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Full Solution Manual - Numerical Methods for Engineers, 7th Edition by Steven Chapra – Step-by-Step Solutions for All Chapters

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Master the mathematical techniques required for engineering excellence with this complete Solution Manual for Numerical Methods for Engineers, 7th Edition by Steven Chapra. This resource provides detailed, step-by-step solutions to complex computational problems, helping you bridge the gap between theoretical math and practical engineering applications. What is Included: • Comprehensive Chapter Coverage: Solutions spanning the entire curriculum, including Curve Fitting (Chapters 18–20), Numerical Integration and Differentiation (Chapters 21–24), Ordinary Differential Equations (Chapters 25–28), and Partial Differential Equations (Chapters 29–32). • Detailed Algorithmic Solutions: Step-by-step breakdowns of essential methods such as Newton’s divided-difference interpolating polynomials, Lagrange polynomials, and Cubic Splines. • Advanced Integration Techniques: Clear guidance on using the Trapezoidal rule, Simpson’s 1/3 and 3/8 rules, Gauss Quadrature, and Romberg Integration. • ODE & PDE Mastery: Expert solutions for Initial-Value Problems using Euler’s method, Heun’s method, and 4th-order Runge-Kutta (RK4), as well as Steady-State PDE solutions via Liebmann’s method and Crank-Nicolson. • Programming Implementations: Includes logic and code snippets for implementing numerical methods in VBA/Excel and MATLAB, covering Discrete Fourier Transforms (DFT) and Fast Fourier Transforms (FFT). • Specialized Engineering Topics: Solutions for the Finite Element Method, Thomas Algorithm for tridiagonal systems, and Power Methods for eigenvalues. This manual is the ultimate companion for engineering students and professionals looking to ace their numerical analysis courses and implement stable, accurate computational models.

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1


CHAPTERV18
18.1 (a)
1.0791812V−V0.90309V
fV(10)V=V0.90309V+V (10V−V8)V=V0.991136
1
12V−V8
1−V0.991136V
V =V 100%V=V0.886%
t
1

(b)
1.0413927V−V0.9542425V
fV(10)V=V0.9542425V+V (10V−V9)V=V0.997818
1
11−V9
1−V0.997818V
V =V 100%V=V0.218%
t
1

18.2 First,VorderVtheVpoints

x0V=V9 f(x0)V=V0.9542425
x1V=V11V f(x1)V=V1.0413927
x2V=V8 f(x2)V=V0.9030900

ApplyingVEq.V(18.4)

b0V=V0.9542425

EquationV(18.5)Vyields

1.0413927V−V0.9542425V
b1V =V =V0.0435751
11−V9

EquationV(18.6)Vgives

0.9030900V−1.0413927V
−V0.0435751
8V−11 0.0461009V−V0.0435751V
b2V =V V
=V =V−0.0025258
8V−V9 8V−V9

SubstitutingVtheseVvaluesVintoVEq.V(18.3)VyieldsVtheVquadraticVformula

f2V(x)V=V0.9542425V+V0.0435751(xV−V9)V−V0.0025258(xV−V9)(xV−11)

whichVcanVbeVevaluatedVatVxV=V10Vfor

f2V(10)V=V0.9542425V+V0.0435751(10V−V9)V−V0.0025258(10V−V9)(10V−11)V=V1.0003434

18.3 First,VorderVtheVpoints

x0V=V9 f(x0)V=V0.9542425
x1V=V11V f(x1)V=V1.0413927
x2V=V8 f(x2)V=V0.9030900
x3V=V12V f(x3)V=V1.0791812

TheVfirstVdividedVdifferencesVcanVbeVcomputedVas

PROPRIETARYVMATERIAL.V ©VTheVMcGraw-
HillVCompanies,VInc.V AllVrightsVreserved.V NoVpartVofVthisVManualVmayVbeVdisplayed,VreproducedVorVdistri
butedVinVanyVformVorVbyVanyVmeans,VwithoutVtheVpriorVwrittenVpermissionVofVtheVpublisher,VorVusedVbey
ondVtheVlimitedVdistributionVtoVteachersVandVeducatorsVpermittedVbyVMcGraw-
HillVforVtheirVindividualVcourseVpreparation.V IfVyouVareVaVstudentVusingVthisVManual,VyouVareVusingVitVwi
thoutVpermission.

, 2



1.0413927V−V0.9542425V
fV[xV ,VxV ]V=V =V0.0435751
1
VV0 11−V9
0.9030900V−1.0413927V
fV[xV ,VxV ]V=V =V0.0461009
2
VV1 8V−V11
1.0791812V−V0.9030900V
fV[xV ,VxV ]V=V =V0.0440228
3
VV2 12V−V8

TheVsecondVdividedVdifferencesVare
0.0461009V−V0.0435751V
fV[xV ,VxV ,VxV ]V=V =V−0.0025258
2V V 1
VV0 8V−V9
0.0440228V−V0.0461009V
fV[xV ,VxV ,VxV ]V=V =V−0.0020781
3V V 2
VV1 12V−V11

TheVthirdVdividedVdifferenceVis

−0.0020781−V(−0.0025258)V
fV[x3VV,VVxV2V,VVxV ,VxV ]V=V =V0.00014924
1V V 0 12V−V9

SubstitutingVtheVappropriateVvaluesVintoVEq.V(18.7)Vgives

f3V(x)V=V0.9542425V+V0.0435751(xV−V9)V−V0.0025258(xV−V9)(xV−11)
+V0.00014924(xV−V9)(xV−11)(xV−V8)

whichVcanVbeVevaluatedVatVxV=V10Vfor

f3V(x)V=V0.9542425V+V0.0435751(10V−V9)V−V0.0025258(10V−V9)(10V−11)
+V0.00014924(10V−V9)(10V−11)(10V−V8)V=V1.0000449

18.4
18.1 V(a):
x0V=V8 f(x0)V=V0.9030900
x1V=V12 f(x1)V=V1.0791812
10V−12V 10V−V8V
fV (10)V=V 0.9030900V+V 1.0791812V=V0.991136
1
8V−12 12V−V8

18.1 (b):
x0V=V9 f(x0)V=V0.9542425
x1V=V11 f(x1)V=V1.0413927
10V−11V 10V−V9V
fV (10)V=V 0.9542425V+V 1.0413927V=V0.997818
1
9V−11 11−V9

18.2 :
x0V=V8 f(x0)V=V0.9030900
x1V=V9 f(x1)V=V0.9542425
x2V=V11V f(x2)V=V1.0413927

PROPRIETARYVMATERIAL.V ©VTheVMcGraw-
HillVCompanies,VInc.V AllVrightsVreserved.V NoVpartVofVthisVManualVmayVbeVdisplayed,VreproducedVorVdistri
butedVinVanyVformVorVbyVanyVmeans,VwithoutVtheVpriorVwrittenVpermissionVofVtheVpublisher,VorVusedVbey
ondVtheVlimitedVdistributionVtoVteachersVandVeducatorsVpermittedVbyVMcGraw-
HillVforVtheirVindividualVcourseVpreparation.V IfVyouVareVaVstudentVusingVthisVManual,VyouVareVusingVitVwi
thoutVpermission.

, 3



(10V−V9)(10V−11)V (10V−V8)(10V−11)V
fV2 (10)V=V 0.9030900V+V 0.9542425
(8V−V9)(8V−11) (9V−V8)(9V−11)
(10V−V8)(10V−V9)V
+V 1.0413927V=V1.0003434
(11−V8)(11−V9)
18.3 :
x0V=V8 f(x0)V=V0.9030900
x1V=V9 f(x1)V=V0.9542425
x2V=V11V f(x2)V=V1.0413927
x3V=V12V f(x3)V=V1.0791812

(10V−V9)(10V−11)(10V−12)V (10V−V8)(10V−11)(10V−12)V
fV3 (10)V=V 0.9030900V+V 0.9542425
(8V−V9)(8V−11)(8V−12) (9V−V8)(9V−11)(9V−12)
(10V−V8)(10V−V9)(10V−12)V (10V−V8)(10V−V9)(10V−11)V
+V 1.0413927V+V 1.0791812V=V1.0000449
(11−V8)(11−V9)(11−12) (12V−V8)(12V−V9)(12V−11)

18.5 First,VorderVtheVpointsVsoVthatVtheyVareVasVcloseVtoVandVasVcenteredVaboutVtheVunknownVasVpossible

x0V=V2.5V f(x0)V=V14
x1V=V3.2V f(x1)V=V15
x2V=V2 f(x2)V=V8
x3V=V4 f(x3)V=V8
x4V=V1.6V f(x4)V=V2

Next,VtheVdividedVdifferencesVcanVbeVcomputedVandVdisplayedVinVtheVformatVofVFig.V18.5,

i xi f(xi) f[xi+1,xi] f[xi+2,xi+1,xi] f[xi+3,xi+2,xi+1,xi] f[xi+4,xi+3,xi+2,xi+1,xi]
0 2.5 14 1.428571 -8.809524 1.011905 1.847718
1 3.2 15 5.833333 -7.291667 -0.651042
2 2 8 0 -6.25
3 4 8 2.5
4 1.6 2

TheVfirstVthroughVthird-orderVinterpolationsVcanVthenVbeVimplementedVas

f1(2.8)V=V14V+1.428571(2.8V−V2.5)V=V14.428571
f2V(2.8)V=V14V+1.428571(2.8V−V2.5)V−V8.809524(2.8V−V2.5)(2.8V−V3.2)V=V15.485714
f3V(2.8)V=V14V+1.428571(2.8V−V2.5)V−V8.809524(2.8V−V2.5)(2.8V−V3.2)
+V1.011905(2.8V−V2.5)(2.8V−V3.2)(2.8V−V2.)V=V15.388571

TheVerrorVestimatesVforVtheVfirstVandVsecond-orderVpredictionsVcanVbeVcomputedVwithVEq.V18.19Vas

R1V=V15.485714V−14.428571V=V1.057143
R2V =V15.388571−15.485714V=V−0.097143

TheVerrorVforVtheVthird-orderVpredictionVcanVbeVcomputedVwithVEq.V18.18Vas

R3V =V1.847718(2.8V−V2.5)(2.8V−V3.2)(2.8V−V2)(2.8V−V4)V=V0.212857

18.6 First,VorderVtheVpointsVsoVthatVtheyVareVasVcloseVtoVandVasVcenteredVaboutVtheVunknownVasVpossible


PROPRIETARYVMATERIAL.V ©VTheVMcGraw-
HillVCompanies,VInc.V AllVrightsVreserved.V NoVpartVofVthisVManualVmayVbeVdisplayed,VreproducedVorVdistri
butedVinVanyVformVorVbyVanyVmeans,VwithoutVtheVpriorVwrittenVpermissionVofVtheVpublisher,VorVusedVbey
ondVtheVlimitedVdistributionVtoVteachersVandVeducatorsVpermittedVbyVMcGraw-
HillVforVtheirVindividualVcourseVpreparation.V IfVyouVareVaVstudentVusingVthisVManual,VyouVareVusingVitVwi
thoutVpermission.

, 4



x0V=V3 f(x0)V=V19
x1V=V5 f(x1)V=V99
x2V=V2 f(x2)V=V6
x3V=V7 f(x3)V=V291
x4V=V1 f(x4)V=V3

Next,VtheVdividedVdifferencesVcanVbeVcomputedVandVdisplayedVinVtheVformatVofVFig.V18.5,

i xi f(xi) f[xi+1,xi] f[xi+2,xi+1,xi] f[xi+3,xi+2,xi+1,xi] f[xi+4,xi+3,xi+2,xi+1,xi]
0 3 19 40 9 1 0
1 5 99 31 13 1
2 2 6 57 9
3 7 291 48
4 1 3

TheVfirstVthroughVfourth-orderVinterpolationsVcanVthenVbeVimplementedVas

f1(4)V=V19V+V40(4V−V3)V=V59
f2V(4)V=V59V+V9(4V−V3)(4V−V5)V=V50
f3V(4)V=V50V +1(4V−V3)(4V−V5)(4V−V2)V=V48
f4V(4)V=V48V+V0(4V−V3)(4V−V5)(4V−V2)(4V−V7)V=V48

ClearlyVthisVdataVwasVgeneratedVwithVaVcubicVpolynomialVsinceVtheVdifferenceVbetweenVthe
V4thVandVtheV3rd-orderVversionsVisVzero.

18.7
FirstVorder:
x0V=V3 f(x0)V=V19
x1V=V5 f(x1)V=V99
4V−V5 4V−V3V
fV (10)V=V 19V+V 99V=V59
1
3V−V5 5V−V3

SecondVorder:
x0V=V3 f(x0)V=V19
x1V=V5 f(x1)V=V99
x2V=V2 f(x2)V=V6
(4V−V5)(4V−V2)V (4V−V3)(4V−V2)V (4V−V3)(4V−V5)V
fV2 (10)V=V 19V+V 99V +V 6V=V50
(3V−V5)(3V−V2) (5V−V3)(5V−V2) (2V−V3)(2V−V5)

ThirdVorder:
x0V=V3 f(x0)V=V19
x1V=V5 f(x1)V=V99
x2V=V2 f(x2)V=V6
x3V=V7 f(x3)V=V291
(4V−V5)(4V−V2)(4V−V7)V (4V−V3)(4V−V2)(4V−V7)V
fV (10)V=V 19V+V 99
3
(3V−V5)(3V−V2)(3V−V7) (5V−V3)(5V−V2)(5V−V7)
(4V−V3)(4V−V5)(4V−V7)V (4V−V3)(4V−V5)(4V−V2)V
+V 6V+V 291V=V48
(2V−V3)(2V−V5)(2V−V7) (7V−V3)(7V−V5)(7V−V2)




PROPRIETARYVMATERIAL.V ©VTheVMcGraw-
HillVCompanies,VInc.V AllVrightsVreserved.V NoVpartVofVthisVManualVmayVbeVdisplayed,VreproducedVorVdistri
butedVinVanyVformVorVbyVanyVmeans,VwithoutVtheVpriorVwrittenVpermissionVofVtheVpublisher,VorVusedVbey
ondVtheVlimitedVdistributionVtoVteachersVandVeducatorsVpermittedVbyVMcGraw-
HillVforVtheirVindividualVcourseVpreparation.V IfVyouVareVaVstudentVusingVthisVManual,VyouVareVusingVitVwi
thoutVpermission.

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