MATH 215/255
Fall 2014
Assignment 1
§1.1, §1.2, §1.3
Solutions to selected exercises can be found in [Lebl], starting from page 303.
• 1.1.3: Solve
dy
= sin(5x), for y(0) = 2.
dx
Answer.
Z
y=
1
sin(5x)dx = − cos(5x) + c.
5
Using y(0) = 2, we have 2 = − 15 + c and hence c =
11
5 .
Thus
11
1
y(x) = − cos(5x) + .
5
5
• 1.1.4: Solve
Answer.
(1)
(2)
dy
1
= 2
, for y(0) = 0.
dx
x −1
By partial fraction,
1
1/2
1/2
=
−
.
x2 − 1
x−1 x+1
Thus
Z y=
1/2
1/2
−
x−1 x+1
dx =
1
1
ln |x − 1| − ln |x + 1| + c.
2
2
(3)
(4)
Since our solution is for x near 0, we have
y=
1
1
ln(1 − x) − ln(x + 1) + c.
2
2
(5)
Using y(0) = 0, we have 0 = 0 − 0 + c and hence c = 0. Thus
y(x) =
1
1
ln(1 − x) − ln(x + 1).
2
2
Remark. It exists for −1 < x < 1.
• 1.1.6: Solve y 0 = (y − 1)(y + 1) for y(0) = 3.
Answer.
Method 1: Consider
⇒
1
dy
=1
(y − 1)(y + 1) dx
1 1
1
dy
(
−
)
= 1.
2 y − 1 y + 1 dx
Integrating this equation, we obtain
1
1
ln |y − 1| − ln |y + 1| = x + c1 .
2
2
(6)
Using y(0) = 3, we have
is for y near 3, we have
1
2
ln 2 − 21 ln 4 = c1 and hence c1 = − 12 ln 2. Since our solution
x=
1
1
1
ln(y − 1) − ln(y + 1) + ln 2.
2
2
2
(7)
2 + e2x
;
2 − e2x
(8)
Therefore,
y=
see below.
Method 2: By considering x as a function of y (inverse function) and integrating
dx
1
dy = (y−1)(y+1) , we have
Z
1
x=
dy
(9)
(y − 1)(y + 1)
By the previous problem with x and y switched,
x=
1
1
ln |y − 1| − ln |y + 1| + c.
2
2
(10)
Since our solution is for y near 3, we have
x=
Using y(0) = 3, we have 0 =
x=
1
2
1
1
ln(y − 1) − ln(y + 1) + c.
2
2
ln 2 − 12 ln 4 + c and hence c =
(11)
1
2
ln 2. Thus
1
1
1
ln(y − 1) − ln(y + 1) + ln 2
2
2
2
(12)
which defines y as an implicit function of x.
We can actually solve y(x) explicitly:
2x = ln(y − 1) − ln(y + 1) + ln 2 = ln
e2x =
and hence
y=
2(y − 1)
y+1
2 + e2x
.
2 − e2x
1
, x(1) = 1.
x2
Denote the independent variable by t.
• 1.1.102: Solve x0 =
Answer.
Method 1: Consider
x2 ·
Then
dx
= 1.
dt
d [x(t)]3
(
) = 1.
dt
3
2
2(y − 1)
y+1
(13)
(14)
(15)
We integrate to get
[x(t)]3
= t + c,
3
i.e.,
x(t) = [3(t + c)]1/3 .
From x(1) = 1, we obtain c = − 32 . The solution is thus x(t) = [3(t − 23 )]1/3 .
Method 2: By considering t as a function of x (inverse function) and integrating
dt
2
dx = x , or using that the equation is separable, we have
Z
x3
t = x2 dx =
+ c.
(16)
3
Using x(1) = 1, we have 1 =
1
3
+ c and hence c = 23 . Thus x = [3(t − 23 )]1/3
• 1.2.2: Sketch slope field for y 0 = x2 .
• 1.2.3: Sketch slope field for y 0 = y 2 .
p
• 1.2.5: Is it possible to solve the equation y 0 = y |x| for y(0) = 0? Is the solution
unique? Justify.
p
Answer. Note that f (x, y) = y |x| is defined and continuous for (x, y) near (0, 0).
Moreover,
p
∂f
= |x|
(17)
∂y
is also defined and continuous for (x, y) near (0, 0). By Picard’s theorem, there is a
solution near (0, 0) and it is unique.
∂f
Remark.
is not defined at x = 0 but it does not matter.
∂x
3
• 1.2.103: Is it possible to solve y 0 =
Answer.
No, because f (x, y) =
x2
x2
x
for y(1) = 0?
−1
x
is not defined at x = 1.
−1
dx
= (x2 − 1)t, for x(0) = 0.
dt
The equation is separable.
• 1.3.3: Solve
Answer.
Formulation 1: Consider
(x2
1
dx
= t,
− 1) dt
i.e.,
[
1
1 dx
−
]
= 2t,
x − 1 x + 1 dt
Then
ln |x − 1| − ln |x + 1| = t2 + c.
Since our solution is for x near 0, we have
ln(1 − x) − ln(x + 1) = t2 + c.
Using x(0) = 0, we have 0 − 0 = 0 + c and hence c = 0. Thus
t2 = ln(1 − x) − ln(x + 1) = ln
2
et =
and hence
1−x
,
1+x
1−x
,
1+x
2
x=
Formulation 2: Consider
1 − et
.
1 + et2
dx
= tdt.
−1
(18)
x2
Integrating and using Exercise 1.1.4,
1
1
1
ln |x − 1| − ln |x + 1| = t2 + c.
2
2
2
Since our solution is for x near 0, we have
1
1
1
ln(1 − x) − ln(x + 1) = t2 + c.
2
2
2
Using x(0) = 0, we have 0 − 0 = 0 + c and hence c = 0. Thus
t2 = ln(1 − x) − ln(x + 1) = ln
2
et =
and hence
1−x
,
1+x
1−x
,
1+x
(19)
(20)
(21)
(22)
2
1 − et
x=
.
1 + et2
4
(23)
dx
= x sin(t), for x(0) = 1.
dt
The equation is separable.
• 1.3.4: Solve
Answer.
Formulation 1:
1 dx
·
= sin(t).
x dt
Thus,
d
ln |x(t)| = sin(t),
dt
and
ln |x(t)| = − cos(t) + c.
Therefore,
|x(t)| = e− cos(t)+c .
Using x(0) = 1, we obtain c = 1. And the solution is
x(t) = e− cos(t)+1 .
Formulation 2:
Consider
dx
= sin(t)dt.
x
(24)
ln |x| = − cos(t) + c.
(25)
Integrating, we get
Since our solution is for x near 1, we have
ln x = − cos(t) + c.
(26)
Using x(0) = 1, we have 0 = −1 + c and hence c = 1. Thus
ln x = − cos(t) + 1,
(27)
x = e1−cos(t) .
(28)
and
• 1.3.102: Solve x0 = 3xt2 − 3t2 , x(0) = 2.
Answer.
The right side can be factored as 3t2 (x − 1) and hence the equation is
separable.
Formulation 1:
1
dx
·
= 3t2 .
x − 1 dt
Thus,
d
ln |x(t) − 1| = 3t2 ,
dt
and
ln |x(t) − 1| = t3 + c.
5
Therefore,
3 +c
|x(t) − 1| = et
.
Using x(0) = 2, we obtain c = 0. And the solution is
3
x(t) = et + 1.
Formulation 2: Consider
dx
= 3t2 dt.
x−1
(29)
ln |x − 1| = t3 + c.
(30)
Integrating, we get
Since our solution is for x near 2, we have
ln(x − 1) = t3 + c.
(31)
Using x(0) = 2, we have 0 = 0 + c and hence c = 0. Thus
ln(x − 1) = t3 ,
and
3
x = 1 + et .
• 1.3.103: Find an implicit solution for x0 =
Answer.
(32)
(33)
1
, x(0) = 1.
+1
3x2
The equation is separable.
Formulation 1: Consider
(3x2 + 1) ·
dx
= 1.
dt
Thus,
d 3
[x (t) + x(t)] = 1,
dt
and
x3 (t) + x(t) = t + c.
Using x(0) = 1, we have 1 + 1 = 0 + c, and thus c = 2. And the solution is
x3 (t) + x(t) = t + 2.
This defines x(t) as an implicit function of t for (t, x) near (0, 1).
Formulation 2: Consider
(3x2 + 1)dx = dt.
(34)
x3 + x = t + c.
(35)
Integrating, we get
Using x(0) = 1, we have 1 + 1 = 0 + c and hence c = 2. Thus
x3 + x = t + 2.
This defines x as an implicit function of t for (t, x) near (0, 1).
6
(36)