1
Spring 2019
Assignment Section 2.2 Separable Equations
2
Solve the initial value problem: 3. (1 point)
Solve the initial value problem:
dy
3
dx = 4y, y(0)= 4 dy = −0.8 ,
y(0)= π dx cos(y)
4
Solution: Separating variables gives y(x)= .
Z dy Z Solution: Separating variables gives
= 4dx, Z Z
y cosy dy = −0.8dx,
ln|y| = 4x+C so
|y| = eCe4x siny = −0.8x+C y = Ke4x
The initial conditon y gives C and
y = 4e4x
Solving for y yields
Answer(s) submitted:
!
• exp(4x+1.387) y
(correct)
Correct Answers: Answer(s) submitted:
• 4 * 2.71828182845905**(4*x) • sinˆ(-1)((-0.8x)+(sqrt(2)/2))
(correct)
Correct Answers:
y0(x)=p−2y(x)+22, y(−2)= 3 • arcsin(-0.8*x + 0.707106781186547)
dy 10x+5
(x)= . dx = 15y2 +14y+9
Solution: Separating variables gives
has an implicit general solution of the form F(x,y)= K, where
dx, K is an arbitary constant.
so In fact, because the differential equation is separable, we can
√define the solution curve implicitly by a function in the form
−1√−2y+22 = x+C
−2y+22 = −1x+C F(x,y)= G(x)+H(y)= K.
Find such a solution and then give the related functions re-
The initial conditon y(−2)= 3 gives C = 2 and √ quested.
4
.
y
Solution: Separating variables gives
Answer(s) submitted: Z 2 Z
(15y +14y+9) dy = (10x+5) dx,
• [((x-2)ˆ2)/-2]+11
(correct) so
—(5x*
+
5x)
+
(5y°
+7y?
+9y)
=K
Answer(s)
submitted:
e@
5y°3+7y*>2+9y-5x*2-5x
(correct)
Correct
Answers:
@
k*((5
*
(x**2)
+
5
*x
)
- (5
*
(y**3)
+
7
*(y**2)
5. (1
point)
The
differential
equation
dy
_
15
dx
y!/84.25x2
yl/8
has
an
implicit
general
solution
of
the
form
F(x,
y)
=
K,
where
K
is
an
arbitrary
constant.
In
fact,
because
the
differential
equation
is
separable,
we
can
define the
solution
curve
implicitly
by
a
function
in
the
form
F
(x,y)
=
G(x)
+H(y)
=K.
Find
such
a
solution
and
then
give
the
related
functions
re-
quested.
F(x,y)
=
G(x)
+H(y)
=
Solution:
We
factor
the
denominator
to
get
dy
15
dx
y/8(1+25x2)’
sO
15
1/8
dy
=
d
ye
Tye
Integrating
both
sides,
yields
8
5)”
8
—
3arctan(5x)
+K.
Isolating
the
arbitrary
constant,
yields
8
97/8
—3
arctan(5x)
+
=
K
Answer(s)
submitted:
e
((8y*
(9/8)
)/9)-3arctan
(5x)
(correct)
Correct
Answers:
e
k(
3*arctan(5
*x)
+
(-8/9
)*(y**(9
/8
))
)
+
m
l
y>—6dx__2y
+ 9
*y))
+m
In|y?>—6|
=
1+
>
_6
K-e73
The
initial
conditon
y(1)=
7
gives
1=Ke
',so
K =e
and
2
el-3
+6.
7.
(1
point)
Solve
the
initial
value
problem:
77
Iny
=
(x8
+255)"/°
@
e*
((x°8+2°8-1)
70.125)
5
6. (1 point) Solve the initial value problem: x2
6
dy √
= , y(1)= 7
y(x)= .
Solution: Separating variables gives
Z 2y Z 1
dy = dx,
y2 −6 x2
so
C
y
where K = ±eC. √
y
Solving for y yields
y
√
Since y(1)= 7 > 0, it must be
y
Answer(s) submitted:
• sqrt(exp((-1/x)+1)+6)
(correct)
Correct Answers:
• sqrt(eˆ(1-1/x)+6)
dy
(ln(y)) = x y, y(1)= e dx
y(x)= .
Solution: Separating variables gives
Z (lny)7 Z 7 dy = x dx, y
so
(lny)8
x
8
C
(lny)8 = x8 +C
The initial condition y(1)= e2 gives 28 = 1+C, so C = 255 and
(lny)8 = x8 +255. Solving for y yields
7
y = e(x8+255)1/8
Answer(s) submitted:
• exp((xˆ8+255)ˆ(1/8))
(correct)
Correct Answers:
8.
(1
point)
Find
f(x)
if
y
=
f(x)
satisfies
dy
4
o
=
45y:
dx
”
and
the
y-intercept
of
the
curve
y
=
f(x)
is
2.
f(x)
=
Solution:
Separating
variables
gives
dy
/
dy=
/
45x4dx,
y
In|y|
=9x°
+C
sO
y=
Ke
Since
the
y-intercept
of
the
curve
is
2,
we
have
that
y(0)
=2.
,
so
K
=2,
and
y=
f(x)
=20"
Answer(s)
submitted:
@
exp
(9x75+0.693)
(correct)
Correct
Answers:
e
2
*
exp(9
*
(x75)
)
9.
(1
point)
Solve
the
initial
value
problem:
dy
24]
wt
Ik
5x
—2y
0,
y(0)
=
—7
y(x)
=
Solution:
Separating
variables
gives
_[s_*
Jro=fh
al
dx
Jiatse
yr
=5Vr2+414€
The
initial
condition y(0)
=—7
gives
49=5+C,so
C=
44
and
y*
=5V/x2+1+444.
Solving
for
y
yields
y=tV5Vx2+1444
Since
the
initial
condition
y(0)
=
—7
<
Answer(s)
submitted:
@
-sgrt
(5(x*2+1)
*
(1/2)
+44)
sO
bolt
y
2
0,
it
must
be
(correct)
Correct
Answers:
e
-
sqrt
(5*(x*2
+1)*.5
-5
+
-7*-7)
10.
(1
point)
Find
the
function
y
=
y(x)
(for
x
>
0
)
which
satisfies
the
separable
differential
equation
dy
8+16x
dx
2
=
>
x>0
dx
xy
*
with
the
initial
condition
y(1)
=
5.
y
=
Solution:
Separating
variables
gives
[va=]
(E
+16)
ax
~
=
8Inx+16x+C
sO
~
ols.
=
24Inx+48x+C
The
initial
condition
v()
=5
gives
55=484C,so
C=77
and
y?
=
24Inx+48x+77.
Solving
for
y
yields
=
(24Inx
+48x+77)!/3
Answer(s)
submitted:
@
(241n(x)+48x+77)
*
(1/3)
(correct)
Correct
Answers:
e
(3
* 8
*
In(abs(x))
+
3
*
16
*
x
+
77)**(1/3)
11.
(1
point)
Find
the
solution
to
the
differential
equation
2
=
0.8(y
—
250)
if
y=
45
when
¢
=
0.
y
=
Solution:
SOLUTION
Separating
variables
gives
IS
250
In|y
—
250]
=
[o.
8dt,
=
0.8t
+C.
sO
Solving
for
y,
y
=
250+
Ae?*,
where
A
=
+e©.
The
initial
condition,
y(0)
=
45,
gives
45
=
250+A,
so
A
=
205,
and
y
=
250
—205¢°*",
Answer(s)
submitted:
@
-—205exp(0.8t)
+250
(correct)
Correct
Answers:
@
(45-250)
*e*
(0.8*t)
+
250
8
9
12. (1 point)
dx
=
(
)
y
+
y
+
y
+
(
,
y
)=
,
(
,
y
)=
(
)+
(
y
)=
.
(
,
y
)=
(
)+
(
y
)=
Separating variables gives
Z
y
+
y
+
y
+
=
Z
,
Z
y
+
(
y
+
)(
y
+
)
=
Z
,
Performing partial fraction decomposition yields
Z
y
+
+
y
+
=
Z
,
(
|
y
+
|
)+
(
|
y
+
|
)=
+
−
+
(
|
y
+
|
)+
(
|
y
+
|
)=
•
2
•
=
,
10
Solve the initial value problem
5u+3t
u(0)= 2
u(t)=.
Solution: Separating variables gives
Z Z
e−5u du = e3t dt,
so
C
e C
The initial condition u(0)= 2 gives C and
e.
Solving for u yields
u Answer(s) submitted:
.
• (-1/5)ln((-5/3)exp(3t)+exp(-10)+(5/3))
(correct)
Correct Answers:
• -ln(exp(-2*5) + (5/3) - (5/3)*exp(3*t))/5
13. (1 point) Find the solution of the initial value problem
dy −3)e−2y, y(3)= ln(3) =(x dx
y(x)= .
Solution: Separating variables gives
Z Z
e2y dy = (x−3)dx,
so
e2y x2
C
e2y = x2 −6x+C
The initial condition y(3) = ln(3) gives 9 = 9 − 18 +C so
C = 18 and
e2y = x2 −6x+18.
Solving for y yields
y
Answer(s) submitted:
11
• (1/2)ln(xˆ2-6x+18)
(correct)
Correct Answers:
• (1/2)*ln((x-3)ˆ2+3ˆ2)
15.
(1
point)
The
differential
equation
dy
a
28
+
35x+
28
y+
35
xy
Xx
has
an
implicit
general
solution
of
the
form
F
(x,y)
=
K,
where
K
is
an
arbitrary
constnat.
In
fact,
because
the
differential
equation
is
separable,
we
can
define the
solution
curve
implicitly
by
a
function
in
the
form
F(x,y)
=
G(x)
+H(y)
=
K.
Find
such
a
solution
and
then
give
the
related
functions
re-
quested.
F(x,y)
=
G(x)
+
A(y)
=
Solution:
Factoring
by
grouping
yields
dy
dx
dy
dx
Separating
variables
gives
dy
_
7T4+7Ty
—
4(7+7y)
+5x(7+7y)
(4+5x)(7+7y).
/
(4-4
5x)dx,
I
3
In(7
+7y)
=
4x+
3°
+K
Isolating
the
constant of
integration
gives
+2In(7+7y)
=K
—(4x+
3x”)
Answer(s)
submitted:
e@
(1/7)
1n(yt1)-
(5/2)
x*2-4x
(correct)
Correct
Answers:
@
k(x*(2*4
+
5*x)/(2*4
+5
)
+
(-
2
/(7*5
+
2*4*7)
16.
(1
point)
Find
the
solution
to
the
differential
equation
du
oT
=
u’,
subject
to
the
initial
conditions
u(0)
=
3.
—
Solution:
SOLUTION
Separating
variables
gives
1 1
|
dum
fet
l l
—t+C.
5
or
u
)*
The
initial
condition
gives
C =
a
and
so,
after
rearranging
slightly,
Answer(s)
submitted:
@
-1/((t/5)-(1/3))
(correct)
Correct
Answers:
@
5*3/(5
-
3*t)
17.
(1
point)
Solve
the
differential
equation
dx
_
2xInx
dt
t
Assume
x,¢
>
0,
and
use
the
initial
condition
x(1)
=
5.
2
Solution:
SOLUTION
Separating
variables
gives
dx
24,
xInx
ft
Ne)
2
[=
=
[=a
J
xinx
t
and
thus
In|
Inx|
=
2Int+C.
Exponentiating
both
sides,
C
2
Int
|Inx|
=
ee?!"
=
&?
=
Ar’,
*
In(abs(7
+
To
)))
+
where
A
=
e.
Then,
with
x(1)
=5,
InS5
=A,
so
Inx
=
In(5)t?,
or
2 ‘2
2
x-e
In(5)
_
eln(5
i
x.
Answer(s)
submitted:
e
exp(iln(5)t*2)
(correct)
Correct
Answers:
@ 5°
(t*(2))
12
44/784
—4428
4
4
13
18. (1 point) (a) Find the explicit solution of the initial value
14
problem:
dy 2x
= , y(0)= −8 dx −8y−8
y(x)=
(b) Determine the interval where the solution is defined.
Interval:
Solution: (a)
We first separate the variables
(−8y−8)dy = 2x dx
and then integrate
−4y2 −8y = x2 +C
Substituting the initial condition yieldsC =−192 so the implicit
solution is
−4y2 −8y = x2 −192
or, equivalently,
4y2 +8y+x2 −192 = 0.
Using the quadratic formula yields
y
y
Substituting x = 0 yields
Since the initial condition is y(0)=−8, we have that the explicit
solution is
y
(b)
The solution is defined when 784−4x2 > 0. Solving
the inequality yields the interval
(−14,14)
Answer(s) submitted:
• [-8-sqrt(64-16(xˆ2-192))]/8
• (-14,14)
(correct)
Correct Answers:
• -(1+[sqrt(3136-16*xˆ2)]/8)
• (-14,14)
19. (1 point) Determine whether each first-order differential
equation is separable, linear, both, or neither.
• choose one
• Separable
• Linear
• Both
• Neither
dy x 2 2
1. +e y = x y dx
• choose one
• Separable
• Linear
• Both
• Neither
2. y+ex sinx = x3y0
• choose one
• Separable
• Linear
• Both
• Neither
3. lnx−x2y = xy0
• choose one
• Separable
• Linear
• Both
• Neither
Note: You only have two attempts at this problem.
Answer(s) submitted:
• Neither
• Linear
• Linear
• Neither
(correct)
Correct Answers:
• Neither
• Linear
• Linear
• Neither
dx
+
y
=
15
20. (1 point) Which of the following are separable
differential equations?
• A. dxdyx+x2y = x
• B. dxdy = xy
• C. dxdy = 1+xyx2
• D. dxdy +yx2 = x2
• E. y0 = 2x−3y+1
16
• F. 2xydx+(x2 −1)dy = 0
y
• H. dx dy = x+y2
Answer(s) submitted:
Generated by c WeBWorK, http://webwork.maa.org, Mathematical Association of America
• ( B, C, D, F )
(correct)
Correct Answers:
• BCDF
