Ordinary Differential Equations, Spring 2015
Solutions to Homework 2
Maximal grade for HW2: 100 points
Section 1.1 2. (10 points) Draw a direction field for the differential
equation y 0 = 2y = 3. Based on the direction field, determine the behavior
of y as t → ∞ and its dependency on the initial value at t = 0.
Solution: The direction field is shown in the following figure:
4
3
2
1
−1
1
2
3
−1
We have y 0 > 0 if an only if y > 3/2, so y(x) insreases for y > 3/2 and
decreases for y < 3/2. There is also a constant solution y = 3/2. We have


+∞ if y(0) > 3/2,
lim y(x) = 3/2 if y(0) = 3/2,
t→∞


−∞ if y(0) < 3/2.
Section 2.1 Solve the initial value problems:
19. (15 points)
t3 y 0 + 4t2 y = e−t , y(−1) = 0.
Solution: Let us divide the equation by t3 and multiply by the integrating factor µ(t):
µ(t)y 0 + 4µ(t)t−1 y = µ(t)t−3 e−t .
1
The left hand side is the exact derivative of (µ(t)y) if
µ0 (t) = 4µ(t)t−1 ,
4
µ0 (t)
= (ln |µ(t)|)0 = ,
µ(t)
t
ln |µ(t)| = 4 ln |t| + C ⇒ µ(t) = ±e4 ln |t|+C = At4 .
We can pick A = 1, so
t4 y 0 + 4t3 y = te−t , (t4 y)0 = te−t .
Therefore:
Z
Z
Z
4
−t
−t
−t
t y = te dt = − td(e ) = −te + e−t dt = −te−t − e−t + C,
and the general solution has the form:
y(t) = −t−3 e−t − t−4 e−t + Ct−4 .
Since y(−1) = e − e + C = C, we have C = 0 and
y(t) = −t−3 e−t − t−4 e−t .
20. (15 points)
ty 0 + (t + 1)y = t, y(ln 2) = 1.
Solution: Let us divide the equation by t and multiply by the integrating
factor µ(t):
1
µ(t)y 0 + µ(t)(1 + )y = µ(t).
t
The left hand side is the exact derivative of (µ(t)y) if
1 µ0 (t)
1
µ0 (t) = µ(t)(1 + ),
= (ln |µ(t)|)0 = (1 + ),
t
µ(t)
t
ln |µ(t)| = t + ln |t| + C ⇒ µ(t) = ±et+ln |t|+C = Atet .
We can pick A = 1, so
tet y 0 + (t + 1)et y = tet , (tet y)0 = tet .
2
Therefore:
Z
t
te y =
Z
t
t
t
Z
td(e ) = te −
te dt =
et dt = tet − et + C,
and the general solution has the form:
y(t) = 1 −
Since y(ln 2) = 1 −
1
ln 2
+
C
2 ln 2
C
1
+ t.
t te
= 1, we have C = 2 and
y(t) = 1 −
2
1
+ t.
t te
30. (20 points) 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 → ∞.
Solution: If µ(t) is the integrating factor then µ0 (t) = −µ(t), so µ(t) =
e−t . We get
(e−t y(t))0 = e−t y 0 − e−t y = (1 + 3 sin t)e−t ,
Z
Z
−t
−t
−t
e y(t) = (1 + 3 sin t)e dt = −e + 3 sin te−t dt
R
Let us compute the integral I = sin te−t dt using integration by parts twice:
Z
Z
Z
−t
−t
−t
I = sin te dt = − sin td(e ) = − sin te + e−t cos tdt =
−t
− sin te
Z
−
−t
−t
cos td(e ) = − sin te
− cos te
−t
Z
−
e−t sin tdt =
− sin te−t − cos te−t − I.
Therefore
1
2I = − sin te−t − cos te−t ⇒ I = − (sin t + cos t)e−t ,
2
so
3
e−t y(t) = −e−t + 3I + C = −e−t − (sin t + cos t)e−t + C.
2
3
Finally,
3
y(t) = −1 − (sin t + cos t) + Cet .
2
This function is bounded for t → ∞ if and only if C = 0, hence
y0 = y(0) = −5/2.
32. (20 points) Show that all solutions of 2y 0 + ty = 2 approach a limit
at t → ∞, and find the limiting value.
Solution: If we divide the equation by 2, we get y 0 + 2t y = 1. The integratR
2
ing factor µ(t) satisfies the equation µ0 (t) = 2t µ(t), so ln µ(t) = 2t dt = t4 +C
t2
and µ(t) = Ae 4 . We can choose A = 1 and get
Z 2
t2
t2
t2
t
0
(e 4 y(t)) = e 4 , e 4 y(t) = e 4 dt + C,
hence
2
− t4
y(t) = e
Z
t2
t2
e 4 dt + Ce− 4 .
(1)
See equations (41)-(47) in the textbook for more detailed explanations.
The second term in (1) tends to 0 when t → ∞, let us find the limit of
the first term using L’Hôpital’s rule:
R t2 0
R t2
e 4 dt
e 4 dt
=
lim
lim
t2 0 =
t2
t→∞
t→∞
e4
e4
t2
lim
e4
t→∞ t
e
2
t2
4
2
= 0.
t→∞ t
= lim
Therefore for all solutions y(t) we have limt→∞ y(t) = 0.
33. (20 points) Show that if a and λ are positive constants, and b is any
real number then every solution of the equation
y 0 + ay = be−λt
has the property that y → 0 as t → ∞.
4
Solution: The integrating factor µ(t) satisfies the equaltion µ0 (t) =
aµ(t), so we can choose µ(t) = eat . Therefore
(eat y)0 = eat y 0 + aeat y = be(a−λ)t .
1) If a 6= λ, we have
at
Z
e y=b
e(a−λ)t dt =
y(t) =
b (a−λ)t
e
+ C,
a−λ
b −λt
e + Ce−at .
a−λ
Since for a, λ > 0 both e−λt and e−at approach 0 when t → ∞, we get
limt→∞ y(t) = 0.
2) If a = λ then a − λ = 0 and
Z
at
e y = b 1dt = bt + C,
so
t
+ Ce−at .
eat
Again, both terms approach 0 as t → ∞ (for the first term this follows from
the L’Hôpital’s rule).
y(t) = b
5