Applied Numerical Methods – Solutions Manual | Complete Worked Solutions and Explanations

Solutions Manual
to accompany

Applied Numerical Methods
With MATLAB for Engineers and Scientists


CHAPTER 1
1.1 You are given the following differential equation with the initial condition, v(t = 0) = 0,

dv cd
g v2
dt m

Multiply both sides by m/cd

m dv m
 g  v2
cd dt cd

Define a  mg / cd

m dv
 a2  v2
cd dt

Integrate by separation of variables,

dv cd
a 2
v 2
  m dt
A table of integrals can be consulted to find that
dx 1 x
 tanh1
a 2
 x2 a a

Therefore, the integration yields

1 v cd
tanh 1  t C
a a m

If v = 0 at t = 0, then because tanh–1(0) = 0, the constant of integration C = 0 and the solution
is
1 v cd
tanh 1  t
a a m

,This result can then be rearranged to yield

gm  gc 
v  tanh d 
t
 m 
cd  

,
1.2 This is a transient computation. For the period from ending June 1:




Balance = Previous Balance + Deposits – Withdrawals

Balance = 1512.33 + 220.13 – 327.26 = 1405.20

The balances for the remainder of the periods can be computed in a similar fashion as
tabulated below:

Date Deposit Withdrawal Balance
1-May $ 1512.33
$ 220.13 $ 327.26
1-Jun $ 1405.20
$ 216.80 $ 378.61
1-Jul $ 1243.39
$ 350.25 $ 106.80
1-Aug $ 1586.84
$ 127.31 $ 450.61
1-Sep $ 1363.54

1.3 At t = 12 s, the analytical solution is 50.6175 (Example 1.1). The numerical results are:

absolute
step v(12) relative error
2 51.6008 1.94%
1 51.2008 1.15%
0.5 50.9259 0.61%

where the relative error is calculated with

analytical  numerical
absolute relative error  100%
analytical

The error versus step size can be plotted as

2.0%




1.0%


relative error
0.0%
0 0.5 1 1.5 2 2.5

, Thus, halving the step size approximately halves the error.

1.4 (a) The force balance is


dv c'
g v
dt m

Applying Laplace transforms,

g c'
sV  v(0)   V
s m

Solve for
g
V 
v(0) (1)
s(s  c' / m) s  c' / m

The first term to the right of the equal sign can be evaluated by a partial fraction expansion,

g A B
  (2)
s(s  c' / m) s s  c' / m

g A(s  c' / m)  Bs

s(s  c' / m) s(s  c' / m)

Equating like terms in the numerators yields

A B0
c'
g A
m

Therefore,

mg
A
mg B
c' c'

These results can be substituted into Eq. (2), and the result can be substituted back into Eq.
(1) to give

mg / c' mg / c' v(0)
V  
s s  c' / m s  c' / m

Applying inverse Laplace transforms yields

mg mg (c'/ m)t
v  e  v(0)e (c'/ m)t
c' c'