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

All Chapters Covered
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SOLUTIONS

,TableofContents qe qe




Chapter 1: First-Order Ordinary Differential Equations 1
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qe Chapter 2: Higher-Order Ordinary Differential Equations
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qe Chapter 3: Linear Algebra
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Chapter 4: Vector Calculus
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qe Chapter 5: Fourier Series
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q e Chapter 6: The Fourier Transform
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Chapter7:TheLaplaceTransform
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qe Chapter 8: The Wave Equation
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qe Chapter 9: The Heat Equation
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qe Chapter 10: Laplace’s Equation
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Chapter 11: The Sturm-Liouville Problem
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qe Chapter 12: Special Functions
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Appendix A: Derivation of the Laplacian in Polar Coordinates Appendix
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qe B: Derivation of the Laplacian in Spherical Polar Coordinates
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, Solution Manual qe




Section 1.1 qe




1. first-order, linear
q e qe 2. first-order, nonlinear
q e qe




3. first-order, nonlinear
q e qe 4. third-order, linear
q e q e




5. second-order, linear
q e q e 6. q e first-order, nonlinear q e




7. q e third-order, nonlinear q e 8. second-order, linear
q e q e




9. second-order, nonlinear
q e q e 10. q e first-order, nonlinear q e




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




13. first-order, nonlinearq e qe 14. third-order, linear q e qe




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




Section 1.2 qe




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




tion immediately gives —e−y =2 1x2 — C, or y = —ln(C —x2/2).
qe qe qe qe qe qe e
q qe q
e qe qe qe qe e
q qe q
e




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




this equation, we find that —
qe tan−1(x)
qe
—
tan−1(y) = tan(C), or (x qe qe qe qe q e qe qe qe qe




q y)/(1+xy) = C.
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3. Because the differential equation can be rewritten ln(x)dx/x = y dy, inte-
qe qe qe qe qe qe qe qe qe qe qe




gration immediately gives
qe
2
1
ln2(x) + C =
2
y , or y2(x) — ln2(x) = 2C.
1 2
qe qe q e qe qe qe qe q e e
q qe qe qe qe qe qe




4. Because the differential equation can be rewritten y2 dy = (x + q e q e q e q e q e q e q e qe q e q e qe




x3) dx, integration immediately gives y3(x)/3 = x2/2 + x4/4 + C.
qe qe qe qe qe qe qe qe qe qe qe qe




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



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


2 2 2
2 + y2(x) = C(1 + x2).
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6. Because the differential equation can be rewritten dy/y1/3 = q e q e q e q e q e q e q e

q e
q e



1
q x1/3 dx, integration immediately gives 3y2/3 = 3x4/3 + 3C, or y(x) =
e qe qe qe qe q e e
q qe q e e
q qe qe e
q qe qe qe


qe
3/2 2 4 2 2
qe
x4/3 + C qe qe .

1

, 2 Advanced Engineering Mathematics with q e q e q e




qMATLAB e




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




tion immediately gives —e−y = ex — C, or y(x) = —ln(C —ex).
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e qe qe qe qe e
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8. Because the differential equation can be rewritten dy/(y2 + 1) = (x3
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+ 5)dx, integration immediately gives tan−1(y) = 1x4 + 5x + C,
qe qe qe q e q e q e q e q e q e e
q qe qe qe qe




q or y(x) =
e q e q e

q e 4 q e


tan 4 1 x4 + 5x + C . qe
qe qe qe qe




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



3 y
integration immediately gives ln[b — ay ] y0 = —3at, 3 — b)/(ay
q e
3
0 — b) = q e q e qe qe qe
q e q e q e
q
e qe q
e qe



or (ay
q e

q e q e




e−3at. q e




10. Because the differential equation can be written du/u = dx/x2, integra-
qe qe qe qe qe qe qe qe qe qe




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 = g
q e q e q e q e q e q e q e q e qe




dz/(RT ). Substituting for T (z),
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dp g
=— dz. qe e
q


p R(T 0 — Γz) q e
q
e




Integrating from 0 to z, qe qe qe qe




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


ln = ln =
qe qe


, or .
p0 RΓ T0 qe p0 T0


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




T0 —ΓH g/(RΓ) qe

qe
q
e
q
e

q e


p(H) = p0 . qe qe


T0
Then we follow the example in the text for an isothermal atmosphere
q e q e q e q e q e q e q e q e q e q e q e




for
q e




z ≥ H. qe qe




13. Separating variables, we find that qe qe qe qe




dV dV R dV dt
— =— .
qe



= q e qe qe e
q


V q e + RV 2/S
qe qe q e V q e q e S(1 + RV/S) RC qe qe q e qe




Integration yields q e




q e q e q e


V t
=—
q e q e q e


ln qe e
q + ln(C).
qe


1 + RV/S qe qe RC

Upon applying the initial conditions,
q e q e q e q e




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

V (t) = qe qe q e
qe q e
q e qe


1 + RV0/S
qe 1 + RV0/S
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