Solution Manual for Advanced Mathematics for Engineering Students The Essential Toolbox 1st Edition By Bre
Lewis, Nihan Onder, Andrew
, Solutions Manual
Brent J. Lewis, E. Nihan Onder and Andrew A. Prudil
,Solution Manual for Advanced Mathematics for Engineering Students The Essential Toolbox 1st Edition By Brent J. Lewis, Nihan
Onder, Andrew
Copyright c⃝ 2022 Brent J. Lewis, E. Nihan Onder and Andrew A. Prudil
All rights reserved
Solution Manual for Advanced Mathematics for Engineering Students The Essential Toolbox 1st Edition By Brent J. Lewis, Nihan
Onder, Andrew
, Problems Chapter 2
2.1 Consider the nonhomogeneous system of first-order, linear differential
equations y1J = y2 + cosh t and y2J = y1, with boundary conditions y1(0) = 0
and y2(0) = − 12 . Solve this initial value problem by the following three
methods:
(a) Matrix methods. Use the method of variation of parameters to deter-
mine the particular solution.
(b) Laplace transform methods.
(c) Convert the two first-order differential equations into a single second-
order differential equation and solve this latter equation.
Solution
(a)
y1J =0 · y1 + 1 · y2 + cosh t
y2J =1 · y1 + 0 · y2 + 0
y 1J 0 1 y1 cosh t y1(0) 0
Therefore, = + and = .
y2J 1 0 y2 0 y2(0) − 21
0 1
Homogeneous equation: yJ = y⇒
1 0
−λ 1
Characteristic equation: det(A — λI) = = λ2 − 1 = 0.
1 −λ
Therefore, λ2 = 1 ⇒ Eigenvalues: λ1 = +1 and λ2 = −1.
The eigenvectors are obtained from: −λ x1 + x2 = 0.
1
For λ 1 = 1 ⇒ x 1 = x2 and we can take .
1
1
For λ2 = −1 ⇒ x1 = −x2 and we can take .
−1
The homogeneous solution is: y(h) = c1x(1)et + c2x(2)e—t, where the eigenvec-
1 1
tors are: x(1) = and x(2) = .
1 −1
Method of Variation of Parameters for y(p):
et e—t cosh t
Y = [y (1) y(2)] = et −e—t and g = .
0
—1 1 y22 −y12 1 −e— −e—
t t 1 e—t e—t
Y = = − = .
detY −y21 y11 2 −e t et 2 et −et
2-1
Lewis, Nihan Onder, Andrew
, Solutions Manual
Brent J. Lewis, E. Nihan Onder and Andrew A. Prudil
,Solution Manual for Advanced Mathematics for Engineering Students The Essential Toolbox 1st Edition By Brent J. Lewis, Nihan
Onder, Andrew
Copyright c⃝ 2022 Brent J. Lewis, E. Nihan Onder and Andrew A. Prudil
All rights reserved
Solution Manual for Advanced Mathematics for Engineering Students The Essential Toolbox 1st Edition By Brent J. Lewis, Nihan
Onder, Andrew
, Problems Chapter 2
2.1 Consider the nonhomogeneous system of first-order, linear differential
equations y1J = y2 + cosh t and y2J = y1, with boundary conditions y1(0) = 0
and y2(0) = − 12 . Solve this initial value problem by the following three
methods:
(a) Matrix methods. Use the method of variation of parameters to deter-
mine the particular solution.
(b) Laplace transform methods.
(c) Convert the two first-order differential equations into a single second-
order differential equation and solve this latter equation.
Solution
(a)
y1J =0 · y1 + 1 · y2 + cosh t
y2J =1 · y1 + 0 · y2 + 0
y 1J 0 1 y1 cosh t y1(0) 0
Therefore, = + and = .
y2J 1 0 y2 0 y2(0) − 21
0 1
Homogeneous equation: yJ = y⇒
1 0
−λ 1
Characteristic equation: det(A — λI) = = λ2 − 1 = 0.
1 −λ
Therefore, λ2 = 1 ⇒ Eigenvalues: λ1 = +1 and λ2 = −1.
The eigenvectors are obtained from: −λ x1 + x2 = 0.
1
For λ 1 = 1 ⇒ x 1 = x2 and we can take .
1
1
For λ2 = −1 ⇒ x1 = −x2 and we can take .
−1
The homogeneous solution is: y(h) = c1x(1)et + c2x(2)e—t, where the eigenvec-
1 1
tors are: x(1) = and x(2) = .
1 −1
Method of Variation of Parameters for y(p):
et e—t cosh t
Y = [y (1) y(2)] = et −e—t and g = .
0
—1 1 y22 −y12 1 −e— −e—
t t 1 e—t e—t
Y = = − = .
detY −y21 y11 2 −e t et 2 et −et
2-1
