1. No category

Sample Solutions of Assignment 2 for MATH3270A: 2.1-2.8
Note: Any problems about the sample solutions, please email Mr.Xiao Yao
a math.cuhk.edu.hk) directly.
(yxiao
September,2013
1. A mass of 0.5kg is dropped from rest offering a resistance of 0.1|v|,
where v is measured in m/sec.
(a) If the mass is dropped from a height of 6m, find its velocity when
it hits the ground.
(b) If the mass is to attain a velocity of no more than 5m/sec, find
the maximum height from which it can be dropped.
(c) Suppose that the resistance force is k|v|, where v is measured in
m/sec and k is a constant. If the mass is dropped from a height of
30m and must hit the ground with a velocity of no more that 10m/sec,
determine the coeffcient of the resistance k that is required.
Answer:
Denote the mass m, the height of the mass x(t),the velocity v = x0 ,
the resistance force k|v|, , the initial height h, then, x0 (0) = 0, x(0) = h.
By Newton’s 2nd Law, we have

00

= −mg − kx0
mx
x0 (0) = 0

x(0) = h
1
2
Hence
x00 +
⇒
⇒
⇒
⇒
⇒
k 0
x = −g
m
k
k
t
t
k
e m (x00 + x0 ) = −ge m
m
k
k
t
t
0
0
(e m x ) = −ge m
k
k
t
t
m
e m x0 = − g(e m − 1)
k
k
t
−
mg
0
x =−
(1 − e m )
k
k
mg
m − t
x=h−
(t + (e m − 1))
k
k
Here g ' 9.8m/s2 .
mg
(t +
(a). From h = 6, when it hits the ground,x = 0, i.e.
k
k
m − t
mg
(e m − 1)) = 6, we have t ' 1.1489s.Then v = x0 = −
(1 −
k
k
k
−
t
e m ) = −10.0593m/s.
(b).
⇒
|x0 | ≤ 5
k
−
t
mg
(1 − e m ) ≤ 5
k
t ≤ 0.53815s
⇒
h ≤ 1.3695m
⇒
3
(c). According to the problem,

k


mg
m − t


x = h−
(t + (e m − 1))



k
k


k

−
t
mg
0
m)
v
=
x
=
−
(1
−
e

k


 x = 0




h = 30


v = −10
which gives t = 3.9523s and k = 0.239438Kg/s. Hence the coefficient
of the resistance k must be 0.239438Kg/s.
2. State the region in the ty-plane where the (existence and uniqueness)
hypotheses of Theorem 2.4.2 are satisfied.
dy
log|ty|
dy
1 + t2
(a) =
(b)
=
dt
1 − t2 + y 2
dt
3y − y 2
Answer:
log |ty|
1
(a). By direct computing, f (t, y) =
, ∂y f (t, y) =
−
2
2
1−t +y
y(1 − t2 + y 2 )
2y log |ty|
. Hence the region of ty-plan where f (t, y), ∂y f (t, y) are
(1 − t2 + y 2 )2
both continuous is {(t, y)|t 6= 0, y 6= 0, t2 − y 2 6= 1}.
1 + t2
2y − 3
(b). Here f (t, y) =
, ∂y f (t, y) = (1 + t2 )
. So
2
3y − y
(y − 3)2 y 2
the region of ty-plan where f (t, y), ∂y f (t, y) are both continuous is
{(t, y)|y 6= 0, y 6= 3}.
3. Solve the following ODE and state the Interval of Existence in terms
of y0
t2
,
y(1 + t2 )
Answer: (a). By separation variable method, we have
(a)y 0 = 2ty 2 ,
y(0) = y0
⇒
⇒
(b)y 0 =
dy
= 2tdt
y2
1
1
− +
= t2
y y0
y0
y=
1 − t2 y0
y(0) = y0
4
So the Interval of Existence is (−∞, +∞) when y0 ≤ 0, the Interval of
1
1
Existence is (− √ , √ ) when y0 > 0.
y0
y0
(b). By separation variable method, we have
t2 dt
1 + t2
y 2 − y02 = 2(t − arctan t)
p
y = sign(y0 ) 2(t − arctan t) + y02
ydy =
⇒
⇒
Denote t∗ the only root of equation t−arctan t = −
of Existence is (t∗ , +∞).
y02
,then the Interval
2
4. Show that
(
(t − C)2
y(t) =
0
for t ≥ C ≥ 0
for t ≤ C
1
satisfies y 0 = 2y 2 , y(0) = 0 for any constant C ≥ 0. So we don’t have
uniqueness, why?
Answer: By direct computing,
2(t − C) for t ≥ C ≥ 0
0
y (t) =
0 for t ≤ C
and
1
(t − C) for t ≥ C ≥ 0
y 2 (t) =
0 for t ≤ C
1
We have y (t) = 2y 2 , y(0) = 0. On the Other hand, y(t) ≡ 0 is
1
another solution. So we don’t have uniqueness. Let f (t, y) = 2y 2 , it
0
is easy to check that f (t, y) is not local Lipschitz continuous in any
neighbourhood of y = 0 respect to y, so we have no uniqueness.
5. Solve the following initial value problem and state the Interval of
Existence
y 0 = 2y − y 2 ,
y(0) = 1
5
Answer: By separate variable method,
y 0 = 2y − y 2
dy
⇒
= dt
2y − y 2
dy
dy
+
= 2dt
⇒
y
2−y
y
⇒
ln
= 2t
2−y
y
= e2t
⇒
2−y
2e2t
2
⇒ y = 2t
=
e +1
1 + e−2t
So the Interval of Existence is (−∞, ∞).
6. In the following problems, sketch the graph of f (y) versus y, determine the critical points and classify each one as asymptotically stable
or unstable
Answer:
6
1.5
1
0.5
-1
1
2
3
-0.5
-1
-1.5
Figure 1. for problem 6
f (y) = y(y − 1)(y − 2)
2.5
2
1.5
1
0.5
-1
-0.5
0.5
1
1.5
2
Figure 2. for problem 6
f (y) = (y − 1)(ey − 1);
(a). From the figure of f (y) = y(y − 1)(y − 2), there are 3 critical
points y = 0, y = 1, y = 2 and y = 1 is stable, y = 0, y = 2 are unstable.
(b).From the figure of f (y) = (y − 1)(ey − 1), there are 2 critical
points y = 0, y = 1 and y = 0 is stable, y = 1 are unstable.
7
7. Consider the following bifurcation equation
dy
= x − x3
dt
Show that for < 0, there exists only one critical point which is
asymptotically stable; while for > 0, there are three critical points,
of which one is unstable and the other two are stable.
Answer:
(a).If < 0, then x − x3 = −x(x2 + ||) = 0 has only one root,so
there exist only one critical point x = 0. Since x > 0, x − x3 < 0 and
x < 0, x − x3 > 0, the critical point x = 0 is stable.
√
√
(b). If > 0, then x − x3 = −x(x − )(x − ) = 0 has 3 root,so
√
√
there exist 3 critical points − , 0, .
√
√
Since x < − , x − x3 > 0 and − < x < 0, x − x3 < 0, the critical
√
point x = − is stable.
√
√
Since − < x < 0, x − x3 < 0 and 0 < x < , x − x3 > 0, the
critical point x = 0 is unstable.
√
√
Since x > , x − x3 < 0 and 0 < x < , x − x3 > 0, the critical
√
point x = is stable.
√
i.e. the critical point x = 0 is unstable and the critical points x = ± is stable.
8. Solve the following Chemical Reactions equations
dx
= α(p − x)(q − x),
dt
x(0) = x0
Answer:
By separate variable method,
dx
= αdt
(p − x)(q − x)
8
Case 1: p = q
⇒
⇒
dx
= αdt
(x − p)2
1
1
−
= αdt
x0 − p x − p
x0 − p
x=p+
α(x0 − p)t + 1
Case 2: p 6= q
⇒
⇒
⇒
⇒
dx
= αdt
(p − x)(q − x)
1
1
−
= (p − q)αdt
x−p x−q
x − p x0 − q
ln(
) = (p − q)αt
∗
x − q x0 − p
x−p
x0 − p (p−q)αt
=
e
x−q
x0 − q
qαe(p−q)αt (p − x0 ) − p(q − x0 )
x=
αe(p−q)αt (p − x0 ) − (q − x0 )
Problem 1 on page 40
Answer:
1.By the graphics of direction field,the solutions will go to infinity for
large t.
Multiply e4t on both sides of the ODE.
e4t y 0 + 4e4t y = te4t + e2t
⇒
(e4t y)0
⇒
e4t y
⇒
y
= te4t + e2t
1
1
1
= e2t + te4t − e4t + C
2
4
16
1 −2t 1
1
= e + t−
+ Ce−4t .
2
4
16
Thus by the general solution, y → ∞ when t → ∞.
9
y ’ = − 4 y + t + exp( − 2 t)
4
3
y
2
1
0
−1
−2
−2
0
2
4
t
6
8
10
Figure 3. for problem 1 on page 40
y 0 + 4y = t + 2e−2t
3.By the graphics of direction field, the solutions will go to 2 for large
t.
Mulitply et on both sides of the ODE:
et y 0 + et y = t + 2et
⇒
(et y)0
⇒
et y
⇒
y
= t + 2et
1
= t2 + 2et + C
2
1 2 −t
= t e + 2 + Ce−t .
2
Thus by the general solution, y → 2 when t → ∞.
10
y ’ = − y + 2 + t exp( − t)
4
3
y
2
1
0
−1
−2
−2
0
2
4
t
6
8
10
Figure 4. for problem 3 on page 40
y 0 + y = 2 + te−t
5.By the graphics of the direction field, the solutions will go to +∞ for
large t if the initial value if large enough,while −∞ otherwise. Multiply
x−et on both sides of the ODE:
e−3t y 0 − 3e−3t y
= 4e−2t
(e−3t y)0
= 4e−2t
⇒
⇒
e−3t y − e−3t0 y(t0 ) = −2e−2t + 2e−2t0
⇒
y
= −2et + (e−3t0 y(t0 ) + 2e−2t0 )e3t
Thus by the general solution,y → ∞ if y(t0 ) > −2et0 ,y → −∞ if
y(t0 ) ≤ −2et0
Problem 13 on page 40
11
y ’ = 3 y + 4 exp(t)
2
1.5
1
y
0.5
0
−0.5
−1
−1.5
−2
−5
−4
−3
−2
−1
t
0
1
2
Figure 5. for problem 5 on page 40
y 0 − 3y = 4et
Answer: Multiply e−t on both sides:
e−t y 0 − ye−t = 4tet
⇒
(e−t y)0
⇒
e−t y − 1
⇒
y
Problem 20 on page 40
Answer:
= 4tet
= 4tet − 4et + 4
= 4te2t − 4e2t + 5et
3
12
Multiply et on both sides
tet y 0 + (t + 1)et y = tet
= tet
⇒
(tet y)0
⇒
tet y − 2 ln 2
⇒
y
= tet − 2 ln 2 − et + 2
1
2
=1− + t
t te
Problem 31 on page 41
Answer: Multiply e−2t on both sides:
(e−2t y)0
⇒
e−2t y − y0
⇒
y
= 3te−2t + 2e−t
3
11
3
− 2e−t
= − te−2t − e−2t +
2
4
4
3
3
11
= − t − − 2et + ( + y0 )e2t
2
4
4
11
The critical value is y0 = − ;
4
11
When y0 = − ,y → −∞ as t → ∞.
4
Problem 39 on page 42
Answer:
Z
A (t) = t e exp( −4dt) = t2
0
2 4t
1
A(t) = t3 + C
3
1 3 4t
y
= t e + Ce4t
3
13
Problem 41 on page 41
Answer:
Z
2
2
A (t) = sin texp(
dt) = 2t sin t
t
t
A(t) = −2t cos t + 2 sin t + C
0
= (C − 2t cos t + 2 sin t)/t2
y
Problem 25 on page 49
Answer:
(3 + 2y)dy = 2 cos 2xdx
3y + y 2 + 2 = sin 2x
Along with the intial value
y=
When x = kπ +
−3 +
√
1 + 4 sin 2x
2
π
, k ∈ Z, y attains its maximum.
4
Problem 27 on page 49
Answer:
dy
t
=
dt
y(4 − y)
3
1
y
1
y0
1 2
ln |
| − ln |
| =
t
4
4−y
4
4 − y0
6
14
Along with the intial condition:
y =4−
4
(2/3)t2
e
y0
4−y0
+1
(a)
If y0 < 0,y → −∞ rapidly as t increases;
If y0 > 0,y → 4 as t increases;
If y0 = 0,y = 0.
(b)
4
y = 4−
2 2
4 + e3t
p
T =
1.5 ln(199 × 7)
∼
= 3.29527
Problem3 on page 76
Answer:
π
π
, 2kπ + ), sin t continous on the real
2
2
line, so the max interval including the initial point where the soulution
3 5
exist is ( π, π).
2 2
tan t continuous on (2kπ −
Problem 5 on page 76
Answer:
Consider the max interval including 1 where
2t
3t2
and
are
16 − t2
16 − t2
continuous.i.e. (−4, 4).
Problem 7 on page 76
Answer: f (t, y) continuous on R2 \{(t, y) : 2t + 5y = 0};
15
∂f
∂y
continuous on R2 \{(t, y) : 2t + 5y = 0};
Hence the region is R2 \{(t, y) : 2t + 5y = 0}
Problem 10 on page 76
Answer:
Everywhere.
Prohlem 15 on page 76
Answer: If y0 = 0, y ≡ 0,otherwise:
dy
= −dt
y5
1
1 1
− ( 4 − ) = −t
4 y
y0
y =
y0
(1 + y0 4 t)1/4
Thus the interval is {t > − y10 4 }
Problem 22 on page 77
Answer:
(a)Trival.
(b)Since the RHS of the ODE is not continuous on the whole t − y
plane.
(c)
y0
=c
−t + |t + 2c|
RHS =
= 2 (t ≥ −2)
2
Hence y 0 = RHS.
There is no constant c s.t. y become a second order linear function of
t.
Problem 23 on page 77
(a) Show that φ(t) = exp3t is a solution of y 0 −3y = 0 and that y = cφ(t)
16
is also a solution of this equation for any value of the constant c.
(b)show that φ(t) = 1/t is a solution of y 0 + y 2 = 0 for t > 0 but that
y = cφ(t) is not a solution of this equation unless c = 0 or c = 1.
Answer:
(a) Let y = exp3t , then y 0 = 2 exp2t , thus y 0 − 3y = 0.
Let y = cφ(t) = c exp3t =⇒ y 0 − 2y = 0
thus y = cφ(t) is also a solution of this equation.
(b) Let y = φ(t) = 1/t, then y 0 = − t12 , and y 0 + y 2 = 0.
Let y = cφ(t) = c/t, then y 0 = − tc2 . If we want y 0 + y 2 = − tc2 +
c2
t2
to
hold, then c = 0 or c = 1.
Problem 24 on page 77
Answer:
Suppse φ0 (t) + p(t)φ(t) = 0,then cφ0 (t) + cp(t)φ(t) = 0 for arbitrary
constant c.
Problem 25 on page 77
Answer:
(y1 (t)+y2 (t))0 +p(t)(y1 (t)+y2 (t)) = (y10 (t)+p(t)y1 (t)+(y20 (t)+p(t)y2 (t)) = 0+g(t) = g(t)
Hence y1 (t) + y2 (t) is a solution of Eq.(ii).
Problem 26 on page 77
Answer: (a) For
y 0 + p(t)y = g(t)
y(0) = y0
the solution is
Z t
1
y(t) =
[ µ(s)g(s)ds + c].
µ(t) t0
1
1 Rt
Hence y1 (t) =
, y2 =
µ(s)g(s)ds.
µ(t)
µ(t) t0
1
µ(t)0
(b) Since y1 (t) =
, then y10 (t) = −
.
µ(t)
µ(t)2
17
From µ(t)0 = µ(t)p(t), we get
y10 + p(t)y1 =
(c) From y2 =
−µ(t)0 + µ(t)p(t)
= 0.
µ(t)2
1 Rt
µ(s)g(s)ds, we obtain that
µ(t) t0
Z
p(t) t
0
y2 = g(t) −
µ(s)g(s)ds.
µ(t) t0
Then we get that
y20 + p(t)y2 = g(t).
Problem 27 on page 77
Answer:
(a)n = 0
y 0 + p(t)y = q(t)
R
e
p(t)dt
y+e
R
p(t)dt
(e
R
p(t)y = q(t)e
R
p(t)dt
R
y)0 = q(t)e p(t)dt
Z
R
R
p(t)dt
e
y =
q(t)e p(t)dt + C
Z
R
R
R
− p(t)dt
y = e
q(t)e p(t)dt + Ce− p(t)dt
p(t)dt
n=1
y 0 + (p(t) − q(t))y = 0
R
(e
(p(t)−q(t))dt
R
e
y)0 = 0
(p(t)−q(t))dt
y = C
R
y = Ce
(q(t)−p(t))dt
18
(b) Multiply y −n on both sides:
y −n y 0 + p(t)y 1−n = q(t)
1
v 0 + p(t)v = q(t)
1−n
Problem 29 on page 78
Answer: n = 2,let v = y −1 ,Then:
−v 0 − rv = −k
(ert v)0 = kert
k
v =
+ Ce−rt
r
r
y =
k + Cre−rt
Problem 2 on page 88
Answer: Since f (y) = ay + by 2 , the critical points are f (y) = 0.
a
Then we get y = 0 or y = − From the graph, we can see that
b
a
y = 0 is stable; y = − is unstable.
b
Problem 4 on page 88
Answer: Since f (y) = y(y − 2)(y − 4), the critical points are f (y) = 0.
Then we get y = 0 or y = 2 or y = 4 From the graph, we can see that
y = 0 and y = 4 are unstable; y = 2 is stable.
Problem 6 on page 88
Answer: Since f (y) = −4(arctan y)/(1 + y 2 ), the critical point is
f (y) = 0.
Then we get y = 0 or y = −
y = 0 is unstable;
a
From the graph, we can see that
b
19
Problem 8 on page 89
Answer: Since f (y) = −k(y − 2)2 ,the critical point is f (y) = 0. Then
we get y = 2. From the graph,we can see that y = 2 is semistable.
Problem 12 on page 89
Answer: Since f (y) = y 2 (9 − y 2 ), the critical points are f (y) = 0.
Then we get y = 0 or y = ±3. From the graph, we can see that
y = 0, semi-stable;
y = 3, asymptotically stable;
y = −3, unstable.
Problem 14 on page89
Answer:
Since f (y1 ) = 0, if f 0 (y1 ) < 0, then for sufficiently small , for
dy
> 0; for 0 < y − y1 < =⇒
− < y − y1 < 0 =⇒ f (y) > 0 =⇒
dt
dy
f (y) < 0 =⇒
< 0.
dt
Then we obtain that φ(t) = y1 is asymptotically stable if f 0 (y1 ) < 0.
Use the same method, we can get that if f 0 (y1 ) > 0, then φ(t) = y1 is
unstable.
Problem 16 on page 90
K
,critical points are f (y) = 0.Then
y
the critical points are y = 0,y = K.From the graph,y = 0 is unstable,y =
Answer: (a)Since f (y) = ry ln
K is stable.
(b)
f 0 (y) = r(ln
K
− 1)
y
20
Thus y concave up in (0,
K
K
),concave down in ( , e)
e
e
(c)Suppose not,then:
K
1
< ry(1 − y)
y
K
1
K
y < 1 − ln
K
y
y
y < K ln(e ) < K
K
ry ln
Since when y is in (0, K],y will finally approach K for large t.Thus
contradiction.
Answer: Problem 1 on page 101
My = 0 = Nx ,hence it’s exact;
d(2x2 + 3x + 3y 2 − y) = 0
2x2 + 3x + 3y 2 − y = C
Problem 4 on page 101
Answer: My = 8xy + 4 = Nx hence it’s exact.
d(2x2 y 2 + 4xy) = 0
2x2 y 2 + 4xy = C
Problem 11 on page 101
x
y
Answer: My = + x, Nx = + y,My 6= Nx ,hence it’s not exact.
y
x
Problem 19 on page 101
Answer: My = 6x2 y 2 ,Nx = 1 + y 2 ,My 6= Nx ,hence it’s not exact.
21
(M µ)y = 0 = (N µ)x ,hence it’s exac now.
2xdx +
1 + y2
dy = 0
y3
1 1
+ ln |y|) = 0
2 y2
1 1
x2 −
+ ln |y| = C
2 y2
d(x2 −
Problem 22 on page 101
Answer: My = (x + 2) cos y,Nx = cos y,My 6= Nx so it’s not exact.
(M µ)y = xex (x + 2) cos y = (N µ)x ,hence it’s exact now.
d(sin yx2 ex ) = 0
x2 ex sin y = C
Problem 24 on page 101
Answer:
M µ(xy)dx + N µ(xy)dy = 0
(M µ(xy))y = My µ(xy) + xM µ0
(N µ(xy))x = Nx µ(xy) + yN µ0
(Nx − My )µ(xy) + (yN − xM )µ0 = (xM − yN )µ(xy)R(xy) + (yN − xM )µ0
= (xM − yN )(µ(xy)R(xy) − µ0 )
= 0
µ0 = µR(xy)
µ = eR(xy)
22
Problem 27 on page 101
Answer: Multiply y on both sides:
ydx + (x − y cos y)dy = 0
d(xy − y sin y − cos y) = 0
xy − y sin y − cos y = C
Problem 2 on page120: dy/dt = 2t2 + y 2 , y(1) = 2
Answer: Set y1 (t) = y(t + 1) − 2, then y(t) = y1 (t − 1) + 2. We obtain,
dy1 (t − 1)
= 2t2 + (y1 (t − 1) + 2)2
dt
=⇒ dy1 /dt = 2(t + 1)2 + (y1 (t) + 2)2
So the given problem are equivalent to the new problem: dy/dt =
2(t + 1)2 + (y + 2)2 , y(0) = 0
Problem 7 on page120
Let φ0 (t) = 0 and use the method of successive approximations to solve
the given initial value problem. (a) Determine φn (t) for an arbitrary
value of n. (b) Plot φn (t) for n = 1, · · ·, 4. Observe whether the iterates
appear to be converging.
0
y = ty + 1, y(0) = 0
Answer: If y = φ(t), then the corresponding integral equation is
Z
φ(t) =
t
(sφ(s) + 1)ds
0
23
If the initial approximation is φ0 (t) = 0, then
Z t
φ1 (t) =
1ds = t
0
Z t
1 3
φ2 (t) =
(s2 + 1)ds =
t +t
1·3
0
Z t
1
t5
t3
(s( s3 + s) + 1)ds =
+
+t
φ3 (t) =
3
1·3·5 1·3
0
Z t
n
X
t2k−1
(sφn−1 (s) + 1)ds =
φn (t) =
1 · 3 · 5 · · · (2k − 1)
0
k=1
By D’Alembert’s criterion,the series converges.
Problem 9 on page 120
Let φ0 (t) = 0 and use the method of successive approximations to solve
the given initial value problem.
(a) Calculate φ1 (t), · · · , φ3 (t).
(b) Plot φ1 (t), · · · , φ3 (t), and observe whether the iterates appear to
be converging.
y 0 = 2t2 + y 2 , y(0) = 0.
Rt
Answer: Since φ(t) = 0 (2s2 + φ(s)2 )ds, then
Z t
2
φ1 (t) =
(2s2 )ds = t3 ,
3
0
Z t
4
2
4 7
φ2 (t) =
(2s2 + s6 )ds = t3 +
t,
9
3
9·7
0
Z t
2
4 7 2
2
4 7
16
16
φ3 (t) =
(2s2 +( s3 +
s ) )ds = t3 +
t+
t11 +
t15 .
2 · 15
3
9
·
7
3
9
·
7
3
·
7
·
9
·
11
(7
·
9)
0
Problem 19 on page 120
In this problem, we deal with the question of uniqueness of the solution
of the integral equation (3),
Z
φ(t) =
t
f (s, φ(s))ds
0
24
(a) suppose that φ and ψ are two solutions of Eq.(3). Show that for
t ≥ 0,
t
Z
{f (s, φ(s)) − f (s, ψ(s))}ds.
φ(t) − ψ(t) =
0
(b) show that
Z
t
|f (s, φ(s)) − f (s, ψ(s))|ds.
|φ(t) − ψ(t)| ≤
0
(c) use the result of problem 15 to show that
Z t
|φ(t) − ψ(t)| ≤ K
|φ(s)) − ψ(s)|ds,
0
∂f
where K is an upper bound for | | in D. Answer: (a) Since φ(t) =
Rt
R t ∂y
f (s, φ(s))ds, and ψ(t) = 0 f (s, ψ(s))ds, then we get
0
Z t
{f (s, φ(s)) − f (s, ψ(s))}ds.
φ(t) − ψ(t) =
0
(b) Since |φ(t)−ψ(t)| = |
Rt
0
(f (s, φ(s))−f (s, ψ(s)))ds| ≤
Rt
0
|f (s, φ(s))−
f (s, ψ(s))|ds.
∂f
∂f
(c) Since
is continuous ⇒ ∃K, such that | | ≤ K. Hence, we can
∂y
∂y
show that
Z
t
|φ(s)) − ψ(s)|ds.
|φ(t) − ψ(t)| ≤ K
0