MATH 2260 Worksheet
Han-Bom Moon
Worksheet Week 5 Model Solution
Section 7.2.
• This worksheet is for improvement of your mathematical writing skill. Writing
using correct mathematical expression and steps is really important part of doing
math. Please print out this worksheet and try to solve problems, following given
steps.
• You don’t need to submit this worksheet. It is not homework.
• The steps in the solution represent important information you have to use in the
solution. It must be mentioned on your own solution.
1. Solve the differential equation
dy
= 3x2 e−y .
dx
(a) Collect all y terms on the left side and all x terms on the right side.
ey dy = 3x2 dx
(b) Integrate both sides of this equation.
Z
Z
ey dy = 3x2 dx
(c) Compute each side separately and get an implicit function.
Z
ey dy = ey + C1
Z
3x2 dx = x3 + C2
ey + C1 = x3 + C2
ey = x3 + C
1
MATH 2260 Worksheet
Han-Bom Moon
2. Solve the differential equation
dy
e2x−y
= x+y .
dx
e
dy
= e2x−y−x−y = ex−2y = ex e−2y ⇒ e2y dy = ex dx
dx
Z
Z
e2y dy = ex dx
Z
1
e2y dy = e2y + C1
2
Z
ex dx = ex + C2
1 2y
e + C1 = e x + C2
2
1 2y
e = ex + C
2
3. Solve the differential equation
dy
= ex−y + ex + e−y + 1.
dx
dy
= (ex + 1)(e−y + 1)
dx
1
dy = ex + 1dx
e−y + 1
Z
Z
1
dy = ex + 1dx
e−y + 1
Z
Z
Z
1
ey
ey
dy
=
dy
=
dy
e−y + 1
ey (e−y + 1)
1 + ey
u = 1 + ey ⇒
Z
ey
dy =
1 + ey
Z
du
= ey , du = ey dy
dy
1
du = ln |u| + C1 = ln |1 + ey | + C1 = ln(1 + ey ) + C1 (1 + ey > 0)
u
Z
ex + 1dx = ex + x + C2
ln(1 + ey ) + C1 = ex + x + C2
ln(1 + ey ) = ex + x + C
2
MATH 2260 Worksheet
Han-Bom Moon
4. The intensity L(x) of light x feet beneath the surface of the ocean satisfies the
differential equation
dL
= −kL.
dx
As a diver, you know from experience that diving to 18 ft in the Caribbean Sea
cuts the intensity in half. You cannot work without artificial light when the intensity falls below one-tenth of the surface value. About how deep can you expect
to work without artificial light?
(a) Write down the quantity we want to describe as a function, if it is not explicitly stated (In this example, it is L(x), the intensity of the light.).
L(x) = intensity of the light at x feet beneath the surface
(b) Write down the differential equation about it, if it is not explicitly stated (In
this example, we already have one.).
dL
= −kL
dx
(c) Find a general solution of given differential equation.
dL
= −kL ⇒ L(x) = L0 e−kx , L0 = intensity of the light on the surface.
dx
(d) Find the constant k, from the information given in the problem.
L(18) =
L0
L0
1
⇒ L0 e−18k =
⇒ e−18k = ⇒ e18k = 2
2
2
2
ln 2 = ln e18k = 18k ⇒ k =
ln 2
18
(e) Find the value of x that makes the intensity as one-tenth of the surface value.
ln 2
L(x) = L0 e− 18 x
ln 2
L0 e− 18 x =
ln 2
ln 2
L0
1
⇒ e− 18 x =
⇒ e 18 x = 10
10
10
ln 2
18 ln 10
x⇒x=
≈ 59.8
18
ln 2
As you can see, the initial value L0 does not have a special role.
ln 2
⇒ ln 10 = ln e 18 x =
3
MATH 2260 Worksheet
Han-Bom Moon
(f) Write down your answer with an appropriate unit.
18 ln 10
ft (≈ 59.8 ft)
ln 2
5. The charcoal from a tree killed in the volcanic eruption that formed Crater Lake
in Oregon contained 44.5% of the carbon-14 found in living matter. About how
old is Crater Lake? (The half-life of the carbon-14 is 5730 years.)
C(x) = ratio of carbon-14 to the ordinary carbon in a matter x years after its death
dC
= −kC
dx
dC
= −kC ⇒ C(x) = C0 e−kx ,
dx
C0 = C(0) = ratio of carbon-14 to ordinary carbon in a living matter
C(5730) =
C0
1
C0
⇒ C0 e−5730k =
⇒ e−5730k =
2
2
2
⇒ e5730k = 2 ⇒ ln 2 = ln e5730k = 5730k ⇒ k =
ln 2
ln 2
5730
ln 2
C(x) = 0.445C0 ⇒ C0 e− 5730 x = 0.445C0 ⇒ e− 5730 x = 0.445
ln 2
ln(0.445) = ln e− 5730 x = −
−
5730 ln(0.445)
ln 2
x⇒x=−
≈ 6693
5730
ln 2
5730 ln(0.445)
years (≈ 6693 years)
ln 2
4
MATH 2260 Worksheet
Han-Bom Moon
Let’s try a little bit different problem.
6. A colony of bacteria is grown under ideal conditions in a laboratory so that the
population increases exponentially with time. At the end of 3 hours there are
10,000 bacteria. At the end of 5 hours there are 40,000. How many bacteria were
present initially?
B(x) = number of bacteria after x hours
dB
= kB
dx
B(x) = B0 ekx , B0 = B(0) = number of initial bacteria
B(3) = 10000 ⇒ B0 e3k = 10000
B(5) = 40000 ⇒ B0 e5k = 40000
4=
B(5)
B0 e5k
e5k
=
=
= e5k−3k = e2k
B(3)
B0 e3k
e3k
4 = e2k ⇒ ln 4 = ln e2k = 2k ⇒ k =
ln 4
ln 22
2 ln 2
=
=
= ln 2
2
2
2
10000 = B(3) = B0 e3 ln 2 ⇒ B0 = 10000e−3 ln 2 = 10000(eln 2 )−3
= 10000 · 2−3 =
5
10000
= 1250
8