HW 5 - Due Sep 26, Wed
1. Find the general solution
(a) y 00 + 2y 0 − 3y = 0
• r2 + 2r − 3 = 0,
(r + 3)(r − 1) = 0,
r = −3, 1
• y = C1 e−3x + C2 ex
(b) 2y 00 − 3y 0 + y = 0
• 2r2 − 3r + 1 = 0,
(2r − 1)(r − 1) = 0,
1
r = ,1
2
• y = C1 e1/2x + C2 ex
(c) y 00 − 2y 0 = 0
• r2 − 2r = 0,
r(r − 2) = 0,
r = 0, 2
• y = C1 e0x + C2 e2x = C1 + C2 e2x
(d) 6y 00 − y 0 − y = 0
• 6r2 − r − 1 = 0,
(3r + 1)(2r − 1) = 0,
1 1
r=− ,
3 2
• y = C1 e−1/3x + C2 e1/2x
2. Solve the initial value problem and describes the behavior of the solution as x
increases
(a) 3y 00 + y 0 − 4y = 0,
• 3r2 + r − 4 = 0,
y(0) = 0,
y 0 (0) = 1
(3r + 4)(r − 1) = 0,
• General solution: y = C1 e−3/4x + C2 ex
3
• y 0 = − C1 e−3/4x + C2 ex
4
• 0 → x, 0 → y, 0 = C1 + C2
3
• 0 → x, 1 → y 0 , 1 = − C1 + C2
4
2
2
• C1 = − , C2 =
3
3
2 −3/4x 2 x
• y=− e
+ e
3
3
• lim y = ∞
x→∞
1
3
r = − ,1
4
(b) 4y 00 − y = 0,
•
•
•
•
•
•
•
•
00
y(−2) = 1, y 0 (−2) = −1
1
1
4r2 − 1 = 0, r2 = , r = ±
4
2
y = C1 e1/2x + C2 e−1/2x
1
1
y 0 = C1 e1/2x − C2 e−1/2x
2
2
−2 → x, 1 → y, C1 e−1 + C2 e1 = 1
1
1
−2 → x, −1 → y 0 ,
C1 e−1 − C2 e1 = −1
2
2
3 −1
C1 = −e, C2 = e
2
3 −1 −1/2x
3
1/2x
y = −e · e
+ e ·e
= −e1/2x+1 + e−1/2x−1
2
2
lim y = −∞
x→∞
0
(c) y + 4y + 3y = 0,
y(0) = 2,
y 0 (0) = −1
r2 + 4r + 3 = 0, (r + 3)(r + 1) = 0, r = −3, −1
y = C1 e−3x + C2 e−x
y 0 = −3C1 e−3x − C2 e−x
0 → x, 2 → y, C1 + C2 = 2
0 → x, −1 → y 0 , −3C1 − C2 = −1
1
5
• C1 = − , C2 =
2
2
1 −3x 5 −x
• y=− e
+ e
2
2
1 5
• lim y = − + = 2
x→∞
2 2
•
•
•
•
•
3. Determine the longest interval in which the given initial value problem has a unique
solution. Do not attempt to find the solution.
(a) ty 00 + 3y = t, y(1) = 1, y 0 (1) = 2
3
• y 00 + y = 1
t
3
• Domain of is t 6= 0, or (−∞, 0) ∪ (0, ∞)
t
• Domain of 1 is (−∞, ∞)
• 1 ∈ (0, ∞), so the longest interval is (0, ∞)
(b) t(t − 4)y 00 + 3ty 0 + 4y = 2, y(3) = 0, y 0 (3) = −1
3t
4
2
• y 00 +
+
=
t(t − 4) t(t − 4)
t(t − 4)
2
• the longest interval of p(t), q(t) and g(t) is t 6= 0 and t 6= 4 or (−∞, 0) ∪
(0, 4) ∪ (4, ∞)
• 3 ∈ (0, 4), so the longest interval is (0, 4)
(c) y 00 + (cos t)y 0 + 3(ln |t|)y = 0,
•
•
•
•
y(1) = 0,
y 0 (1) = 1
Domain of cos t is (−∞, ∞)
Domain of ln |t| is t 6= 0 or (−∞, 0) ∪ (0, ∞).
the overlap is (−∞, 0) ∪ (0, ∞).
1 ∈ (0, ∞), so the longest interval is (0, ∞)
4. Show that if y = φ(t) is a solution of the differential equation y 00 +p(t)y 0 +q(t)y = g(t),
where g(t) 6= 0, then y = Cφ(t), C 6= 1 is not a solution. And if g(t) = 0, then
y = Cφ(t) is a solution.
• cφ → y,
cφ00 + cp(t)φ0 + cq(t)φ = g(t).
• since φ is a solution, φ00 + p(t)φ0 + q(t)φ = g(t),
• so we have cg(t) = g(t), ⇒ c = 1. ↔
5. Find the Wronskian without solve the equaiton
(a) t2 y 00 − t(t + 2)y 0 + (t + 2)y = 0
t(t + 2) 0 t + 2
y + 2 y=0
• y 00 −
t2
t
t(t + 2)
t+2
• p(t) =
=
t2Z
t
Z
2
•
p(t)dt = 1 + dt = t + 2 ln t
t
• W (t) = Ce−
R
p(t)dt
= Ce−t−2 ln t =
C
et t2
(b) (cos t)y 00 + (sin t)y 0 − ty = 0
sin t 0
t
• y 00 +
y −
y=0
cos t
cos t
sin t
• p(t) =
cos t Z
Z
Z
1
sin t
•
p(t)dt =
dt = −
d cos t = − ln(cos t)
cos
t
cos
t
R
• W (t) = Ce− p(t)dt = Celn(cos t) = C cos t
(c) x2 y 00 + xy 0 + (x2 − k 2 )y = 0,
• y 00 +
Bessel’s equation
x 0 x2 − k 2
y +
=0
x2
x2
3
• p(x) =
1
x
• W (x) = Ce−
R
1/xdx
= Ce− ln x =
C
x
6. If cos2 θ and 1 + cos 2θ are two particular solutions of y 00 + p(t)y 0 + q(t)y = 0, can
they form a fundamental set of solutions? Explain your answer.
• 1 + cos 2θ = 2 cos2 θ
2
2
cos
θ
2
cos
θ
• W = −2 cos θ sin θ −4 cos θ sin θ cos2 θ
2 cos2 θ
•
and
are linear dependent, so W = 0
−2 cos θ sin θ
−4 cos θ sin θ
• cos2 θ and 1 + cos 2θ can not form a fundamental set of solutions.
4