MATH 215/255
Fall 2015
Assignment 1
Due date: September 25, 2015
§1.1, §1.2, §1.3, §1.4
dy
= x2 + x for y(1) = 3.
dx
Rewrite the DE as
1. (1.1.2) Solve
Answer.
dy = (x2 + x)dx.
Integrating,
1
1
y = x3 + x2 + c.
3
2
Using y(1) = 3, we get
3=
thus c =
13
6 ,
and hence
2. (1.1.7) Solve
Answer.
1 1
+ + c,
3 2
1
1
13
y = x3 + x2 + .
3
2
6
dy
1
=
for y(0) = 0.
dx
y+1
Rewrite the DE as
(y + 1)dy = dx.
Integrating,
1 2
y + y = x + c.
2
Using y(0) = 0, we get
0 + 0 = 0 + c,
thus c = 0, and 12 y 2 + y = x. Completing square,
(y + 1)2 = 2x + 1.
Thus
√
y + 1 = ± 2x + 1.
Since (x, y) is near (0, 0), we take positive sign, and
√
y = −1 + 2x + 1.
Remark. The maximal interval of existence is − 12 < x < ∞.
Exercise 1.2.3: Sketch slope field for y0 = y2 .
Exercise 1.2.4: Is it possible to solve the equation y0 =
xy
cos x
p
for y(0) = 1? Justify.
Exercise 1.2.5: Is it possible to solve the equation y0 = y |x| for y(0) = 0? Is the solution unique?
Justify.
0 = 1 − x, y 0 0= x − 2y, y 00 = x(1 − y) to
3. (1.2.6)
Match
equations
Exercise
1.2.6:
Matchyequations
y = 1 x, y = x 2y, y0 =slope
x(1 fields.
y) toJustify.
slope fields. Justify.
a)
b)
c)
Exercise 1.2.7 (challenging):
Take y0 = f (x, y), y(0) = 0, where f (x, y) > 1 for all x and y. If the
(b) is for y 0 = 1 − x, since the slope field is independent of y.
solution exists for all x, can you say what happens to y(x) as x goes to positive infinity? Explain.
Answer.
(c) is for y 0 = x − 2y, since the slopes are the same
on every level set x − 2y = c.
0
Exercise 1.2.8 (challenging): Take (y x)y = 0, y(0) = 0. a) Find two distinct solutions. b) Explain
(a) is for y 0 = x(1 − y) since it is the remaining case, and the picture is symmetric
why this does not violate Picard’s theorem.
with respect to the center (0, 1).
Exercise 1.2.101: Sketch the slope field of y0 = y3 . Can you visually find the solution that satisfies
4. (§1.2)
y(0) Solve
= 0? the problem
0
3
y + y = 0,
y(0) = y0 ,
Exercise 1.2.102: Is it possible to solve y0 = xy for y(0) = 0? Is the solution unique?
and determine how the interval in which the0 solution
exists depends on the initial
x
Exercise
1.2.103:
Is
it
possible
to
solve
y
=
for
y(1) = 0?
2
value y0 .
x 1
0
Exercise 1.2.104:
equations
y0 0=exists
sin x,for
y0 all
= cos
Answer.
If y0 = 0,Match
the solution
y(t) =
t. y, y = y cos(x) to slope fields. Justify.
If y0 6= 0, the equation is separable, with dy/y 3 = −dt. Integrating both sides and
1
1
invoking the initial condition, − y −2 = −t− y0−2 and thus y −2 = 2t+y0−2 . If y0 > 0,
2
a)
b)2
c)
1
then y(t) > 0 for all t in the domain. So y(t) = q
for t > − 12 y0−2 . If y0 < 0,
−2
2t + y0
1
for t > − 12 y0−2 .
then y(t) < 0 for all t in the domain. So y(t) = − q
−2
2t + y0
5. (§1.2) (a) Verify that both y1 (t) = 1 − t and y2 (t) = −t2 /4 are solutions of the initial
value problem
−t + (t2 + 4y)1/2
y0 =
, y(2) = −1.
2
Where are these solutions valid?
(b) Explain why the existence of two solutions of the given problem does not contradict
the uniqueness part of Theorem 1.2.1.
Answer.
side)
(a) For y = y1 (t) = 1 − t, y(2) = −1, y 0 = −1 while (RHS = right hand
1
1
RHS(DE) = (−t + (t2 + 4 − 4t)1/2 ) = (−t + |t − 2|)
2
2
which is −1 if t ≥ 2 and 1 − t if t < 2. Thus y1 satisfies the DE for t ≥ 2.
For y = y2 (t) = −t2 /4, y(2) = −1, y 0 = −t/2 while
1
RHS(DE) = (−t + (t2 − t2 )1/2 ) = −t/2
2
which is y 0 . Thus y2 satisfies the DE for all t ∈ R.
2
(b) By Theorem 2.4.2 we are guaranteed a unique solution only where f (t, y) =
1
2
1/2 ) and f (t, y) = (t2 + 4y)−1/2 are continuous. In this case the initial
y
2 (−t + (t + 4y)
point (2, −1) lies in the region t2 + 4y ≤ 0 so fy is not continuous and hence the
theorem is not applicable and there is no contradiction.
6. (§1.3) Solve y 0 + y 2 sin x = 0.
Answer.
If y(x0 ) = 0 for some x0 , then y(x) = 0 for all x. If y 6= 0, the DE may
be written as y −2 dy = − sin x dx. Integrating both sides, we get −y −1 = cos x + c, or
y = −1/(cos x + c).
Note: For any given initial data y(x0 ) = y0 6= 0, we have cos x + c 6= 0 for x near x0 .
The interval of existence is finite if |c| ≤ 1.
7. (§1.3) Consider
y 0 = (1 − 2x)/y,
y(1) = −2.
(a) Find the solution in explicit form.
(b) Plot the graph of the solution.
(c) Determine the interval in which the solution is defined.
Answer.
(a) Rewrite as ydy = (1 − 2x)dx. √Integrating we get y 2 /2 = x − x2 + c.
Using y(1) = −2 we get c = 2. Thus y(x) = − 4 + 2x − 2x2 , where the negative sign
is due to y(1) < 0.
(b) Answer.
(c) The solution is defined if 4+2x−2x2 > 0. Thus the interval of existence is (−1, 2).
8. (§1.3) Solve (an implicit solution of)
y 0 = (1 + 3x2 )/(3y 2 − 6y),
y(0) = 1.
Answer. We have (3y 2 − 6y)dy = (1 + 3x2 )dx so that y 3 − 3y 2 = x + x3 + c. Using
y(0) = 1 we get c = −2, and
y 3 − 3y 2 = x + x3 − 2,
which defines y as an implicit function of x near (x, y) = (0, 1).
3
9. (§1.4) Solve y 0 − y = 2te2t ,
Answer.
y(0) = 1.
µ(t) = e−t so that (e−t y)0 = 2tet and thus
Z
Z
−t
t
t
e y(t) = 2te dt = 2te − 2 et dt = 2(t − 1)et + c.
Setting t = 0 and y = 1 we get 1 = −2 + c, or c = 3. Hence y(t) = 2(t − 1)e2t + 3et .
10. (§1.4) Solve ty 0 + 2y = t2 − t + 1,
y(1) = 21 ,
t > 0.
0
Answer. Rewrite
with p(t) = 2/t and g(t) = t − 1 + 1/t. Take
R 2dt as 2y + p(t)y =2 g(t)
0
µ(t) = exp( t ) = t so that (t y) = t3 − t2 + t. Integrating and dividing by t2
gives y = t2 /4 − t/3 + 1/2 + c/t2 . Setting t = 1 and y = 1/2 we have c = 1/12. Thus
1
y = t2 /4 − t/3 + 1/2 + 12t
2.
11. (§1.4) Find the value of y0 for which the solution of the initial value problem
y 0 − y = 1 + 3 sin t,
y(0) = y0
remains finite as t → ∞ (i.e., uniformly bounded by some constant for all t).
Answer.
(e−t y)0 = e−t + 3e−t sin t so e−t y(t) = −e−t − 23 e−t (sin t + cos t) + c, or
3
y(t) = −1 − (sin t + cos t) + cet .
2
Setting t = 0 we get y0 = −1 − 3/2 + c. If y(t) is to remain bounded as t → ∞, we
must have c = 0 and thus y0 = −5/2.
12. (§1.4) A tank initially contains 120 L of pure water. A mixture containing a concentration of γ g/L of salt enters the tank at a rate of 2 L/min, and the well-stirred
mixture leaves the tank at the same rate. Find an expression in terms of γ for the
amount of salt in the tank at any time t. Also find the limiting amount of salt in the
tank as t → ∞.
Answer. Let S(t) be the amount of salt at time t. Then S(0) = 0. We have rate in
= 2γ and rate out = 2(S/120), thus
dS/dt = 2γ − S/60,
S(0) = 0.
This is a linear equation and we can take integrating factor et/60 , resulting in (et/60 S)0 =
2γet/60 . Integrating and dividing by et/60 , we get S(t) = 120γ + ce−t/60 . The condition S(0) = 0 gives c = −120γ, so S(t) = 120γ(1 − e−t/60 ) and hence S(t) → 120γ as
t → ∞.
4