solutions manual to Advanced Engineering Mathematics with MATLAB, 5th Edition. Duffy

AllChaptersCovered
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SOLUTIONS

,TableofContents e e




Chapter 1: First-Order Ordinary Differential Equations 1
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Chapter 2: Higher-Order Ordinary Differential Equations
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Chapter 3: Linear Algebra
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Chapter 4: Vector Calculus
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Chapter 5: Fourier Series Chapter
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6: The Fourier Transform
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Chapter7:TheLaplaceTransform
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Chapter 8: The Wave Equation
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Chapter 9: The Heat Equation
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Chapter 10: Laplace’s Equation
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Chapter11:The Sturm-LiouvilleProblem
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Chapter 12: Special Functions
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Appendix A: Derivation of the Laplacian in Polar Coordinates Appendix B:
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Derivation of the Laplacian in Spherical Polar Coordinates
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, Solution Manual e




Section 1.1 e




1. first-order, linear
e e 2. first-order, nonlinear
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3. first-order, nonlinear
e e 4. third-order, linear
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5. second-order, linear
e e 6. first-order, nonlinear
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7. third-order, nonlinear
e e 8. second-order, linear
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9. second-order, nonlinear
e e 10. first-order, nonlinear e e




11. first-order, nonlinear
e e 12. second-order, nonlinear e e




13. first-order, nonlinear
e e 14. third-order, linear e e




15. second-order, nonlinear
e e 16. third-order, nonlinear e e




Section 1.2 e




1. Because the differential equation can be rewritten e−y dy = xdx, integra-
e e e e e e e e e e e




tion immediately gives —e−y = 21x2 —C, or y = —ln(C —x2/2).
e e e e e e e e e e e e e e e e




2. Separating variables, we have that dx/(1 + x2) = dy/(1 + y2). Integrating this
e e e e e e e e e e e e e




equation, we find that tan−1(x) tan— − (y) = tan(C), or (x y)/(1+xy)
—
1
e e e = C. e e e e e e e e e e




3. Because the differential equation can be rewritten ln(x)dx/x = y dy, inte-
e e e e e e e e e e e




gration immediately gives2 1 ln2(x) + C = 12y2, or y2(x) — ln2(x) = 2C.
e e e e e e e e e e e e e e e e




4. Because the differential equation can be rewritten y2 dy = (x + x3)dx,
e e e e e e e e e e e e e




integration immediately gives y3(x)/3 = x2/2 + x4/4 + C.
e e e e e e e e e e




5. Because the differential equation can be rewritten y dy/(2+y2) = xdx/(1+
e e e e e e e e e e



x2), integration immediately gives 1 ln(2 + y2) = 1 ln(1 + x2) + 1 ln(C), or
e e e e e e e e e e e e e e e e


2 2 2
2 2
2 + y (x) = C(1 + x ).
e e e e e e




6. Because the differential equation can be rewritten dy/y1/3 = x1/3
e e e e e e e e e




3 2/3 3 4/3 3
e

1 4/3 3/2 e


dx, integration immediately gives
e e
2 y4 = x e e e e e e e
2 + C, or y(x) =
e e e e e e
2 x +C e
. e e




1

, 2 Advanced Engineering Mathematics with MATLAB e e e e




7. Because the differential equation can be rewritten e−y dy = ex dx, integra-
e e e e e e e e e e e e




tion immediately gives —e−y = ex —C, or y(x) = —ln(C —ex).
e e e e e e e e e e e e e e e




8. Because the differential equation can be rewritten dy/(y2 + 1) = (x3 +
e e e e e e e e e e e e




5)dx, integration immediately gives tan−1(y) = 1x4 + 5x + C, or y(x) =
e e e e e e e e e e e e e e e e

e 4 e


tan 41 x4 + 5x + C . e
e e e e




9. Because the differential equation can be rewritten y2 dy/(b — ay3) = dt,
e e e e
e
e e e e e e e


y
integration immediately gives ln[b — ay 3 ] y0 = —3at, or 3 —b)/(ay30 —b) =
e e e e e e ee e e
e e e e



(ay
e

e




e−3at. e




10. Because the differential equation can be written du/u = dx/x2, integra-
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tion immediately gives u = Ce−1/x or y(x) = x + Ce−1/x.
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11. From the hydrostatic equation and ideal gas law, dp/p—=
e e e e e e e e e g
dz/(RT ). Substituting for T (z),
e e e e e e


dp g
=— dz. e e


p R(T 0 —Γz) e
e




Integrating from 0 to z, e e e e




p(z) g T0 —Γz p(z) T0 — Γ z g/(RΓ)
e e e e e e
e e
e e e


ln = ln =
e e


, or .
p0 RΓ T0 e p0 T0


12. For 0 < z < H, we simply use the previous problem. At z = H,
e e e e e e e e e e e e e e e




ethe pressure ise e




T0 —ΓH g/(RΓ) e

e
e
e

e


p(H) = p0 . e e


T0
Then we follow the example in the text for an isothermal atmosphere for
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z ≥ H.
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13. Separating variables, we find that e e e e




dV dV R dV dt
— =—
e



= e e e e .
V e + RV 2/Se e e V e e S(1 + RV/S) e e e RC e




Integration yields e




e e e


V t
=—
e e e


ln e e + ln(C).e


1 + RV/Se e RC

Upon applying the initial conditions,
e e e e




V0 RV0/S
e−t/(RC) + e−t/(RC)V (t).
e e e e

V (t) = e e e
e e
e e


1 + RV0/S
e e1 + RV0/S e e