Math 16B: Homework 4 Solutions
Due: July 23
1. Verify that the given solution is a solution of the corresponding differential equation:
(a) y = t + 1t for ty 0 = 2t − y:
Plugging in the solution into the left side of the differential equation gives
1
0
ty = t 1 − 2
t
1
= t−
t
Plugging it into the right side gives
1
2t − y = 2t − t +
t
1
= t−
t
Since both expressions are identical, y = t +
1
t
is a solution of ty 0 = 2t − y.
(b) y = sin(t) cos(t) − cos(t) for y 0 + (tan(t))y = cos2 (t):
Plugging in the solution into the left side of the differential equation gives
y 0 + (tan(t))y = [cos(t)(cos(t)) + sin(t)(− sin(t)) + sin(t)] +
tan(t)(sin(t) cos(t) − cos(t))
= cos2 (t) − sin2 (t) + sin(t) + sin2 (t) − sin(t)
= cos2 (t)
which is the same as the right side of the differential equation. Thus, y =
sin(t) cos(t) − cos(t) is a solution of y 0 + (tan(t))y = cos2 (t).
2. Sketch the slope field for the differential equation y 0 = y(y − 1)(y + 1).
Note that:
at y = 2, the slope is y 0 = 6
at y = 1.5, the slope is y 0 = 1.875
at y = 1, the slope is y 0 = 0
at y = 0.5, the slope is y 0 = −0.375
1
at y = 0, the slope is y 0 = 0
at y = −1, the slope is y 0 = 0
at y = −1.5, the slope is y 0 = −1.875
at y = −2, the slope is y 0 = −6
The resulting slope field is shown in Figure 1.
Figure 1: Slope field for y 0 = y(y − 1)(y + 1)
3. Solve the following differential equations using the appropriate technique. In the cases
with initial conditions, find the particular solutions.
(a) y 0 = e2t+4y :
We use separation of variables. We have
dy
= e2t+4y
dt
dy
= e2t e4y
dt
Z
Z
−4y
e
dy =
e2t dt
e−4y
e2t
=
+C
−4
2
e−4y = −2e2t − 4C
−4y = ln(−2e2t − 4C)
1
y = − ln(−2e2t − 4C)
4
2
(b) xy 0 + (5x + 1)y = e−5x :
We use the integrating factor method. The standard first-order linear form of the
above differential equation is
e−5x
5x + 1
0
y=
(1)
y +
x
x
Then, a(x) =
5x+1
x
=5+
1
x
so
Z
A(x) =
Z
a(x) dx =
5+
1
dx = 5x + ln |x|.
x
Finally, the integrating factor is I(x) = eA(x) = e5x+ln |x| = e5x eln |x| = e5x |x|.
Multiply (1) throughout by I(x) to get
e−5x
5x + 1
5x
0
5x
y = (e5x |x|)
e |x|y + (e |x|)
x
x
d
ye5x |x| = ±1
dx
Z
5x
ye |x| =
±1dx
ye5x |x| = ±x + C
±x + C
y =
e5x |x|
(c) ty 0 = t2 sin(t) + y, t > 0:
We use the integrating factor method. The standard first-order linear form of the
above differential equation is
1
0
y −
y = t sin(t)
(2)
t
Then, a(t) =
−1
t
so
Z
A(t) =
a(t) dt =
−1
dt = − ln(t).
t
Note that because t > 0, we do not need the absolute values inside the logarithm
above. Finally, the integrating factor is I(t) = eA(t) = e− ln(t) = 1t . Multiply (2)
throughout by I(t) to get
1 0 1
y − y
t t2
d 1
y
dt t
y
t
y
t
y
3
= sin(t)
= sin(t)
Z
=
sin(t) dt
= − cos(t) + C
= −t cos(t) + Ct
(d) x2 y 0 = y − xy with initial condition y(−1) = −1:
Solution 1: We use separation of variables. We have
dy
= y(1 − x)
dx
Z
Z
1−x
1
dy =
dx
y
x2
Z
Z
1
1
1
dy =
− dx
2
y
x
x
−1
ln |y| =
− ln |x| + C
x
−1
|y| = e x −ln |x|+C
x2
1
|y| = eC e− x e− ln |x|
−1
C
− x1
y = (±e )(e )
|x|
1
Ae− x
y = −
|x|
where A = ±eC is a real constant. Using the initial condition y(−1) = −1, we
have
1
Ae− (−1)
−1 = −
| − 1|
1
Ae
1 =
1
1
A =
= e−1 .
e
Thus, the specific solution of this initial value problem is
1
1
e−1 e− x
e−1− x
y=−
=−
|x|
|x|
Solution 2: We use the integrating factor method. Writing the differential equation in the standard first-order linear form,
x−1
0
y = 0.
(3)
y +
x2
Then, a(x) =
x−1
x2
so
x−1
dx
x2
Z
1
1
=
− 2 dx
x x
1
= ln |x| +
x
Z
A(x) =
4
1
1
Thus, the integrating factor is I(x) = eA(x) = eln |x|+ x = |x|e x . Multiply (3) by
I(x) to get
1
1
x−1
0
y |x|e x + |x|e x
y = 0
x2
d 1 ye x |x| = 0
dx
Z
1
ye x |x| =
0 dx
1
ye x |x| = C
1
Ce− x
y =
|x|
where C is any real constant. This is the same general solution as in Solution 1
above. Plugging in the initial condition therefore gives the same specific solution
1
e−1− x
y=−
|x|
(e) t ln(t)y 0 + y = tet , t > e, with initial condition y(e2 ) = 0:
We use the integrating factor method. The standard first order linear form of the
differential equation is
y0 +
1
et
y =
t ln(t)
ln(t)
(4)
1
Then, a(t) = t ln(t)
. To integrate this and determine A(t), we use the substitution
= 1t ⇒ du = 1t dt. Thus,
u = ln(t) ⇒ du
dt
Z
Z
1
1
dt =
du
t ln(t)
u
= ln |u|
= ln | ln(t)|
Note that because we are taking t > e, we have ln(t) > ln(e) = 1 > 0 so
actually we can do away with the absolute value bars inside the logarithm to get
A(t) = ln(ln(t)). Thus, the integrating factor is I(t) = eA(t) = eln(ln(t)) = ln(t).
Multiply (4) throughout by I(t) to get
1
y 0 ln(t) + y = et
t
d
(y ln(t)) = et
dt
Z
y ln(t) =
et dt
y ln(t) = et + C
et + C
y =
.
ln(t)
5
Finally, use the initial condition y(e2 ) = 0 to get
2
ee + C
0 =
ln(e2 )
2
ee + C
0 =
2
e2
C = −e
The specific solution therefore is
et − ee
y=
ln(t)
6
2