Solution Manual For Applied Numerical Methods with MATLAB: for Engineers & Scientists 3rd Edition by Steven Chapra . Latest 2025.

Solution Manual For
Applied Numerical Methods W/MATLAB: for Engineers & Scientists
Author: Steven Chapra
3rd Edition




1

,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 − V 2
=
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 C
Tanh −1 = d 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 C
Tanh −1 = d 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:




2

, 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



3

, Dv C'
=G− V
Dt M

Applying Laplace Transforms,
G C'
Sv − V(0) = − V
S M

Solve For

V= +
V(0) (1)
G
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' / S S + C' /
M) M

G A(S + C' / M) +
=
S(S + C' / Bs S(S + C' /
M) M)

Equating Like Terms In The Numerators Yields

A+ B=0
C'
G=
AM

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 + C' / M S + C' /
S M

Applying Inverse Laplace Transforms Yields

Mg Mg −(C'/ M)T
V= − E + V(0)E −(C'/ M)T
C' C'


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