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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′ = 2t− y:
Plugging in the solution into the left side of the differential equation gives
ty′ = t
(1− 1
t2
)= t− 1
t
Plugging it into the right side gives
2t− y = 2t−(t +
1
t
)= t− 1
t
Since both expressions are identical, y = t + 1t
is a solution of ty′ = 2t− y.
(b) y = sin(t) cos(t)− cos(t) for y′ + (tan(t))y = cos2(t):
Plugging in the solution into the left side of the differential equation gives
y′ + (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′ + (tan(t))y = cos2(t).
2. Sketch the slope field for the differential equation y′ = y(y − 1)(y + 1).
Note that:at y = 2, the slope is y′ = 6
at y = 1.5, the slope is y′ = 1.875
at y = 1, the slope is y′ = 0
at y = 0.5, the slope is y′ = −0.375
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at y = 0, the slope is y′ = 0
at y = −1, the slope is y′ = 0
at y = −1.5, the slope is y′ = −1.875
at y = −2, the slope is y′ = −6
The resulting slope field is shown in Figure 1.
Figure 1: Slope field for y′ = y(y − 1)(y + 1)
3. Solve the following differential equations using the appropriate technique. In the caseswith initial conditions, find the particular solutions.
(a) y′ = e2t+4y:
We use separation of variables. We have
dy
dt= e2t+4y
dy
dt= e2te4y∫
e−4y dy =
∫e2t dt
e−4y
−4=
e2t
2+ C
e−4y = −2e2t − 4C
−4y = ln(−2e2t − 4C)
y = −1
4ln(−2e2t − 4C)
2
(b) xy′ + (5x + 1)y = e−5x:
We use the integrating factor method. The standard first-order linear form of theabove differential equation is
y′ +
(5x + 1
x
)y =
e−5x
x(1)
Then, a(x) = 5x+1x
= 5 + 1x
so
A(x) =
∫a(x) dx =
∫5 +
1
xdx = 5x + ln |x|.
Finally, the integrating factor is I(x) = eA(x) = e5x+ln |x| = e5xeln |x| = e5x|x|.Multiply (1) throughout by I(x) to get
e5x|x|y′ + (e5x|x|)(
5x + 1
x
)y = (e5x|x|)e
−5x
x
d
dx
(ye5x|x|
)= ±1
ye5x|x| =
∫±1dx
ye5x|x| = ±x + C
y =±x + C
e5x|x|
(c) ty′ = t2 sin(t) + y, t > 0:
We use the integrating factor method. The standard first-order linear form of theabove differential equation is
y′ −(
1
t
)y = t sin(t) (2)
Then, a(t) = −1t
so
A(t) =
∫a(t) dt =
−1
tdt = − ln(t).
Note that because t > 0, we do not need the absolute values inside the logarithmabove. Finally, the integrating factor is I(t) = eA(t) = e− ln(t) = 1
t. Multiply (2)
throughout by I(t) to get
1
ty′ − 1
t2y = sin(t)
d
dt
(1
ty
)= sin(t)
y
t=
∫sin(t) dt
y
t= − cos(t) + C
y = −t cos(t) + Ct
3
(d) x2y′ = y − xy with initial condition y(−1) = −1:
Solution 1: We use separation of variables. We have
x2 dy
dx= y(1− x)∫
1
ydy =
∫1− x
x2dx∫
1
ydy =
∫1
x2− 1
xdx
ln |y| =−1
x− ln |x|+ C
|y| = e−1x−ln |x|+C
|y| = eCe−1x e− ln |x|
y = (±eC)(e−1x )
(−1
|x|
)y = −Ae−
1x
|x|
where A = ±eC is a real constant. Using the initial condition y(−1) = −1, wehave
−1 = −Ae−1
(−1)
| − 1|
1 =Ae1
1
A =1
e= e−1.
Thus, the specific solution of this initial value problem is
y = −e−1e−1x
|x|= −e−1−
1x
|x|
Solution 2: We use the integrating factor method. Writing the differential equa-tion in the standard first-order linear form,
y′ +
(x− 1
x2
)y = 0. (3)
Then, a(x) = x−1x2 so
A(x) =
∫x− 1
x2dx
=
∫1
x− 1
x2dx
= ln |x|+ 1
x
4
Thus, the integrating factor is I(x) = eA(x) = eln |x|+1x = |x|e 1
x . Multiply (3) byI(x) to get
y′|x|e1x + |x|e
1x
(x− 1
x2
)y = 0
d
dx
(ye
1x |x|
)= 0
ye1x |x| =
∫0 dx
ye1x |x| = C
y =Ce−
1x
|x|where C is any real constant. This is the same general solution as in Solution 1above. Plugging in the initial condition therefore gives the same specific solution
y = −e−1−1x
|x|
(e) t ln(t)y′ + y = tet, t > e, with initial condition y(e2) = 0:
We use the integrating factor method. The standard first order linear form of thedifferential equation is
y′ +1
t ln(t)y =
et
ln(t)(4)
Then, a(t) = 1t ln(t)
. To integrate this and determine A(t), we use the substitution
u = ln(t)⇒ dudt
= 1t⇒ du = 1
tdt. Thus,∫
1
t ln(t)dt =
∫1
udu
= ln |u|= ln | ln(t)|
Note that because we are taking t > e, we have ln(t) > ln(e) = 1 > 0 soactually we can do away with the absolute value bars inside the logarithm to getA(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
y′ ln(t) +1
ty = et
d
dt(y ln(t)) = et
y ln(t) =
∫et dt
y ln(t) = et + C
y =et + C
ln(t).
5
Finally, use the initial condition y(e2) = 0 to get
0 =ee
2+ C
ln(e2)
0 =ee
2+ C
2
C = −ee2
The specific solution therefore is
y =et − ee
2
ln(t)
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