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Solutions Manual
to accompany
Applied Numerical Methods
With MATLAB for Engineers and Scientists
Steven C. Chapra
Tufts University
CHAPTER 1
1.1 You are given the following differential equation with the initial condition, v(t = 0) = 0,
dv c
= g − d v2
dt m
Multiply both sides by m/cd
m dv m
= g − v2
c d dt c d
Define a = mg / c d
m dv
= a2 − v2
c

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Solutions Manual
to accompany
Applied Numerical Methods
With MATLAB for Engineers and Scientists
Steven C. Chapra
Tufts University
1
CHAPTER 1
1.1
You are given the following differential equation with the initial condition,
v
(
t
= 0) = 0,
2
vmcgdt dv
d
−=
Multiply both sides by
m
/
c
d
2
vgcmdt dvcm
d d
−=
Define
d
cmga
/
=
22
vadt dvcm
d
−=
Integrate by separation of variables,
dt mcvadv
d
∫∫
=−
22
A table of integrals can be consulted to find that
a xa xadx
122
tanh1
−
=−
∫
Therefore, the integration yields
C t mcava
d
+=
−
1
tanh1 If
v
= 0 at
t
= 0, then because tanh
–1
(0) = 0, the constant of integration
C
= 0 and the solution is
t mcava
d
=
−
1
tanh1 This result can then be rearranged to yield
⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ =
t mgccgmv
d d
tanh
1.2
This is a transient computation. For the period from ending June 1:
2Balance = 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 WithdrawalBalance
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:
step v(12) absolute 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
%100analyticalnumericalanalytical error relativeabsolute
×−=
The error versus step size can be plotted as
0.0%1.0%2.0%0 0.5 1 1.5 2 2.5relative error
Thus, halving the step size approximately halves the error.
1.4
(a)
The force balance is
3
vmcgdt dv
'
−=
Applying Laplace transforms,
V mcsgvsV
')0(
−=−
Solve for
mcsvmcss
gV
/')0()/'(
+++=
(1) The first term to the right of the equal sign can be evaluated by a partial fraction expansion,
mcs Bs Amcss
g
/')/'(
++=+
(2) )/'(
)/'(
)/'(
mcss
Bsmcs A
mcss
g
+++=+
Equating like terms in the numerators yields
Amcg B A
'0
==+
Therefore, ' '
cmg Bcmg A
−==
These results can be substituted into Eq. (2), and the result can be substituted back into Eq. (1) to give
mcsvmcscmgscmgV
/')0(/''/'/
+++−=
Applying inverse Laplace transforms yields
t mct mc
evecmgcmgv
)/'()/'(
)0(''
−−
+−=
or

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