1. No category

Sample Solutions of Assignment 1 for MAT3270A:
1.1-1.3,2.1-2.2.2
Note: Any problems about the sample solutions, please email Ms.Zhang
a math.cuhk.edu.hk) directly.
Rong (rzhang
1. Solve the following intial value problms:
(a).
(b).
(c).
dy
dt
dy
dt
dy
dt
= 2y − 1, y(0) = 1.
= y − 4, y(0) = 2.
= 3y, y(0) = 1.
Answer:
(a). Multiply e−2t on both sides of the equation
e−2t dy
− 2e−2t y = −e−2t
dt
⇒
d
(e−2t y)
dt
⇒
e−2t y = 12 e−2t + C,
⇒
y=
1
2
= −e−2t ,
+ Ce2t ,
y(0) = 1,
C = 12 ,
⇒
⇒
y(t) =
1
2
+ 12 e2t .
(b).Mutiply e−t on both sides:
e−t dy
− e−t y = −4e−t ,
dt
⇒
d
(e−t y)
dt
= −4e−t ,
⇒
e−t y = 4e−t + C,
⇒
y = 4 + Cet .
y(0) = 2,
⇒
C = −2,
⇒
y(t) = 4 − 2et .
1
2
(c).Multiply e−3t on both sides:
e−3t dy
− 3ye−3t = 0,
dt
d
(e−3t y)
dt
⇒
= 0,
⇒
e−3t y = C,
⇒
y = Ce3t ,
y(0) = 1,
⇒
C = 1,
⇒
y(t) = e3t .
2. Determine the order of the given ODE and state whether the equation is linear or nonlinear
3
(a). ddt3y + t2 dy
+ (cos2 t)y = t4 − t2 ,
dt
2
(b). ddt2y + sin(t +
d4 y
)
dt4
= cos t,
(c). dy
+ ey = 2t + 1.
dt
Answer:
(a) is a 3rd order linear ODE,
(b) is a 4th order nonlinear ODE,
(c) is a 1st order nonlinear ODE.
0
3. Solve ty + 2y = t2 − t + 1, y(1) = 21 , t > 0
Answer:
Multiply t on both sides of the equation, we have
t2 y 0 + 2ty = t3 − t2 + t
⇒
(t2 y)0 = t3 − t2 + t
⇒ t2 y − y(1) = 41 (t4 − 1) − 13 (t3 − 1) + 12 (t2 − 1)
⇒
y=
t2
4
− 3t +
1
12t2
+
1
2
3
0
2
4. Solve y + 2ty = 2te−t , y(−1) = 5
Answer:
2
Multiply et on both sides of the equation, we have
2
2
et y 0 + 2tet y = 2t
2
(et y)0 = 2t
⇒
2
⇒ et y − ey(−1) = t2 − 1
⇒
0
5. Solve ty + 2y =
2
y = e−t (t2 − 1 + 5e)
cos t
, y(π)
t
= 21 , t > 0.
Answer: Multiply t on both sides of the equation, we have
t2 y 0 + 2ty = cos t
⇒
(t2 y)0 = cos t
⇒ t2 y − π 2 y(π) = sin t
⇒ y = (sin t + 21 π 2 )t−2
6. Prove that y(t) → 0 as t → ∞ if y solves
0
y + ay = be−λt
where a > 0, λ > 0 and b are constants.
Answer:
Multiply eat on both sides of the equation, we have
eat y 0 + aeat y = be(a−λ)t
⇒
(eat y)0 = be(a−λ)t
4
case I: If a = λ, we have
(eat y)0 = b
⇒
eat y = bt + c
⇒ y = e−at (bt + c)
Then y(t) → 0 as t → ∞ since a > 0.
case II If a 6= λ, we get
(eat y)0 = be(a−λ)t
eat y = y0 +
⇒
⇒ y = y0 e−at +
b
(e(a−λ)t
(a−λ)
b
(e(a−λ)t
(a−λ)
− 1)
− 1)e−at
we also has y(t) → 0 as t → ∞ since a > 0 and λ.
7. Solve the following separable equations
0
(a). y + 3y 2 sin x = 0
0
(b). y =
0
2x2
1+y 2
1
(c). xy = (1 − y 2 ) 2
Answer: (a). The original equation can be rewritten as
dy
y2
= −3 sin xdx
⇒ − y1 = 3 cos x + C
⇒
y=
1
−3 cos x−C
(b). The original equation can be rewritten as
(1 + y 2 )dy = 2x2 dx
⇒
y + 13 y 3 = 23 x3 + C
⇒ y 3 + 3y − 2x3 + C = 0
5
(c). The original equation can be rewritten as
1
(1 − y 2 )− 2 dy =
dx
x
⇒ arcsin y = ln |x| + C
⇒
y = sin(ln |x| + C)
8. Find the solutions to the IVP and determine the Interval of Existence
0
(a) y =
0
(b) y =
2x
, y(2) = 0.
1+2y
2x
, y(1) = −2
y+x2 y
(c). sin(2x)dx + cos(3y)dy = 0, y( π2 ) = π3 .
0
(d). y =
0
(e). y =
1+3x2
, y(0) = 1.
3y 2 −6y
3x2
, y(1) = 0
3y 2 −4
Answer:
(a). By separation variable method, we have
(1 + 2y)dy = 2xdx
⇒ y + y 2 − y(2) − y(2)2 = x2 − 4
⇒
⇒
y 2 + y = x2 − 4
√
y = 12 (−1 ± 4x2 − 15)
Since y(2) = 0, y = 12 (−1 +
√
4x2 − 15) and the existence interval is
√
(
15
, +∞).
2
(b). By separate variable method, we have
ydy =
⇒
⇒
1 2
(y
2
2xdx
1+x2
− 4) = ln(1 + x2 ) − ln 2
q
2
y = ± 2 ln 1+x
+4
2
6
2
2 ln 1+x
+4>0
2
2
> −2
ln 1+x
2
⇔
1+x2
2
⇔
q
> e−2
2
+ 4 and the existence interval is
Since y(1) = −2, y = − 2 ln 1+x
2
(−∞, ∞).
(c). By separate variable method, we have
− sin 2xdx = cos 3ydy
⇒
⇒
1
(cos 2x
2
+ 1) = 13 sin 3y
sin 3y = 23 (cos 2x + 1)
Since y( π2 ) = π3 , y = 13 (π − arcsin( 23 (cos 2x + 1))), Further more,
−1 ≤ 32 (cos 2x + 1)) ≤ 1
−1 ≤ cos 2x ≤ − 31
⇔
⇔
1
2
arccos(− 13 ) ≤ x ≤ π − 21 arccos(− 13 )
then the existence interval is [ 12 arccos(− 31 ), π − 12 arccos(− 31 )]
(d). By separate variable method, we have
(3y 2 − 6y)dy = (1 + 3x2 )dx
y 3 − 3y 2 = x3 + x − 2
⇒
Since 3y(0)2 − 6y(0) = 3 − 6 = −3 < 0, 3y 2 − 6y < 0 must be true in
the existence interval,otherwise there exists some x such that 3y 2 − 6y
vanishes. That is, 0 < y < 2. Then
−4 < y 3 − 3y 2 < 0
⇒ −4 < x3 + x − 2 < 0
⇒
−1 < x < 1
Hence, the interval of existence is (−1, 1).
7
(e). By separate variable method, we have
(3y 2 − 4)dy = 3x2 dx
⇒
y 3 − 4y = x3 − 1
Since 3y(1)2 − 4 = 0 − 4 < 0, 3y 2 − 4 < 0 must be true in the existence
√
interval. that is, − 2 3 3 < y <
√
2 3
.
3
Then
√
16 3
9
√
√
− 169 3 < x3 − 1 < 169 3
√
√
1 − 169 3 < x3 < 1 + 169 3
√
− 169 3 < y 3 − 4y <
⇒
⇒
Hence, the interval of existence is ((1 −
√
16 3 13
) , (1
9
+
√
16 3 13
) ).
9
9. Solve the following homogeneous equation
dy
ax + by
=
dx
cx + dy
where a, b, c, d are constants.
Answer:
Let w = xy , that is y = xw, we have
y 0 = xw0 + w =
a + bw
c + dw
hence
c + dw
1
w0 =
2
a + (b − c)w − dw
x
this is separable equation and can be solved, we omit the details here.
The solution is
−b + c + 2 xd y
(b + c)
√
√
arctan(
)
−b2 + 2 b c − c2 − 4 a d
−b2 + 2 b c − c2 − 4 a d
1
+ log(−a x2 + (−b + c) x y + d y 2 ) = C
2
8
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
−0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Figure 1. Problem2-The direction field of y 0 = 2y − 4
10.Solve the following equation
dy
x2 − 3y 2
=
dx
2xy
Answer:
Let w = xy , that is y = xw, we have
xw0 + w =
1
2w
xw0 =
=
⇒
1
2w
2wdw
1−5w2
⇒
=
−
3w
2
5w
2
dx
x
⇒ ln |1 − 5w2 | = −5 ln |x| + C
hence
s
1 ± C|x|−5
5
s
1 ± C|x|−5
y = xw = ±x
5
w=±
Page 7:2,3,6,9,11,15,17,25.
2. y 0 = 2y − 4
Answer: By the direction field,y → 2,as t → ∞
9
0.5
0
−0.5
−1
−1.5
−2
−2.5
−3
−3.5
−4
−4.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Figure 2. Problem3-the direction field of y 0 = 2y + 4
If y(0) < 2, the slopes are negative,and hence the solutions decrease.
If y(0) > 2, the slopes are positive,and hence the solutions increase.
All solutions appear to diverge away from the equilibrium solution
y(t) = 2.
3. y 0 = 2y + 4
Answer: By the direction field,y → −2,as t → ∞
If y(0) < −2, the slopes are negative,and hence the solutions decrease.
If y(0) > −2, the slopes are positive,and hence the solutions increase.
All solutions appear to diverge away from the equilibrium solution
y(t) = −2.
6. y 0 = y + 3
Answer: By the direction field,y → −3,as t → ∞
If y(0) < −3, the slopes are negative,and hence the solutions decrease.
If y(0) > −3, the slopes are positive,and hence the solutions increase.
10
0.5
0
−0.5
−1
−1.5
−2
−2.5
−3
−3.5
−4
−4.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Figure 3. Problem6-the direction field ofy 0 = y + 3
4
3.5
3
2.5
2
1.5
1
0.5
0
−0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Figure 4. Problem11-the direction field of y = y(3 − y)
All solutions appear to diverge away from the equilibrium solution
y(t) = −3.
9.All other solutions diverge from y = 3
Answer: For example,y 0 = y − 3
11.y = y(3 − y)
Answer: By the direction field,y = 0 and y = 3 are equilibrium
solutions;
11
60
58
56
54
52
50
48
46
44
42
40
0
2
4
6
8
10
12
Figure 5. Problem25
If y(0) > 0, y → 3.
If y(0) < 0, y diverge away from the equilibrium solution y(t) = 0.
15.The direction field of Figure1.1.5.
Answer: By the direction field,we know that y 0 = 0 when y = 2,y 0 < 0
when y > 2,y 0 > 0 when y < 2.So it corresponds to (j).
17.The direction field of Figure1.1.7.
Answer: By the direction field,we know that y 0 = 0 when y = −2,y 0 <
0 when y > −2,y 0 > 0 when y < −2.So it corresponds to (g).
25. Answer: (a)mv 0 = mg − kv 2
(b)Let v 0 = 0,we have mg − kv 2 = 0,thus v →
(c)By
q
10g/k = 49,we have k = 2/49
(d)See Figure5:Problem25
Page15-19:1(b),2(a),3,5,7,10
1(b). dy
= −2y + 3,y(0) = y0
dt
q
mg/k, as t → ∞
12
2
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
0
1
2
3
4
5
Figure 6. Problem1(b)
Answer: (b). Multiply e2t on both sides of the equation
e2t dy
+ 2e2t y = 3e2t
dt
⇒
d
(e2t y)
dt
= 3e2t ,
⇒
e2t y = 32 e2t + C,
⇒
y=
3
2
+ Ce−2t ,
y(0) = y0 ,
⇒
C = y0 − 23 ,
⇒ y(t) =
3
2
+ (y0 − 23 )e−2t .
From the Figure Problem1(b),we know that y0 =
3
,y
2
≡
3
;Other
2
solutions are converging to the equilibrium solutions as t increase.
2(a). dy
= y − 3,y(0) = y0
dt
13
300
200
100
0
−100
−200
−300
0
1
2
3
4
5
Figure 7. Problem2(a)
Answer: (b). Multiply e−t on both sides of the equation
e−t dy
− e−t y = −3e−t
dt
⇒
d
(e−t y)
dt
= −3e−t ,
⇒
e−t y = 3e−t + C,
⇒
y = 3 + Cet ,
y(0) = y0 ,
⇒
C = y0 − 3,
⇒ y(t) = 3 + (y0 − 3)et .
From the Figure Problem2(a),we know that y0 = 3,y ≡ 3;Other solutions are diverging away from the equilibrium solutions as t increase.
3. dy
= −ay + b
dt
Answer:
14
(a) Multiply eat on both sides of the equation
eat dy
+ eat y = beat
dt
⇒
⇒
⇒
d
(eat y)
dt
= beat ,
eat y = ab eat + C,
y=
b
a
+ Ceat .
(b)
y(0) = y0 ,
⇒
C = y0 − ab ,
⇒ y(t) =
b
a
+ (y0 − ab )eat .
(c)(i)Equilibrium is lower and is approached more rapidly.
(ii) Equilibrium is higher.
(iii)Equilibrium remains the same and is approached more rapidly.
4.Consider the differential equation dy/dt = ay − b
Answer:
(a)Find the equilibrium solution ye . If a 6= 0,ay − b = 0,thus ye = ab .
If a = 0,y 6= 0,there’s no equilibrium solution.
If a = 0, b = 0,ye = C for arbitrary constant C.
(b) For simplicity,let’s assume a 6= 0,ye =
b
.Then dYdt(t)
a
=
y
t
=
ay − b = a(y − ye ) + aye − b = aY (t)
5.Undetermined Coefficients.
Answer: (a) y1 (t) = ceat .
(b) Substitute y = y1 (t)+k into (i),we have dy/dt = aceat = a(ceat +
k) − b,hence k = ab .
15
(c) This solution coincides with Eq.(17) in the text.
7.The field mouse population in Example 1
Answer:
First solve the ODE dp/dt = 0.5p − 450,we have p(t) = 900 + (p(0) −
900)e0.5t .
(a)If p(0) = 800,p(t) = 900 + 100e0.5t ,let p(t) = 0,we have 900 +
100e0.5t = 0,then t = 2 ln 9 ≈ 4.39months.
(b) If 0 < p0 < 900,letp(t) = 0,we have 900 = (900 − p0 )e0.5t ,then
900
t = 2 ln 900−p
.
0
900
< 6,thus 0 < p0 < 900(1 − e−6 ) ∼
(c)t < 12,hence ln 900−p
= 897.8
0
10.Modify Example 2 so that the falling object experiences no air
resistance.
Answer:
(a)dv/dt = 9.8, v(0) = 0.
(b)Solve the above initial value problem ,we have v = 9.8t,since
dx/dt = v, x(0) = 0,thus x = 4.9t2 .There,x = 300,hence t =
q
300/4.9 ≈
7.28s.
q
(c)By v = 9.8t, the velocity at the time of impact is v = 9.8 300/4.9 ≈
76.68m/s.
Page 24-26:1,4,6,9,12,15,19
1.Answer: It’s a 2nd order linear ODE.
4.Answer: It’s a 1st order nonlinear ODE.
6.Answer: It’s a 3rd order linear ODE.
9.ty 0 − y = t2 ; y = 4t + t2 .
Answer:
16
Since ty 0 −y = t(4+2t)−(4t+t2 ) = t2 ,hence y = 4t+t2 is a solution.
12.t2 y 00 + 5ty 0 + 4y = 0, t > 0; y1 (t) = t−2 , y2 (t) = t−2 ln t.
Answer:
Since y10 = −2t−3 ,y100 = 6t−4 ,hence t2 y 00 + 5ty 0 + 4y = 0,so y1 is a
solution.
Since y20 = −2t−3 ln t+t−1 ,y200 = 6t−4 ln t−3t−2 ,hence t2 y 00 +5ty 0 +4y =
2,so y2 isn’t a solution.
Revise:y20 = −2t−3 ln t + t−3 ; thus y200 = 6t−4 ln t − 5t−4 , then t2 y 00 +
5ty 0 + 4y = 0 ,so y2 indeed is a solution.
15. Determine for what values of r the equation y 0 + 3y = 0 has
solutions of the form y = ert .
Answer: Substitute y = erx into the equation, we have
ert (r + 3) = 0
⇒
(r + 3) = 0
⇒
r = −3
hence, y = e−3t satisfies the given ODE.
19. Determine for what values of r the equation t2 y 00 + 5ty 0 + 3y = 0
has solutions of the form y = tr for t > 0.
Answer:
Substitute y = tr into the equation, we have
tr {r(r − 1) + 5r + 3} = 0, t > 0
⇒
(r + 3)(r + 1) = 0,
⇒
r = −1 or r = −3.
hence, y = t−1 and y = t−3 and satisfies the given ODE.