MATH 315 PS #1
Summer 2010
§2.1. Differential Equations and Solutions
#3, 4, 17, 20, 24, 35
PS1 §2.1#3 Show that y(t)=C e − (1 / 2)t is a general solution of the differential equation y ′ = –ty. Use
2
a computer or calculator to sketch the solutions for the given values of the arbitrary constant C= –3,
–2, …, 3. Experiment with different intervals for t until you have a plot that shows what you
consider to be the most important behavior of the family.
y (t ) =C e −0.5t
2
⇒
y ′(t ) = –Ct e −0.5t = t y (t )
2
⇒
y ′(t ) =t y (t )
PS1 §2.1#4 Show that y(t)=2t – 2 + C e − t is a general solution of the differential equation y ′ + y =
2t. Use a computer or calculator to sketch the solutions for the given values of the arbitrary constant
C= –3, –2, …, 3. Experiment with different intervals for t until you have a plot that shows what you
consider to be the most important behavior of the family.
y (t ) = 2t – 2 + C e − t
⇒
DIFFERENTIAL EQUATIONS
y ′(t ) = 2 – C e − t
⇒
y ′(t ) + y (t ) = 2t
Page 1
MATH 315 PS #1
Summer 2010
PS1 §2.1#17 Plot the direction field for the differential equation y ′ = y + t by hand. Do this by
drawing short lines of the appropriate slope centered at each of the integer valued coordinates (t,y),
where –2≤t≤2 and –1≤y≤1.
Table and graph:
y ′ =y+t
-1
0
1
y
t
-2
-3
-2
-1
-1
-2
-1
0
0
-1
0
1
1
0
1
2
2
1
2
3
PS1 §2.1#20 Plot the direction field for the differential equation y ′ = (t2y)/(1+y2) by hand. Do this
by drawing short lines of the appropriate slope centered at each of the integer valued coordinates
(t,y), where –2≤t≤2 and –1≤y≤1.
Table and graph:
y ′ =(t2y)/(1+y2)
y
-1
0
1
-2
-2
0
2
-1
-0.5
0
0.5
t
0
0
0
0
1
-0.5
0
0.5
2
-2
0
2
PS1 §2.1#24 Use a computer to draw a direction field for the given first-order differential equation
y ′ = (y+t)/(y – t), R={(t,y): –5≤t≤5 and –5≤y≤5}. Use the indicated bounds for your display
window. Obtain a printout and use pencil to draw a number of possible solution trajectories on the
direction field. If possible, check your solutions with a computer.
DIFFERENTIAL EQUATIONS
Page 2
MATH 315 PS #1
Summer 2010
dP
=0.44P, where P is
dt
the mass of the accumulated bacteria (measured in milligrams) after t days. Suppose that the initial
mass of the bacterial sample is 1.5 mg. Use a numerical solver to estimate the amount of bacteria
after 10 days.
PS1 §2.1#35 Bacteria in a petri dish is growing according to the equation
The initial differential equation is
P ′ = 0.44 P , P (0) = 1.5
Base on the graph, we estimate that P(10) ≈ 124. Thus, there are approximately
124 mg of bateria present after 10 days.
DIFFERENTIAL EQUATIONS
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MATH 315 PS #1
Summer 2010
§2.2. Solutions to Separable Equations
#3, 6, 10, 22
PS1 §2.2#3 Find the general solution of the differential equation y ′ = e x − y . If possible find an
explicit solution.
dy
x
−y
x
=e e
⇒
e y dy= e x dx ⇒
e y = e x +C
⇒
y(x)=ln( e +C)
dx
PS1 §2.2#6 Find the general solution of the differential equation y ′ = y e x –2 e x +y–2. If possible
find an explicit solution.
⇒
⇔
dy
x
x
x
x
= y e –2 e + y – 2 = e (y-2)+(y-2) = ( e +1)(y-2)
dx
dy
dy
x
x
x
=( e +1)dx
⇒
= (e + 1) dx ⇔
ln|y-2|= e +x+C
∫
y−2
y−2 ∫
|y-2|= e
x 2 + x +C
⇔
y-2=A e
x2 + x
⇔
y = Ae
x2 + x
+ 2
PS1 §2.2#10 Find the general solution of the differential equation x y ′ –y =2x2y. If possible find an
explicit solution.
dy
2
2
=2x y + y = y(2x + 1)
dx
dy 2 x 2 + 1
dy
1
=
dx
⇒
= ∫ ( 2 x + )dx
∫
y
x
y
x
x
⇒
⇔
2
ln|y|= x +ln|x|+C
x + ln | x|+ C
x
⇔
|y|= e
⇔
y=A xe
PS1 §2.2#22 Find exact solutions for the differential equation y ′ =(y2+1)/y, y(1)=2. State the
2
2
interval of existence. Plot each exact solution on the interval of existence. Use a numerical solver to
duplicate the solution curve for the initial value problem.
dy y 2 + 1
=
dx
y
⇔
ln|
y 2 +1|=2x + C
⇒
y
dy =dx
2
y +1
⇒
1 d ( y 2 + 1)
= dx
2 ∫ y2 +1 ∫
⇔
y 2 +1= e 2 x +C
⇔
y 2 = A e 2 x -1
⇔
Ae 2 x − 1
y(x)= ±
Initial condition y(1)=2
⇔
±
Ae 2 − 1 =2
⇔
Ae 2 − 1 =2
So, the particular solution is y(x)=
The interval of existence:
5e 2 x −2 − 1 > 0
⇔
DIFFERENTIAL EQUATIONS
⇔
2
A e -1=4
−2
⇔
A=5 e
⇔
2x> 2-ln5
5e 2 x − 2 − 1
5e 2 x −2 > 1
⇔
e2x >
e2
5
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MATH 315 PS #1
Summer 2010
⇔
x>1-
ln 5
2
⇔
x ∈ (1-
ln 5
,+ ∞ )
2
Plot of the exact solution:
Use numerical solver to duplicate the solution curve:
DIFFERENTIAL EQUATIONS
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MATH 315 PS #1
Summer 2010
§2.4. Linear Equations
#4, 5, 6, 14, 19, 28, 29
PS1 §2.4#4 Find the general solution of the first-order, linear equation y ′ +2ty=5t.
Multiplying
equation becomes
both
sides
by
an
e t y ′ +y(2t e t ) = 5t e t
2
2
⇔
(e t y )′ =5t e t
⇔
ye t = ∫ 5te t dt =
2
2
integrating
factor,
e∫
u(t)=
2 tdt
t2
=e ,
the
2
2
2
5 t2
5 2
e d (t 2 ) = e t + C
∫
2
2
2
5
+ Ce −t
2
PS1 §2.4#5 Find the general solution of the first-order, linear equation x ′ –2x/(t+1)=(t+1)2.
⇔
y=
−2
Multiplying both sides by an integrating factor, u(t)= e
= (t
∫ t +1dt
=e
−2 ln t +1
=
t +1
−2
+ 1) −2 , the equation becomes
⇔
(t + 1) −2 x ′ -2 (t + 1) −3 x =1
((t + 1) −2 x)′ =1
⇔
(t + 1) −2 x = ∫ dt = t + C
⇔
x = ∫ dt =( t + C ) (t + 1) 2
PS1 §2.4#6 Find the general solution of the first-order, linear equation t x ′ =4x+t4.
Rewrite the differential equation as
x ′ -4 t −1 x = t 3
Multiplying both sides by an integrating factor, u(t)= e
∫
−4
dt
t
=e
−4 ln t
=
t
−4
=t
−4
the equation becomes
⇔
t −4 x ′ -4 t −5 x = t −1
(t −4 x)′ = t −1
⇔
t −4 x = ∫ t −1dt = ln t + C
⇔
x = t 4 ln t + Ct 4
PS1 §2.4#14 Find the solution of the initial value problem y ′ =y+2x e 2 x , y(0)=3.
Rewrite the differential equation as
y ′ − y = 2 xe 2 x
DIFFERENTIAL EQUATIONS
Page 6
,
MATH 315 PS #1
Multiplying
equation becomes
Summer 2010
both
sides
by
an
integrating
factor,
u(x)= e ∫
⇔
e − x y ′ - e − x y = 2 xe x
(e − x y )′ = 2 xe x
⇔
e − x y = ∫ 2 xe x dx = 2 ∫ xd (e x ) = 2( xe x − ∫ e x dx) = 2( xe x − e x ) + C
⇔
y = 2e 2 x ( x − 1) + e x C
−dx
=e
−x
,
the
⇔
C = 5.
2x
x
So, the particular solution is y(x)= 2e ( x − 1) + 5e
Initial condition 3=y(0)=-2+C
PS1 §2.4#19 Find the solution of the initial value problem (2x+3) y ′ =y+(2x+3)1/2, y(–1)=0.
Discuss the interval of existence and provide the sketch of your solution.
Rewrite the differential equation as
y′ −
Multiplying
u(x)= e
∫
−
1
dx
2 x +3
=e
1
− ln 2 x + 3
2
1
y = (2 x + 3) −1 / 2
2x + 3
both
=
2x + 3
sides
−1 / 2
= ( 2 x + 3)
by
−1 / 2
an
(assuming
integrating
2x+3>0),
the
factor,
equation
becomes
(2 x + 3) −1 / 2 y ′ - (2 x + 3) −3 / 2 y = (2 x + 3) −1
⇔
((2 x + 3) −1 / 2 y )′ = (2 x + 3) −1
1
⇔
(2 x + 3) −1 / 2 y = ∫ (2 x + 3) −1 dx = ln(2 x + 3) + C
2
1/ 2
(2 x + 3)
⇔
y=
ln(2 x + 3) + C (2 x + 3)1 / 2
2
Initial condition 0=y(-1)= C .
(2 x + 3)1 / 2
So, the particular solution is y(x)=
ln(2 x + 3) .
2
3
The interval of existence: 2x+3>0 ⇔ x ∈ ( − ,+ ∞ ).
2
DIFFERENTIAL EQUATIONS
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MATH 315 PS #1
Summer 2010
PS1 §2.4#28 Suppose that you have a closed system containing 1000 individuals. A flu epidemic
starts. Let N(t) represent the number of infected individuals in the closed system at time t. Assume
that the rate at which the number of infected individuals is changing is jointly proportional to the
number of infected individuals and to the number of uninfected individuals. Furthermore, suppose
that when 100 individuals are infected, the rate at which individuals are becoming infected is 90
individuals per day. If 20 individuals are infected at time t=0, when will 90% of the population be
infected? Hint: The assumption here is that there are only healthy individuals and sick individuals.
Furthermore, the resulting model can be solved using the technique introduced in Exercise 22.
N ′ = kN (1000 − N ) . When N=100, the rate of infection
is 90/day. So k × 100 × 900=90
⇒
k = 0.001.
The model equation is
Therefore,
N ′ = N − N 2 0.001
Let x = N
−1
. Then
x ′ = − N ′ / N 2 = − ( N − N 2 0.001) / N 2 = −
1
+ 0.001 = − x + 0.001
N
Rewrite the differential equation as
x ′ + x = 0.001
u(t)= e ∫ = e ,
dt
Multiplying
equation becomes
both
sides
by
an
integrating
⇔
e t x ′ + e t x =0.001 e t
(e t x)′ =0.001 e t
⇔
e t x = 0.001∫ e t dt = 0.001(e t + C )
⇔
x = 0.001(1 + Ce − t )
factor,
1
−t
= 1000 /(1 + Ce )
x
Initial condition 20=N(0)= 1000 /(1 + C ) ⇔ 1 + C = 50
⇔
t
the
N (t ) =
So, N(t)= 1000 /(1 + 49e
N(t)= 0.9 × 1000
−t
⇔
C = 49 .
).
⇔
⇔
⇔
⇔
1000 /(1 + 49e − t ) =900
10
−t
= 1 + 49e
9
e t =441
t =ln(441)= 6.089 days.
PS1 §2.4#29 In Exercise 33 of Section 2.2, the time of death of a murder victim is determined
using Newton’s law of cooling. In particular it was discovered that the proportionally constant in
Newton’s law was k=ln(5/4)≈0.223. Suppose we discover another murder victim at midnight with a
body temperature of 31oC. However, this time the air temperature at midnight is 0oC, and is falling
at a constant rate of 1oC per hour. At what time did the victim die? (Remember that the normal
body temperature is 37oC.)
The model equation is T ′ = − k (T + t ) . At midnight T(0)=31.
Rewrite the differential equation as
DIFFERENTIAL EQUATIONS
Page 8
MATH 315 PS #1
Summer 2010
T ′ + kT = − kt
Multiplying
equation becomes
both
sides
by
an
integrating
factor,
u(t)= e ∫
⇔
e kt T ′ + k e kt T = − kte kt
(e kt T )′ = − kte kt
⇔
e kt T = − ∫ t (ke kt )dt = − ∫ td (e kt ) = − (te kt − ∫ e kt dt ) = − te kt +
kdt
kt
=e ,
the
1 kt
e +C
k
1
− t + Ce − kt
k
1
1
+ C ⇔ C = 31 − .
At midnight 31=T(0)=
k
k
1
1 − kt
So, T (t ) = − t + (31 − )e .
k
k
5
1
1
At t = t o , we have T (t o ) =37
⇔
− t o + (31 − )e − kto =37, where k = ln
k
k
4
1
1
⇔
− t o + (31 −
)e −to ln( 5 / 4) =37
ln(5 / 4)
ln(5 / 4)
4
4.48 − t o + 26.52 × ( ) to =37
⇔
5
The approximation of t o is -0.8022
⇔
T (t ) =
Thus, the time of death is approximately 11:12PM.
DIFFERENTIAL EQUATIONS
Page 9
MATH 315 PS #1
Summer 2010
§2.5. Mixing Problems
#1, 4, 6
PS1 §2.5#1 A tank contains 100 gal of pure water. At time zero, a sugar-water solution containing
0.2lb of sugar per gal enters the tank at a rate of 3 gal per minute. Simultaneously, a drain is opened
at the bottom of the tank allowing the sugar solution to leave the tank at 3 gal per minute. Assume
that the solution in the tank is kept perfectly mixed at all times.
(a) What will be the sugar content in the tank after 20 minutes?
(b) How long will it take the sugar content in the tank to reach 15lb?
(c) What will be the eventual sugar content in the tank?
S(t)= the amount of sugar in the tank.
The rate in = 3gal/min × 0.2 lb/gal = 0.6 lb/min.
The rate out = 3gal/min ×
S
lb/gal = 0.03S lb/min.
100
Hence,
⇔
dS
=rate in – rate out= 0.6 – 0.03 S
dt
S ′ + 0.03S =0.6
Multiplying both sides by an integrating factor, u(t)= e ∫
equation becomes
e 0.03t S ′ + 0.03S e 0.03t = 0.6e 0.03t
⇔
(e 0.03t S )′ = 0.6e 0.03t
⇔
e 0.03t S = 0.6∫ e 0.03t dt = 20 e 0.03t + C
S (t ) = 20 + Ce −0.03t
The initial condition 0= S (0)
0.03 dt
=e
0.03t
, the
⇔
So,
S (t ) = 20 − 20e
−0.03t
⇔ 20 + C = 0
⇔
C = −20
.
−0.6
⇒
S (20) = 20 − 20e ≈ 9.024
(a) The sugar content in the tank after 20 minutes is 9.024 lb
(b)
(c)
S (t ) =15 ⇔
20 − 20e −0.03t =15
⇔
e −0.03t =
1
4
⇔
t=
ln 4
≈ 46.2098 min
0.03
lim S (t ) = lim (20 − 20e −0.03t ) =20-0=20.
t → +∞
t → +∞
PS1 §2.5#4 A tank contains 500 gal of a salt-water solution containing 0.05 lb of salt per gallon of
water. Pure water is poured into the tank and a drain at the bottom of the tank is adjusted so as to
keep the volume of solution in the tank constant. At what rate (gal/min) should the water be poured
into the tank to lower the salt concentration to 0.01 lb/gal of water in under one hour?
x(t ) = the amount of salt in the solution at time t .
r = the rate that water enters (and leaves) the tank.
DIFFERENTIAL EQUATIONS
Page 10
MATH 315 PS #1
Summer 2010
Consider salt in the tank:
⇒
The rate in = 0 gal/min.
The rate out = r gal/min ×
⇔
⇔
x(t )
lb/gal = 0.002rx (t ) lb/min. Hence,
500
dx
=rate in – rate out= – 0.002rx
dt
dx
dx
=– 0.002rx
⇔
=– 0.002rdt
dt
x
ln x(t ) =– 0.002rt + C
⇔
x(t ) = e −0.002 rt + C
The initial condition 0.05 = x (0)
⇔
x(t ) = Ae −0.002 rt
⇔ A = 0.05
−0.002 rt
So, x (t ) = 0.05e
.
The concentration reachs 1% in 60 minutes, so
e −0.12 r = 5 −1
ln(5)
⇔ − 0.12r = − ln(5)
⇔
r=
≈ 13.4 gal/min
0.12
PS1 §2.5#6 A tank initially contains 100 gal of a salt-water solution containing 0.05 lb of salt for
each gallon of water. At time zero, pure water is poured into the tank at a rate of 3 gal/min.
Simultaneously, a drain is opened at the bottom of the tank that allows the salt-water solution to
leave the tank at a rate of 4 gal/min. What will be the salt content in the tank when precisely 50 gal
of salt solution remain?
0.01= x (60)
⇔ 0.05e −0.12 r = 0.01
⇔
The volume of salt solution the tank is decreasing at (4-3)gal/min. So at time
t, the volume of salt solution in the tank is V (t ) = 100 − t .
S (t ) = the amount of salt in the tank. The initial condition S (0)
The rate (Salt) in = 0
The rate (Salt) out = 3 gal/min ×
= 100 × 0.05 = 5 .
S (t )
3S (t ) 3S (t )
lb/gal=
=
V (t )
V (t ) 100 − t
Hence,
⇔
dS
3S
=rate in – rate out= −
dt
100 − t
dS
3dt
=
S t − 100
ln S (t ) = 3 ln t −100 + C
⇔
S (t ) = e
⇔
S (t ) = A(100 − t ) 3
⇔
3 ln t −100 + C
S ( 0)
−6
3
So, S (t ) = 5 × 10 (100 − t ) .
When V (t ) = 50 , we have 100 − t = 50
The initial condition 5 =
⇔ 100 3 A = 5
⇔
t = 50
So, the salt content in the tank at that time
lb.
DIFFERENTIAL EQUATIONS
⇔ A = 5 × 10 −6
t = 50 is S (50) = 5 × 10 −6 × 10 3 =0.625
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