y = t2 · (2c1 + 2c2 t−3 ) − 2 · (c1 t2 + c2 t−1 )
= (2c1 t2 + 2c2 t−1 ) − (2c1 t2 + 2c2 t−1 ) = 0
⇒ y is a solution.
14. Consider yy 00 + (y 0 )2 = 0 for t > 0.
y1 (t) = 1
⇒ c1 = (2 + α)/3, c2 = (2α − 2)/3,
⇒ y=
2+α 2t
3 e
+
⇒ y10 = 0, y100 = 0
2α−2 −t
3 e
⇒ y1 y100 + (y10 )2 = 0 + 0 = 0
y approaches zero as t → ∞
⇒ y1 is a solution.
⇔ The coefficient of e2t is 0
⇔ (2 + α)/3 = 0
y2 (t) = t1/2
⇒ y20 = 21 t−1/2 , y200 =
⇔ α = −2 .
⇒
00
0
23. y − (2α − 1)y + α(α − 1)y = 0.
−1 −3/2
4 t
−3/2
y2 y200 + (y20 )2 = (t1/2 ) ( −1
)
4 t
1 −1
1 −1
= (4t ) − (4t ) = 0
+ ( 12 t−1/2 )2
⇒ y2 is a solution.
r2 − (2α − 1)r + α(α − 1) = 0
⇒ (r − α)(r − (α − 1)) = 0
⇒ r = α, α − 1,
y(t) = c1 + c2 t1/2
⇒ y = c1 eα + c2 e(α−1)t .
⇒ y 0 = 21 c1 12 t−1/2 , y 00 =
⇒ yy 00 + (y 0 )2 = (c1 +
=
All solutions tend to zero as t → ∞
=
⇔ α < 0 and α − 1 < 0
−3/2
−1
4 c1 c2 t
−3/2
−1
4 c1 c2 t
+
−3/2
−1
4 c2 t
−3/2
c2 t1/2 ) ( −1
)
4 c2 t
−1 2 −1
4 c2 t
+ ( 12 c2 t−1/2 )2
+ 14 c22 t−1
6≡ 0
⇒ y is not a solution if c1 c2 6= 0.
⇔ α < 0 and α < 1
⇔ α<0.
Theorem 3.2.2 does not apply because the differential
equation cannot be written as y 00 + p(t)y 0 + q(t)y = 0.
All nonzero solutions become unbounded as t → ∞
⇔ α > 0 and α − 1 > 0
§3.3
⇔ α > 0 and α > 1
1.
⇔ α>1.
exp(1 + 2i) = e cos 2t + i e sin 2t .
2
23. 3u00 − u0 + 2u = 0, u(0) = 2, u0 (0) = 0.
5.
3r2 − r + 2 = 0
p
1 ± (−1)2 − 4 · 3 · 2
⇒ r=
2·3 √
√
1
1 ± −23
23
= ±
i
=
6
6√
6
√
⇒ u = et/6 (c1 cos( 23 t/6) + c2 sin( 23 t/6))
√
√
⇒ u0 = 16 et/6 (c1 cos( 23 t/6) + c2 sin( 23 t/6))
√
√
√
√
+ et/6 ( − 6 23 c1 sin( 23 t/6) + 623 c2 cos( 23 t/6))
(
2 = u(0) = c1
√
⇒
0 = u0 (0) = 16 c1 + 623 c2
21−i = e(1−i) ln 2 = eln 2+i(− ln 2)
= eln 2 cos(− ln 2) + ieln 2 sin(− ln 2)
= 2 cos(ln 2) − i 2 sin(ln 2) .
7. y 00 − 2y 0 + 2y = 0.
r2 − 2r + 2 = 0
p
√
2 ± (−2)2 − 4 · 2
2 ± −4
⇒ r=
=
=1±i
2
2
⇒ y = et (c1 cos t + c2 sin t) .
11. y 00 + 6y 0 + 13y = 0.
⇒ c1 = 2,
c2 =
√6 −1 c1
23 6
=
−2
√
23
r2 + 6r + 13 = 0
√
√
−6 ± 62 − 4 · 13
−6 ± −16
⇒ r=
=
= −3 ± 2i
2
2
√
⇒ u = et/6 2 cos 623 t −
⇒ y = e−3t (c1 cos 2t + c2 sin 2t) .
(Numerical parts skipped.)
√2
23
√
cos
23
6 t
.
25. y 00 + 2y 0 + 6y = 0, y(0) = 2, y 0 (0) = α ≥ 0.
18. y 00 + 4y 0 + 5y = 0, y(0) = 1, y 0 (0) = 0.
+ e−2t (−c1 sin t + c2 cos t)
(
1 = y(0) = c1
⇒
0 = y 0 (0) = −2c1 + c2
r2 + 2r + 6 = 0
√
√
√
−2 ± 22 − 4 · 6
−2 ± −20
⇒ r=
=
= −1 ± i 5
2 √
√ 2
⇒ y = e−t (c1 cos 5 t + c2 sin 5 t)
√
√
⇒ y 0 = −e−t (c1 cos 5 t + c2 sin 5 t)
√
√
√
√
+ e−t (− 5 c1 sin 5 t + 5 c2 cos 5 t)
(
2 = y(0) = c1
√
⇒
α = y 0 (0) = −c1 + 5 c2
⇒ c1 = 1, c2 = 2c1 = 2
⇒ c1 = 2,
r2 + 4r + 5 = 0
√
√
−4 ± 42 − 4 · 5
−2 ± −4
⇒ r=
=
= −2 ± i
2
2
⇒ y = e−2t (c1 cos t + c2 sin t)
⇒ y 0 = −2e−2t (c1 cos t + c2 sin t)
√
√
c2 = (α − c1 )/ 5 = (α + 2)/ 5
√
√
√ sin 5 t) .
⇒ y = e−t (2 cos 5 t + α+2
5
−t
⇒ y = e (cos t + 2 sin t) .
22. y 00 + 2y 0 + 2y = 0, y(π/4) = 2, y 0 (π/4) = −2.
r2 + 2r + 2 = 0
√
√
−2 ± 22 − 4 · 2
−2 ± −4
⇒ r=
=
= −1 ± i
2
2
−t
⇒ y = e (c1 cos t + c2 sin t).
y(t) = 0 with smallest positive t
√
√
√ sin 5 t) = 0
⇒ e−t (2 cos 5 t + α+2
5
√
⇒ e−t R sin(θ + 5 t) = 0,
q
√
2
2 5
where R = 22 + (α+2)
, θ = arctan( 2+α
)
5
√
⇒ θ + 5t = π
√ √
√
2 5
⇒ t = (π − θ)/ 5 = π − arctan( 2+α
) / 5.
⇒ y 0 = −e−t (c1 cos t + c2 sin t)
⇒
+ e−t (−c1 sin t + c2 cos t)
(
2 = y(π/4) = e−π/4 √12 (c1 + c2 )
−2 = y 0 (π/4) = −e−π/4 √12 2c2
√
c1 + c2 = 2 2 eπ/4
√ π/4
⇒
c2 = 2 e
√
⇒ c2 = 2 eπ/4 ,
√
√
c1 = 2 2 eπ/4 − c2 = 2 eπ/4
√
√
⇒ y = e−t 2 eπ/4 cos t + 2 eπ/4 sin t .
√
As α → ∞, the last expression tends to π/ 5 .
(
(Numerical parts skipped.)
3