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

t AllChaptersCovered
t t




SOLUTIONS

,TableofContents t t




Chapter 1: First-Order Ordinary Differential Equations 1
t t t t t t




t Chapter 2: Higher-Order Ordinary Differential Equations
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t Chapter 3: Linear Algebra
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Chapter 4: Vector Calculus
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t Chapter 5: Fourier Series Chapter
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6: The Fourier Transform
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Chapter7:TheLaplaceTransform
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t Chapter 8: The Wave Equation
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t Chapter 9: The Heat Equation
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t Chapter 10: Laplace’s Equation
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Chapter11:TheSturm-LiouvilleProblem
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t Chapter 12: Special Functions
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Appendix A: Derivation of the Laplacian in Polar Coordinates AppendixB:
t t t t t t t t t t




DerivationoftheLaplacian in SphericalPolarCoordinates
t t t t t t t t

, Solution Manual t




Section 1.1 t




1. first-order, linear
t t 2. first-order, nonlinear
t t




3. first-order, nonlinear
t t 4. third-order, linear
t t




5. second-order, linear
t t 6. first-order, nonlinear
t t




7. third-order, nonlinear
t t 8. second-order, linear
t t




9. second-order, nonlinear
t t 10. first-order, nonlinear t t




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




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




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




Section 1.2 t




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




tion immediately gives —e−y = 12x2 —C, or y = —ln(C —x2/2).
t t t t t t t t t t t t t t t t




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




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




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




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




4. Because the differential equation can be rewritten y2dy = (x + x3)dx,
t t t t t t t t t t t t t




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




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



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


2 2 2
2 + y2(x) = C(1 + x2).
t t t t t t




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




3 2/3 3 4/3 3
t

1 4/3 3/2 t


dx, integration immediately gives
t t
2 y t t t t
4 t = x t t
2+ t t C, or y(x) =
t t t t
2x +C t
. t t




1

, 2 Advanced Engineering Mathematics with MATLAB t t t t




7. Becausethe differential equation can be rewritten e−ydy =exdx, integra- tion
t t t t t t t t t t t t t




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




8. Because the differential equation can be rewritten dy/(y2 +1) = (x3 + 5) t t t t t t t t t t t t t




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

t 4 t


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




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


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

t
t t t
t t t t




e−3at. t




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




tion immediately gives u = Ce−1/x or y(x) = x + Ce−1/x.
t t t t t t t t t t t t




— g dz/(RT).
11. From the hydrostatic equation and ideal gas law, dp/p = t t t t t t t t t t t




Substituting for T (z),
t t t t


dp g
=— dz. t t


p R(T 0 —Γz) t
t




Integrating from 0 to z, t t t t




p(z) g T0 —Γz p(z) T0 —Γz g/(RΓ)
t t t t t t
t t
t t t


ln = ln =
t t


, or .
p0 RΓ T0 t p0 T0


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




tthe pressure ist t




T0 —ΓH g/(RΓ) t

t
t t
t




p(H) = p0 . t t


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




z ≥H.
t t




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




dV dV R dV dt
— =—
t



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




Integration yields t




t t t


V t
=—
t t t


ln t t + ln(C). t


1+ RV/S t t RC

Upon applying the initial conditions,
t t t t




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

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


1+ RV0/S t t 1+ RV0/S t t