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Solution Manual for Advanced Engineering Mathematics By Erwin Kreyszig 10th Edition .Latest Edition
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Solution Manual for Advanced Engineering Mathematics By Erwin Kreyszig 10TH EDITION .Latest Edition
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Solution Manuals OfADVANCED ENGINEERINGMATHEMATICS
By
ERWIN KREYSZIG
10TH EDITION
,Part A. ORDINARY DIFFERENTIAL
EQUATIONS (Odes)
CHAPTER 1 First-Order Odes
Major Changes
There Is More Material On Modeling In The Text As Well As In The
Problem Set. Some Additions On Population Dynamics Appear In Sec.
1.5.
Electric Circuits Are Shifted To Chap. 2, Where Second-Order Odes Will Be
Available. This Avoids Repetitions That Are Unnecessary And Practically Irrelevant.
Team Projects, CAS Projects, And CAS Experiments Are Included In Most Problem Sets.
SECTION 1.1. Basic Concepts. Modeling, Page 2
Purpose. To Give The Students A First Impression What An ODE Is And What We Mean By
Solving It.
Background Material. For The Whole Chapter We Need Integration Formulas And
Techniques, Which The Student Should Review.
General Comments
This Section Should Be Covered Relatively Rapidly To Get Quickly To The Actual
Solution Methods In The Next Sections.
Equations (1)–(3) Are Just Examples, Not For Solution, But The Student Will See
That Solutions Of (1) And (2) Can Be Found By Calculus, And A Solution Y = Ex Of (3)
By Inspection.
Problem Set 1.1 Will Help The Student With The Tasks
Of Solving Y' = Ƒ(X) By Calculus
Finding Particular Solutions From Given General Solutions
Setting Up An ODE For A Given Function As Solution
Gaining A First Experience In Modeling, By Doing One Or Two
Problems Gaining A First Impression Of The Importance Of Odes
Without Wasting Time On Matters That Can Be Done Much Faster, Once Systematic
Methods Are Available.
Comment On “General Solution” And “Singular Solution”
Usage Of The Term “General Solution” Is Not Uniform In The Literature. Some Books
Use The Term To Mean A Solution That Includes All Solutions, That Is, Both The
Particular And The Singular Ones. We Do Not Adopt This Definition For Two Reasons.
First, It Is Frequently Quite Difficult To Prove That A Formula Includes All Solutions;
Hence, This Definition Of A General Solution Is Rather Useless In Practice. Second,
Linear Differential Equations (Satisfying Rather General Conditions On The Coefficients)
Have No Singular Solutions (As Mentioned In The Text), So That For These Equations A
General Solution As Defined Does Include All Solutions. For The Latter Reason, Some
Books Use The Term “General Solution” For Linear Equations Only; But This Seems
Very Unfortunate.
1
,2 Instructor’s Manual
SOLUTIONS TO PROBLEM SET 1.1, Page 8
2. Y = —E—3x/3 + C 4. Y = (Sinh 4x) /4 + C
6. Second Order. 8. First Order.
10. Y = Ce0.5x, Y(2) = Ce = 2, C = 2/E, Y = (2/E)E0.5x = 0.736e0.5x
12. Y = Cex + X + 1, Y(0) = C + 1 = 3, C = 2, Y = 2ex + X + 1
14. Y = C Sec X, Y(0) = C/Cos 0 = C = 1_π, Y = 1_π Sec X
2 2
16. Substitution Of Y = Cx — C2 Into The ODE Gives
Y'2 — Xy' + Y = C2 — Xc + (Cx — C2) = 0.
Similarly,
Y = 1_x 2, Y' = 1_x, _1 x 2 — X(_1 x) + 1_x 2 = 0.
Thus
4 2 4 2 4
18. In Prob. 17 The Constants Of Integration Were Set To Zero. Here, By Two Integrations,
Y ” = G, V = Y' = Gt + C1, Y = _1gt 2 + C1t + C2, Y(0) = C2 = Y0,
2
And, Furthermore,
V(0) = C1 = V0, Hence Y = 2_1 gt 2 + V0 t + Y0,
As Claimed. Times Of Fall Are 4.5 And 6.4 Sec, From T = √¯
100/4.9̄ And √2
¯00/4.9̄.
20. Y' = Ky. Solution Y = Y0ekx, Where Y0 Is The Pressure At Sea Level X = 0. Now
Y(18000) = Y Ek·18000 = _1 y (Given). From This,
0 2 0
E k·18000
= 2_1 , Y(36000) = Y0 Ek·2·18000 = Y0 (Ek·18000)2 = 0Y 2(_1 )2 =4 _10y .
22. For 1 Year And Annual, Daily, And Continuous Compounding We Obtain The Values
Ya(1) = 1060.00, Yd(1) = 1000(1 + 0.06/365)365 = 1061.83,
Yc(1) = 1000e0.06 = 1061.84,
Respectively. Similarly For 5 Years,
Ya(5) = 1000 · 1.065 = 1338.23, Yd(5) = 1000(1 + 0.06/365)365·5 = 1349.83,
Yc(5) = 1000e0.06·5 = 1349.86.
We See That The Difference Between Daily Compounding And Continuous
CompoundingIs Very Small.
The ODE For Continuous Compounding Is Yc' = R Yc.
SECTION 1.2. Geometric Meaning Of Y' = Ƒ(X, Y ). Direction Fields, Page 9
Purpose. To Give The Student A Feel For The Nature Of Odes And The General
Behavior Of Fields Of Solutions. This Amounts To A Conceptual Clarification Before
Entering Into Formal
Manipulations Of Solution Methods, The Latter Being Restricted To Relatively Small—
Albeit Important—Classes Of Odes. This Approach Is Becoming Increasingly Important,
Especially Because Of The Graphical Power Of Computer Software. It Is The Analog
Of Conceptual Studies Of The Derivative And Integral In Calculus As Opposed To
Formal Techniques Of Differentiation And Integration.
Comment On Isoclines
These Could Be Omitted Because Students Sometimes Confuse Them With Solutions. In
The Computer Approach To Direction Fields They No Longer Play A Role.
, Instructor’s Manual 3
Comment On Order Of Sections
This Section Could Equally Well Be Presented Later In Chap. 1, Perhaps After One
Or Two Formal Methods Of Solution Have Been Studied.
SOLUTIONS TO PROBLEM SET 1.2, Page 11
2. Semi-Ellipse X2/4 + Y2/9 = 13/9, Y > 0. To Graph It, Choose The Y-Interval
LargeEnough, At Least 0 ÷ Y ÷ 4.
4. Logistic Equation (Verhulst Equation; Sec. 1.5). Constant Solutions Y = 0 And Y =2 _1 .
_
1 _1
For These, Y' = 0. Increasing Solutions For 0 < Y(0) < 2, Decreasing For Y(0) > 2.
6. The Solution (Not Of Interest For Doing The Problem) Is Obtained By Using
Dy/Dx = 1/(Dx/Dy) And Solving Dx/Dy = 1/(1 + Sin Y) By Integration,
X + C = —2/(Tan2 _1 Y + 1); Thus Y = —2 Arctan ((X + 2 + C) /(X + C)).
8. Linear ODE. The Solution Involves The Error Function.
12. By Integration, Y = C — 1/X.
16. The Solution (Not Needed For Doing The Problem) Of Y' = 1/Y Can Be
Obtained By Separating Variables And Using The Initial Condition; Y 2/2 = T + C,
Y = √2¯ T — 1.
18. The Solution Of This Initial Value Problem Involving The Linear ODE Y' + Y = T2 Is
Y = 4e—T + T2 — 2t + 2.
20. CAS Project. (A) Verify By Substitution That The General Solution Is Y = 1 +
Ce—X. Limit Y = 1 (Y(X) = 1 For All X), Increasing For Y(0) < 1,
Decreasing For Y(0) > 1.
(b) Verify By Substitution That The General Solution Is X4 + Y4 = C. More
“Square- Shaped,” Isoclines Y = Kx. Without The Minus On The Right You Get
“Hyperbola-Like” Curves Y4 — X4 = Const As Solutions (Verify!). The Direction
Fields Should Turn Out InPerfect Shape.
(c) The Computer May Be Better If The Isoclines Are Complicated; But The
Computer May Give You Nonsense Even In Simpler Cases, For Instance When Y(X)
Becomes Imaginary. Much Will Depend On The Choice Of X- And Y-Intervals, A
Method Of Trial And Error. Isoclines May Be Preferable If The Explicit Form Of
The ODE Contains Roots On The Right.
SECTION 1.3. Separable Odes. Modeling, Page 12
Purpose. To Familiarize The Student With The First “Big” Method Of Solving Odes,
The Separation Of Variables, And An Extension Of It, The Reduction To Separable Form
By A Transformation Of The ODE, Namely, By Introducing A New Unknown
Function.
The Section Includes Standard Applications That Lead To Separable Odes, Namely,
1. The ODE Giving Tan X As Solution
2. The ODE Of The Exponential Function, Having Various Applications, Such As In
Radiocarbon Dating
3. A Mixing Problem For A Single Tank
4. Newton’s Law Of Cooling
5. Torricelli’s Law Of Outflow.
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