grandi (rg38778) – Homework 12 – grandi – (11111)
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001 (part 1 of 3) 10.0 points
For the differential equation
1. y0 (x) = 2e3x
2. y0 (x) = 2ex
2
2
3. y0 (x) = 2e−3x
4. y0 (x) = 2e−x
dy
= 9x2 y,
dx
1
3
3
3
5. y0 (x) = 2e3x correct
(i) first find its general solution.
Explanation:
In the general solution
3
1. y(x) = Ae3x correct
2. y(x) = Aex
3
3. y(x) = Aex
2
y = Ae3x
3
of
4. y(x) = Ae3x
4
5. y(x) = Ae3x
2
dy
= 9x2 y,
dx
the value of A is determined by the condition
y(0) = 2 since
y(0) = 2
Explanation:
The differential equation
=⇒
A = 2.
Consequently,
y0 (x) = 2e3x
dy
= 9x2 y
dx
becomes
3
.
003 (part 3 of 3) 10.0 points
Z
dy
= 3
y
Z
3x2 dx
after separating variables and integrating.
Thus
ln y = 3x3 + C,
and so the general solution of the given differential equation is
3
y = eC e3x = Ae3x
(iii) For the particular solution y0 in (ii), determine the value of y0 (2).
1. y0 (2) = 2e4
2. y0 (2) = 2e−24
3. y0 (2) = 2e12
3
with A an arbitrary constant.
002 (part 2 of 3) 10.0 points
(ii) Then find the particular solution y0 such
that y0 (0) = 2.
4. y0 (2) = 2e−8
5. y0 (2) = 2e24 correct
Explanation:
As
y0 (x) = 2e3x
3
grandi (rg38778) – Homework 12 – grandi – (11111)
is the unique solution of
dy
= 9 x2 y,
dx
y0 (0) = 2 ,
with A an arbitrary constant. For the particular solution y0 the condition y(0) = 3
determines A uniquely since
y(0) = 3
the required value of y0 is given by
y0 (2) = 2e24 .
2
=⇒
A = 1.
Consequently,
y0 (x) = 1(x2 + 9)1/2 ,
and so at x = 9,
004
10.0 points
y0 (9) = 101/2 3.
If y0 satisfies the equations
(x2 + 9)
dy
= xy,
dx
y(0) = 3,
If y satisfies the equations
1. y(4) =
3 correct
7
2
3. y0 (9) = 101/2 4
2. y(4) = −1
4. y0 (9) = 30
3. y(4) =
5. y0 (9) = 40
4. y(4) = 3
Explanation:
The differential equation
(x2 + 9)
dy
= xy,
dx
5. y(4) =
y(0) = 3
becomes
Z
dy
=
y
y(1) = 2 ,
find the value of y(4).
1. y0 (9) = 171/2 3
2. y0 (9) = 10
10.0 points
(x + 3)y ′ = y − 1 ,
determine the value of y0 (9).
1/2
005
Z
x2
x
dx
+9
after separating variables and integrating.
Thus the general solution of this differential
equation is given by
1
ln y = ln(x2 + 9) + C,
2
which in explicit form is
y(x) = A(x2 + 9)1/2
−1
4
11
correct
4
Explanation:
The differential equation becomes
Z
Z
dx
dy
=
y−1
x+3
after separating variables and integrating.
Thus its general solution is
ln(y − 1) = ln(x + 3) + C,
which can be written in explicit form as
y = eC (x + 3) + 1 = A(x + 3) + 1
with A an arbitrary constant. The value of A
is determined by the condition y(1) = 2 since
y(1) = 2
=⇒
1
A= .
4
grandi (rg38778) – Homework 12 – grandi – (11111)
3
Consequently,
1
y(x) = (x + 3) + 1 ,
4
so that
y(4) =
11
4
.
3. 2 x2 + 9 y 2 = C
4. 9 x2 − 2 y 2 = C correct
5. 9 x2 + 2 y 2 = C
Explanation:
006
10.0 points
2 yy ′
dy
2y
dx
2 ydy
Z
2 ydy
Solve the differential equation
dy
e5x
=
.
dx
6 y5
1. y = ±
r
1 5x
e +C
5
2. y = ±
r
1 5x
e +C
5
3. y = ±
r
1 5x
e +C
5
4. y = ±
r
1 5x
e + C correct
5
2. y = e
5. y = ±
p
6
e5x
3. y = e
5
5
6
6
008
= 9 xdx
Z
= 9 xdx
10.0 points
Solve the differential equation y ′ =
1. y = e±
e5x
dy
=
dx
6 y5
6y 5 dy = e5x dx
Z
Z
5
6y dy = e5x dx
1
y 6 = e5x + C
5r
1
y = ± 6 e5x + C
5
007 10.0 points
Solve the differential equation 2yy ′ = 9x.
√
x5 +C
√
± 2 x5 +C
√
4. y = e±
Explanation:
2. 2 x2 + 9 y 2 = 11
= 9x
9 x2 − 2 y 2 = C
+C
1. 2 x2 − 9 y 2 = C
= 9x
2 x5 +C
√
√
5. y = e
correct
2 x4 +C
x5 +C
Explanation:
y′ =
5 x4 y
ln y
ln y
dy = 5x4 dx
y
Z
Z
ln y
dy = 5x4 dx
y
1
Let ln y = u ⇒ dy = du So,
y
Z
Z
udu = 5 x4 dx
u2
= x4 + C
2
p
u = ± 2 2 x5 + C
5 x4 y
.
ln y
grandi (rg38778) – Homework 12 – grandi – (11111)
Since, ln y = u ⇒ y = eu So, y = e±
√
2 x5 +C
3. y =
q
513 −
p
x2 + 1 correct
4. y =
q
513 +
p
x2 + 1
5. y =
q
512 −
p
x2 + 1
009 10.0 points
Solve the differential equation
du
= 42 + 6u + 7t + ut .
dt
1. u = −7 + C e
1 2
t +6t
2
1
2. u = −7 + C e 2 t+6t
1 2
3. u = −7 + C e 2 t
correct
2
2
+6t
5. u = −7 + C et
2
−6t
3
3
p
dy
=0
x2 + 1
dx
p
3y 2 x2 + 1 dy = x dx
x dx
3y 2 dy = √
2
Z
Z x +1
x dx
√
3y 2 dy =
x2 + 1
Z
x dx
√
y 3 + c1 =
x2 + 1
x + 3y 2
Explanation:
du
= 42 + 6u + 7t + ut
dt
du
= (7 + u)(6 + t)
dt
1
du = 6 + t dt
Z 7+u
Z
1
du = 6 + t dt [u 6= −7]
7+u
t2
ln |7 + u| = 6t + + C
2
2
6t+ t2 +C
|7 + u| = e
3
Explanation:
−6t
4. u = −7 + C et
4
Let
p
x2 + 1 = u ⇒
3
y + c1 =
Z
x
dx = du So,
x2 + 1
du
y 3 + c1 = u + c2
y 3 = u + C where C = c1 + c2
x
y3 = √
+C
2
x +1
2
6t+ t2 +C
= Ke
t2
7 + u = ±Ke6t+ 2 +C
t2
u = −7 ± Ke6t+ 2 +C where K > 0
So, u = −7 +C e
constant.
1 2
t +6t
2
, where C is an arbitary
010 10.0 points
Find the solution of the differential equation
p
dy
= 0, that satisfies the inix − 3y 2 x2 + 1
dx
tial condition y (0) = 8.
q
p
3
1. y = 512 + x2 + 1
q
p
3
2. y = 513 − x2 − 1
When x = q
0, y = 1 so, C = 1 Therefore, the
p
3
solution is 513 − x2 + 1
011 10.0 points
Find the solution of the differential equation
dy
y2
= 8 that satisfies the initial condition
dx
x
y (1) = 1.
1. No solution exists for this initial condition.
2. y =
6 x7
1 − 6 x7
grandi (rg38778) – Homework 12 – grandi – (11111)
8 x7
3. y =
7 + 6 x7
4. y =
8 x7
1 + 7 x7
7 x7
5. y =
8 − x7
6. None of these
7. y =
7 x8
6 + 42 x7
7 x7
8. y =
correct
1 + 6 x7
Explanation:
y2
dy
= 8 , y(1) = 1.
Z dx Zx
dx
dy
=
2
y
x8
1
1
− =− 7 +C
y
7x
y(1) = 1
1
⇒ −1 = − + C
7
6
⇒C=−
7
1
1
6
So =
+
7
y
7x
7
7
1+ 6x
1
=
y
7 x7
7 x7
y=
1 + 6 x7
012 10.0 points
A certain small country has $10 billion in
paper currency in circulation, and each day
$60 million comes into the country’s banks.
The government decides to introduce new currency by having the banks replace old bills
with new ones whenever old currency comes
into the banks. Let x = x (t) denote the
amount of new currency in circulation at time
t with x (0) = 0.
5
Formulate and solve a mathematical model
in the form of an initial-value problem that
represents the “flow” of the new currency into
circulation (in billions per day).
1. x(t) = 10 1 − e−0.006 t correct
2. x(t) = 10 1 + e−0.06 t
3. x(t) = 60 1 − e−0.006 t
4. x(t) = 10 1 − e−0.06 t
5. x(t) = 10 1 + e−0.006 t
Explanation:
Use 1 billion dollars as the x-unit and 1 day
as the t-unit. Initially, there is $10 billion of
old currency in circulation, so all of the $60
million returned to the bank is old.
At time t, the amount of new currency is
x(t) billion dollars, so 10 − x(t) billion dollars
of currency is old.
The fraction of circulating money that is
old is [10 − x(t)]/10, and the amount of old
currency being returned to the banks each
10 − x(t)
· 0.06 billion dollars.
day is
10
This amount of new currency per day is
10 − x
dx
introduced into circulation, so
=
·
dt
10
0.06 = 0.006(10 − x) billion dollars per day.
dx
= 0.006 dt
10 − x
−dx
= −0.006 dt
10 − x
ln(10 − x) = −0.006 t + c
10 − x = Ce−0.006 t
Where C = ec ⇒ x(t) = 10 − Ce−0.006 t .
From x(0) = 0, we get C = 10, so x(t) =
10 1 − e−0.006 t .
013 (part 1 of 3) 10.0 points
For the differential equation
2y
1
dy
+
= 2 + 6,
dx
x
x
grandi (rg38778) – Homework 12 – grandi – (11111)
(i) first find its general solution.
1
1. y = 2 x + 2x3 + C correct
x
1
2. y = 2 x − 2x3 + C
x
2
3
3. y = x x + 2x + C
4. y = x2 x − 2x3 + C
1 3
x
+
2x
+
1
2x2
3
2
5. y = x x + 2x + 1
4. y =
Explanation:
The value of C in
1
(x + 2x3 + C)
2
x
y =
1 3
5. y = 2 x + 2x + C
2x
6
is determined by the condition y(1) = 4. For
Explanation:
The integrating factor needed for the first
order differential equation
2y
1
dy
+
= 2 +6
dx
x
x
y(1) = 4
=⇒
4 = 3 + C.
Consequently,
y(x) =
1
3
x
+
2x
+
1
.
x2
is
R
µ=e
2
x
dx
015 (part 3 of 3) 10.0 points
2
=x .
(iii) For the particular solution in (ii) determine
the value of y(2).
After multiplying both sides of this differen2
tial equation by x we can thus rewrite the
19
correct
1. y(2) =
equation as
4
d 2 23
x y = 1 + 6x2 .
2. y(2) =
dx
4
Consequently, the general solution is
y=
1
(x + 2x3 + C)
x2
3. y(2) =
25
4
4. y(2) =
27
4
5. y(2) =
21
4
with C an arbitrary constant.
014 (part 2 of 3) 10.0 points
(ii)
Find the particular solution such that
y(1) = 4.
1
3
1. y = 2 x − 2x + 1
x
2. y = x2 x − 2x3 + 1
1
3
3. y = 2 x + 2x + 1 correct
x
Explanation:
At x = 2, therefore,
y(2) =
19
4
.
016 10.0 points
Consider the graph
grandi (rg38778) – Homework 12 – grandi – (11111)
y
7
y
20
2
15
2.
(0, 1)
−2
2
x
2
x
2
x
2
x
10
5
4 −1.6 −0.8
0.8
−2
1.6
y
x
Choose the equation whose solution is
graphed and satisfies the initial condition
y(0) = 4.
2
3.
(0, 1)
−2
1. y ′ = y − x
−2
2. y ′ = y 2 − x
y
3. y ′ = y − 1 correct
2
4. y ′ = y 2 − x2
4.
(0, 1)
−2
5. y ′ = y 3 − x2
Explanation:
−2
017 10.0 points
Use the direction field of the differential
equation y ′ = y 5 to sketch a solution curve
that passes through the point (0, 1).
y
2
y
5.
2
1.
(0, 1)
−2
2
−2
(0, 1)
−2
x
−2
Explanation:
018 10.0 points
Consider the graph
correct
grandi (rg38778) – Homework 12 – grandi – (11111)
y
9
6
3
−2
2
x
Choose the equation whose solution is
graphed and satisfies the initial condition
y(0) = 4.
1. y ′ = cos(x + y)
2. y ′ = y sin 2x correct
3. y ′ = y cos 2x
4. y ′ = sin(x + y)
5. y ′ = sin(x − y)
Explanation:
8

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