MA1S12: SOLUTIONS TO TUTORIAL 6
1. Find the first three approximations given by Euler’s method with increment 4 x = 0 .
1 for the initial value problem y
0
= x
2
√ y y (0) = 1
Solution: From the initial condition we have x
0
= 0 , y
0
= y ( x
0
) = 1
The sample points are x
1
= x
0
+ 4 x = 0 .
1 x
2
= x
1
+ 4 x = 0 .
2 x
3
= x
2
+ 4 x = 0 .
3
The approximations y
1
, y
2 and y
3 are given by the formula y n +1
= y n
+ f ( x n
, y n
) 4 x where we write f ( x, y ) = x 2
√ y .
y
1
= y
0
+ f ( x
0
, y
0
) 4 x
= 1 + f (0 , 1)(0 .
1)
= 1 + (0)
2
√
1(0 .
1)
= 1
1

2 MA1S12: SOLUTIONS TO TUTORIAL 6 y
2
= y
1
+ f ( x
1
, y
1
) 4 x
= 1 + f (0 .
1 , 1)(0 .
1)
= 1 + (0 .
1)
2
√
1(0 .
1)
= 1 .
001 y
3
= y
2
+ f ( x
2
, y
2
) 4 x
= 1 .
001 + f (0 .
2 , 1 .
001)(0 .
1)
= 1 .
001 + (0 .
2)
2
√
1 .
001(0 .
1)
= 1 .
005
2. The particular solution for the ODE in Q.1. is y ( x ) = x
3
+ 1
6
2
Compute the absolute error for the Euler approximations in Q.1.
Solution: Using the exact solution y ( x ) we have y ( x
1
) = y (0 .
1) =
(0 .
1) 3
+ 1
6
2
= 1 .
0003 y y
(
( x x
2
3
) =
) = y y
(0
(0
.
.
2) =
3) =
(0 .
2) 3
+ 1
6
2
= 1 .
0027
(0 .
3) 3
+ 1
6
2
= 1 .
009
The absolute error for the n th approximation is | y ( x n
) − y n
| .

MA1S12: SOLUTIONS TO TUTORIAL 6 n x n y ( x n
) y n
0 0 1 1
| y ( x n
) − y n
|
0
1 0.1 1.0003 1 0.0003
2 0.2 1.0027 1.001 0.0017
3 0.3 1.009
1.005 0.004
3
3. A petri dish contains bacteria with initial population 100. After 2 hours the population is 110. Assuming an exponential growth model find the population of the bacteria after 5 hours.
Solution: Let y ( t ) be the population of the bacteria at time t . Assuming an exponential growth model we have y ( t ) = Ae kt where A and k are both constants. We can find A using the initial condition y (0) = 100.
y (0) = Ae k (0)
= A = 100
= ⇒ y ( t ) = 100 e kt
We can find the growth constant k using the condition y (2) = 110.
y (2) = 100 e k (2)
= 110
= ⇒ e
2 k
=
110
100
= ⇒ 2 k = ln
110
100
= ⇒ k =
1
2 ln
110
100
≈ 0 .
05
= ⇒ y ( t ) ≈ 100 e
0 .
05 t

4 MA1S12: SOLUTIONS TO TUTORIAL 6
After 5 hours the population is approximately y (5) ≈ 100 e
0 .
05(5) ≈ 128
4. A mixing tank contains 300 gallons of water in which 100lbs of salt has been dissolved. Another mixture of salt and water is pumped into the tank at a rate of 2 gal/min. The concentration of salt in this inflow is 1lb per gallon. The resulting mixture is pumped out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank after 10 minutes.
Solution: Let y ( t ) be the amount of salt in the tank at time t . Let
V ( t ) be the volume of the mixture in the tank at time t . Note that in this example the volume is constant
V ( t ) = 300 + 2 t − 2 t = 300
The rate at which salt enters the tank is
(ConcentrationIn)(FlowIn) = 1(2) = 2
The rate at which salt leaves the tank is
(ConcentrationOut)(FlowOut) = y ( t )
V ( t )
(2) =
2
300 y ( t )
The difference between these two values is the rate of change of the amount of salt in the tank. We can model this with a first order linear ODE y
0
( t ) = 2 −
2
300 y ( t )
In standard form we have y
0
( t ) +
2
300 y ( t ) = 2

MA1S12: SOLUTIONS TO TUTORIAL 6
Let p ( t ) =
2
300
=
1
150 and q ( t ) = 2. Then
Z p ( t ) dt =
Z 1
150 dt = t
150
+ constant
The integrating factor is
µ ( t ) = e
R p ( t ) dt
= e t
150
The general solution is y ( t ) =
1
µ ( t )
Z
= e
− t
150
Z
µ ( t ) q ( t ) dt
2 e t
150 dt t
= e
−
150 h
2 (150) e t
150
+ c i t
= 300 + e
−
150 c
Using the initial condition y (0) = 100 we have y (0) = 300 + e
− 0
150 c = 100
= ⇒ 300 + c = 100
= ⇒ c = − 200
The particular solution is y ( t ) = 300 − 200 e
− t
150
After 10 minutes the amount of salt in the tank is y (10) = 300 − 200 e
− 10
150
≈ 112 .
9 lbs
5