SOLUTIONS MANUAL
** All Chapters solutions
,Table of Contents are given below
1. Mathematical Modeling, Numerical Methods, and Problem Solving
2. MATLAB Fundamentals
3. Programming with MATLAB
4. Roundoff and Truncation Errors
5. Roots: Bracketing Methods
6. Roots: Open Methods
7. Optimization
8. Linear Algebraic Equations and Matrices
9. Gauss Elimination
10. LU Factorization
11. Matrix Inverse and Condition
12. Iterative Methods
13. Eigenvalues
14. Linear Regression
15. General Linear Least-Squares and Nonlinear Regression
16. Fourier Analysis
17. Polynomial Interpolation
18. Splines and Piecewise Interpolation
19. Numerical Integration Formulas
20. Numerical Integration of Functions
21. Numerical Differentiation
22. Initial-Value Problems
23. Adaptive Methods and Stiff Systems
24. Boundary-Value Problems
, 1
CHAPTER 1
1.1 You are given the following differential equation with the initial condition, v(t = 0) = 0,
dv c
= g − d v2
dt m
Multiply both sides by m/cd
m dv m
= g − v2
cd dt cd
Define a = mg / cd
m dv
= a2 − v2
cd dt
Integrate by separation of variables,
dv cd
a 2
−v 2
= m dt
A table of integrals can be consulted to find that
dx 1 x
a 2
−x 2
=
a
tanh −1
a
Therefore, the integration yields
1 v c
tanh −1 = d t + C
a a m
If v = 0 at t = 0, then because tanh–1(0) = 0, the constant of integration C = 0 and the solution is
1 v c
tanh −1 = d t
a a m
This result can then be rearranged to yield
gm gcd
v= tanh
m
t
cd
, 2
1.2 (a) For the case where the initial velocity is positive (downward), Eq. (1.21) is
dv c
= g − d v2
dt m
Multiply both sides by m/cd
m dv m
= g − v2
cd dt cd
Define a = mg / cd
m dv
= a2 − v2
cd dt
Integrate by separation of variables,
dv cd
a 2
−v 2
= m dt
A table of integrals can be consulted to find that
dx 1 x
a 2
−x 2
=
a
tanh −1
a
Therefore, the integration yields
1 v c
tanh −1 = d t + C
a a m
If v = v0 at t = 0, then
1 v
C= tanh −1 0
a a
Substitute back into the solution
1 v c 1 v
tanh −1 = d t + tanh −1 0
a a m a a
Multiply both sides by a, taking the hyperbolic tangent of each side and substituting a gives,
mg gcd cd
v= tanh t + tanh −1 v0 (1)
cd m mg
(b) For the case where the initial velocity is negative (upward), Eq. (1.21) is
dv c
= g + d v2
dt m
** All Chapters solutions
,Table of Contents are given below
1. Mathematical Modeling, Numerical Methods, and Problem Solving
2. MATLAB Fundamentals
3. Programming with MATLAB
4. Roundoff and Truncation Errors
5. Roots: Bracketing Methods
6. Roots: Open Methods
7. Optimization
8. Linear Algebraic Equations and Matrices
9. Gauss Elimination
10. LU Factorization
11. Matrix Inverse and Condition
12. Iterative Methods
13. Eigenvalues
14. Linear Regression
15. General Linear Least-Squares and Nonlinear Regression
16. Fourier Analysis
17. Polynomial Interpolation
18. Splines and Piecewise Interpolation
19. Numerical Integration Formulas
20. Numerical Integration of Functions
21. Numerical Differentiation
22. Initial-Value Problems
23. Adaptive Methods and Stiff Systems
24. Boundary-Value Problems
, 1
CHAPTER 1
1.1 You are given the following differential equation with the initial condition, v(t = 0) = 0,
dv c
= g − d v2
dt m
Multiply both sides by m/cd
m dv m
= g − v2
cd dt cd
Define a = mg / cd
m dv
= a2 − v2
cd dt
Integrate by separation of variables,
dv cd
a 2
−v 2
= m dt
A table of integrals can be consulted to find that
dx 1 x
a 2
−x 2
=
a
tanh −1
a
Therefore, the integration yields
1 v c
tanh −1 = d t + C
a a m
If v = 0 at t = 0, then because tanh–1(0) = 0, the constant of integration C = 0 and the solution is
1 v c
tanh −1 = d t
a a m
This result can then be rearranged to yield
gm gcd
v= tanh
m
t
cd
, 2
1.2 (a) For the case where the initial velocity is positive (downward), Eq. (1.21) is
dv c
= g − d v2
dt m
Multiply both sides by m/cd
m dv m
= g − v2
cd dt cd
Define a = mg / cd
m dv
= a2 − v2
cd dt
Integrate by separation of variables,
dv cd
a 2
−v 2
= m dt
A table of integrals can be consulted to find that
dx 1 x
a 2
−x 2
=
a
tanh −1
a
Therefore, the integration yields
1 v c
tanh −1 = d t + C
a a m
If v = v0 at t = 0, then
1 v
C= tanh −1 0
a a
Substitute back into the solution
1 v c 1 v
tanh −1 = d t + tanh −1 0
a a m a a
Multiply both sides by a, taking the hyperbolic tangent of each side and substituting a gives,
mg gcd cd
v= tanh t + tanh −1 v0 (1)
cd m mg
(b) For the case where the initial velocity is negative (upward), Eq. (1.21) is
dv c
= g + d v2
dt m
