Euler’s Method and Applications
1. For the differential equation
y(0) = 1.
dy
dx
= x2 + sin(x)y 2 , and a given initial value condition
(a) Can you solve this equation?
dy
(b) Let f (x, y) = dx
= x2 + sin(x)y 2 , x0 = 0, y0 = y(x0 ) = 1. Compute f (x0 , y0 ).
What is the meaning of this number.
(c) Set step size h = π2 . Use the Euler’s Method to approximate y(π).
(d) Use the Euler’s Method to approximate y(π) for the step size h = π4 .
2. Let P (t) denote the population of algae in Lake Mendota, where t is measured in years.
Say we want to find a model for the population. Researchers have found that since the
year 2000, the change in population from year to year is proportional to the current
population(with the proporitionality constant called k). Given that there were 104
algae in the lake in 2000 and that the population is know to double every 7 years, find
an explicit equation for P (t) in the following way:
(a) Set up a differential equation describing P . (Hint: This will involve k)
(b) Find the general solution to this differential equation.
(c) Use the initial value to solve for the variable other than k.
(d) Given the initial value and the fact that the population doubles every 7 years,
find k.
3. Rabbits in Madison have a birth rate of 5% per year and a death rate of 2% per year.
Each year 1000 rabbits get run over and 700 rabbits move in from San Prairie. If there
were 12000 rabbits in Madison in 1991, how many rabbits there were in 1994?
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4. Salt mixture A tank initially contains 100 gal of brine in which 50 lb of salt are
dissolved. A brine containing 2 lb/gal of salt runs into the tank at the rate of 5
gal/min. The mixture is kept uniform by stirring and flows out of the tank at the rate
of 4 gal/min.
(a) At what rate (pounds per minute) does salt enter the tank at time t?
(b) What is the volume of brine in the tank at time t?
(c) At what rate (pounds per minute) does salt leave the tank at time t?
(d) Write down and solve the initial value problem describing the mixing process.
(e) Find the concentration of salt in the tank 25 min after the process starts.
5. Mixture problem A 200-gal tank is half full of distilled water. At time t = 0, a
solution containing 0.5 lb/gal of concentrate enters the tank at the rate of 5 gal/min,
and the well-stirred mixture is withdrawn at the rate of 3 gal/min.
(a) At what time will the tank be full?
(b) At the time the tank is full, how many pounds of concentrate will it contain?
6. Fertilizer mixture A tank contains 100 gal of fresh water. A solution containing 1
lb/gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal/min, and the
mixtures is pumped out of the tank at the rate of 3 gal/min. Find the maximum
amount of fertilizer in the tank and the time required to reach the maximum.
7. Carbon monoxide pollution An executive conference room of a corporation contains
4500 ft3 of air initially free of carbon monoxide. Starting at time t = 0, cigarette smoke
containing 4% carbon monoxide is blown into the room at the rate of 0.3 ft3 /min. A
ceiling fan keeps the air in the room well circulated and the air leaves the room at the
same rate 0.3 ft3 /min. Find the time when the concentration of carbon monoxide in
the room reaches 0.01%.
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Solutions:
1. (a) No.
(b) f (x0 , y0 ) = 0, the slope at the point (0, 0).
(c)
π3
8
+
π
2
+1
(d) −17.56
2. (a)
dP
dt
= kP
(b) P (t) = Cekt
(c) C = 4 ln 10
(d) k =
3.
dP
dt
ln 2
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= (0.05 − 0.02)P − 1000 + 700, P (0) = 12000. Find the solution and compute P (3)
4. (a) 10 lb/min
(b) 100 + t
(c) Let S(t) denote the mass of salt in the tank at time t.
(d)
dS(t)
dt
= 10 −
4S(t)
,
100+t
4S(t)
100+t
lb/min.
S(0) = 50.
(e) 188.56 lb
5. (a) 50 min
M
(b) eqn: dM
= 0.05 ∗ 5 − 200+2t
∗ 3 where M (t) denotes the mass of the concentrate
dt
at time t. Initial value: M (0) = 0. Answer: M (50) = 191.13 lb.
F
6. Let F (t) denote the mass of fertilizer in the time at time t. Eqn: dF
= 1 − 100−2t
∗ 3,
dt
1.5
initial value F (0) = 0. Solutions: F (t) = (100 − 2t) − 0.1(100 − 2t) . So, we need to
find t to make dF/dt = 0. When t = 27.78, F (t) gets its maximum 14.81.
S(t)
7. Let S(t) be the volume of cigarette smoke at time t. Eqn: dS
= 0.04 ∗ 0.3 − 4500
∗ 0.3,
dt
−1
−15000 t
initial value S(0) = 0. Solution: S(t) = 180(1 − e
). We can get: when
t = 37.55, F (t) = 4500 ∗ 0.01% = 0.45.
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