Download Solutions 1.5-Page 51

Solutions 1.5-Page 51
Problem 27
Solve the differential equation by regarding y as the independent variable rather than x.
( x + ye y )
dy
=1
dx
The differential equation does not yet follow the general form given on pg.43. The
dy
derivative term ( ) is differentiating y with respect to x. Since y is the independent
dx
dy
variable, it must be the other way around. Division by
, and rearranging terms yields
dx
dx
− x = ye y . This equation follows the form given on pg.43. The integrating factor is
dy
−1dy
ρ ( y) = e ∫
= e−y
Multiplying both sides by ρ ( y ) gives
dx
e−y
− e−y x = y
dy
The left hand side is
d (e − y x )
, so the equation becomes
dy
d (e − y x )
=y
dy
Integrating both sides gives
d (e − y x )
∫ dy dy = ∫ ydy
y2
+C
2
Dividing by e − y gives
e−y x =

 y2
x( y ) = e y 
+ C 

 2
Problem 33
A tank contains 1000 liters (L) of a solution consisting of 100 kg of salt dissolved in
water. Pure water is pumped into the tank at the rate of 5 L/s, and the mixture - kept
uniform by stirring - is pumped out at the same rate. How long will it be until only 10 kg
of salt remains in the tank?
Using the terminology consistent with pg.49,
ri = 5
ci = 0
ro = 5
x(t )
V (t )
V (t ) = 1000 − (5 − 5)t = 1000
x(0) = 100
co (t ) =
dx
5x
x
= ri ci − ro co = −
=−
. The differential equation
dt
1000
200
states that the rate of change of salt (x) is equal to the inflow minus the outflow. Since
there is no inflow of salt (pure water is being pumped in) the rate of change of salt is
dt
dx
equal to the outflow. Separating variables yields
. Integrating both
= −
x
200
sides gives x(t).
dx
dt
∫ x = ∫ − 200
−t
+C
ln x =
200
−t
~
x = Ce 200
~
Where C = e C
~
The initial condition is used to find C .
−t
~ 200
x(0) = 100 = Ce
~
C = 100
The differential equation is
x(t ) = 100e
−t
200
x = 10 = 100e
t = 460.517 s
−t
200
t = 7 minutes and 41 seconds
Problem 41
A 30-year-old woman accepts an engineering position with a starting salary of $30000
per year. Her salary S(t) increases exponentially, with S (t ) = 30e t / 20 thousand dollars
after t years. Meanwhile, 12% of her salary is deposited continuously in a retirement
account, which accumulates interest at a continuous annual rate of 6%.
(a) Estimate ∆A in terms of ∆t to derive the differential equation satisfied by the amount
A(t ) in her retirement account after t years. (b) Compute A(40), the amount available for
her retirement at age 70.
(a) Her retirement increases by 6% per year of what is in the account, plus 12% of her
yearly salary. In equation form, this becomes ∆A = 0.06 A∆t + 0.12 S∆t . Dividing by
∆t and taking the limit as time goes to infinite gives the differential equation.
dA
= 0.06 A + 0.12S
dt
Substituting in for S gives
dA
= 0.06 A + 3.6e t / 20
dt
(b) Rearranging the above differential equation gives
dA
− 0.06 A = 3.6e t / 20 . This is a
dt
linear first-order equation.
− 0.06 dt
ρ (t ) = e ∫
= e −0.06t
Multiplying both sides by ρ (t ) gives
dA
e −0.06t
− .06e −0.06t A = 3.6e −0.01t
dt
d (e −0.06t A)
The left hand side is
, so the equation becomes
dt
d (e −0.06t A)
= 3.6e −0.01t
dt
Integrating both sides gives
d (e −0.06t A)
dt = ∫ 3.6e −0.01t dt
∫
dt
− 0.06 t
e
A = −360e −0.01t + C
A(t ) = −360e 0.05t + Ce 0.06t
The initial condition of having nothing in the retirement account is used to solve for C.
A(0) = 0 = −360 + C
∴
C = 360
A(t ) = 360(e 0.06t − e 0.05t )
Substituting t=40 will give the amount in the account when she is 70.
A(40) = $1,308,283