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SOLUTION MANUAL FOR ADVANCED
ENGINEERING MATHEMATICS 10TH
EDITION,2020 BY ERWIN KREYSZIG
.LATEST EDITION 2026
, 2 instructor’s m a n u a l
Solutions to problem set 1.1, page 8
2. Y = —e—3x /3 + c 4. Y = (sinh 4x) /4 + c
6. Second order. 8. First order.
10. Y = ce0.5x , y(2) = ce = 2, c = 2/e, y = (2/e)e0.5x = 0.736e0.5x
12. Y = cex + x + 1, y(0) = c + 1 = 3, c = 2, y = 2ex + x + 1
14. Y = c sec x, y(0) = c/cos 0 = c = _ 1π, y = _ 1π sec x
2 2
16. Substitution of y = cx — c2 into the ode gives
Y'2 — xy ' + y = c2 — xc + (cx — c2 ) = 0.
similarly,
Y=
_ 1x2, y' = _ 1x, thus _1 2
= 40 x. — x(
_1x)+ _1x2
4 2 2 4
18. In prob. 17 the constants of integration were set to zero. Here, by two integrations,
y ” = g, v = y ' = gt + c1 , y = _21gt2 + c1 t + c2 , y(0) = c2 = y 0 ,
And, furthermore,
v(0) = c1 = v 0 , hence y = _ 1 gt 2 + v 0 t + y 0 , 2
As claimed. Times of fall are 4.5 and 6.4 sec, from t = √ 1¯0 . ¯9 and √2¯
0 .9̄ .
20. Y' = ky. Solution y = y 0 e , where y 0 is the pressure at sea level x = 0. Now
kx
y(18000) = y 0 ek·18000 = _21y 0 (given). From this,
Ek·18000 = _ 1, y(36000) = y ek·2·18000 = y (ek·18000 ) 2 = y (_ 1)2 = _ 1y .
0 2 0 0 0 2 4
22. For 1 year and annual, daily, and continuous compounding we obtain the values
Ya(1) = 1060.00, y d (1) = 1000(1 + 0.06/365)365 = 1061.83,
Yc(1) = 1000e0.06 = 1061.84,
Respectively. Similarly for 5 years,
Ya(5) = 1000 · 1.06 5 = 1338.23, y d (5) = 1000(1 + 0.06/365)365·5 = 1349.83,
Yc(5) = 1000e0.06·5 = 1349.86.
We see that the difference between daily compounding and continuous compounding
is very small.
The ode for continuous compounding is y c' = ry c.
Section 1.2. Geometric meaning of y ' = ƒ(x, y ). Direction fields, page 9
Purpose. To give the student a feel for the nature of odes and the general behavior of
fields of solutions. This amounts to a conceptual clarification before entering into formal
manipulations of solution methods, the latter being restricted to relatively small—albeit
important—classes of odes. This approach is becoming increasingly important, especially
because of the graphical power of computer software. It is the analog of conceptual
studies of the derivative and integral in calculus as opposed to formal techniques of
differentiation and integration.
Comment on isoclines
These could be omitted because students sometimes confuse them with solutions. In the
computer approach to direction fields they no longer play a role.
, Instructor’s manual 3
Comment on order of sections
This section could equally well be presented later in chap. 1, perhaps after one or two
formal methods of solution have been studied.
Solutions to problem set 1.2, page 11
2. Semi-ellipse x 2 /4 + y 2 /9 = 13/9, y > 0. To graph it, choose the y-interval large
enough, at least 0 % y % 4.
4. Logistic equation (verhulst equation; sec. 1.5). Constant solutions y = 0 and y = _ 12.
_1
for these, y ' = 0. Increasing solutions for 0 < < _21, decreasing for > 2.
y(0) y(0)
6. The solution (not of interest for doing the problem) is obtained by using
Dy/dx = 1/(dx/dy) and solving dx/dy = 1/(1 + sin y) by integration,
x + c = —2/(tan 2_ 1 y + 1); thus y = —2 arctan ((x + 2 + c) /(x + c)).
8. Linear ode. The solution involves the error function.
12. By integration, y = c — 1/x.
16. The solution (not needed for doing the problem) of y ' = 1/y can be obtained by
separating variables and using the initial condition; y 2 /2 = t + c, y = √2¯
t — 1.
18. The solution of this initial value problem involving the linear ode y ' + y = t2 is
Y = 4e—t + t2 — 2t + 2.
20. Cas project. (a) verify by substitution that the general solution is y = 1 + ce—x .
Limit y = 1 (y(x) = 1 for all x), increasing for y(0) < 1, decreasing
for y(0) > 1.
(b) Verify by substitution that the general solution is x 4 + y 4 = c. More “square-
shaped,” isoclines y = kx. Without the minus on the right you get “hyperbola -
like” curves y 4 — x 4 = const as solutions (verify!). The direction fields should turn
out in perfect shape.
(c) The computer m ay be better if the isoclines are complicated; but the computer
m a y give you nonsense even in simpler cases, for instance when y(x) becomes
imaginary. Much will depend on the choice of x- and y-intervals, a method of trial
and error. Isoclines m a y be preferable if the explicit form of the ode contains roots
on the right.
Section 1.3. Separable odes. Modeling, page 12
Purpose. To familiarize the student with the first “big” method of solving odes, the
separation of variables, and an extension of it, the reduction to separable form by a
transformation of the ode, namely, by introducing a new unknown function.
The section includes standard applications that lead to separable odes, namely,
1. The ode giving tan x as solution
2. The ode of the exponential function, having various applications, such as in
radiocarbon dating
3. A mixing problem for a single tank
4. Newton’s law of cooling
5. Torricelli’s law of outflow.
, 4 instructor’s m a n u a l
In reducing to separability we consider
6. The transformation u = y/x, giving perhaps the most important reducible class
of odes.
Ince’s classical book [a 11] contains m a n y further reductions as well as a systematic
theory of reduction for certain classes of odes.
Comment on problem 5
From the implicit solution we can get two explicit solutions
Y = +√ c¯
— ( 6 x¯
)2
representing semi-ellipses in the upper half-plane, and
Y = —√ c¯
— ( 6 x¯
)2
Representing semi-ellipses in the lower half-plane. [similarly, we can get two explicit
solutions x(y) representing semi-ellipses in the left and right half-planes, respectively.]
On the x-axis, the tangents to the ellipses are vertical, so that y '(x) does not exist. Similarly
for x '(y) on the y-axis.
This also illustrates that it is natural to consider solutions of odes on open rather than
on closed intervals.
Comment on separability
An analytic function ƒ(x, y) in a domain d of the xy-plane can be factored in d,
ƒ(x, y) = g(x)h(y), if and only if in d,
Ƒxyƒ = ƒxƒy
[d. Scott, american math. Monthly 92 (1985), 422–423]. Simple cases are easy to decide,
but this m ay save time in cases of more complicated odes, some of which m ay perhaps be
of practical interest. You m ay perhaps ask your students to derive such a criterion.
Comments on application
Each of those examples can be modified in various ways, for example, by changing the
application or by taking another form of the tank, so that each example characterizes a
whole class of applications.
The m any odes in the problem set, much more than one would ordinarily be willing
and have the time to consider, should serve to convince the student of the practical
importance of odes; so these are odes to choose from, depending on the students’ interest
and background.
Comment on footnote 3
Newton conceived his method of fluxions (calculus) in 1665–1666, at the age of 22.
Philosophiae naturalis principia mathematica was his most influential work.
Leibniz invented calculus independently in 1675 and introduced notations that were
essential to the rapid development in this field. His first publication on differentialcalculus
appeared in 1684.
Solutions to problem set 1.3, page 18
2. Dy/y 2 = —(x + 2) dx. The variables are now separated. Integration on both sides gives
1 _1 2 2
— — = —2x — 2x + c*. Hence y = — — .
2
Y X + 4x + c
Stuvia.com - The Marketplace to Buy and S e l l y o u r Study Material
SOLUTION MANUAL FOR ADVANCED
ENGINEERING MATHEMATICS 10TH
EDITION,2020 BY ERWIN KREYSZIG
.LATEST EDITION 2026
, 2 instructor’s m a n u a l
Solutions to problem set 1.1, page 8
2. Y = —e—3x /3 + c 4. Y = (sinh 4x) /4 + c
6. Second order. 8. First order.
10. Y = ce0.5x , y(2) = ce = 2, c = 2/e, y = (2/e)e0.5x = 0.736e0.5x
12. Y = cex + x + 1, y(0) = c + 1 = 3, c = 2, y = 2ex + x + 1
14. Y = c sec x, y(0) = c/cos 0 = c = _ 1π, y = _ 1π sec x
2 2
16. Substitution of y = cx — c2 into the ode gives
Y'2 — xy ' + y = c2 — xc + (cx — c2 ) = 0.
similarly,
Y=
_ 1x2, y' = _ 1x, thus _1 2
= 40 x. — x(
_1x)+ _1x2
4 2 2 4
18. In prob. 17 the constants of integration were set to zero. Here, by two integrations,
y ” = g, v = y ' = gt + c1 , y = _21gt2 + c1 t + c2 , y(0) = c2 = y 0 ,
And, furthermore,
v(0) = c1 = v 0 , hence y = _ 1 gt 2 + v 0 t + y 0 , 2
As claimed. Times of fall are 4.5 and 6.4 sec, from t = √ 1¯0 . ¯9 and √2¯
0 .9̄ .
20. Y' = ky. Solution y = y 0 e , where y 0 is the pressure at sea level x = 0. Now
kx
y(18000) = y 0 ek·18000 = _21y 0 (given). From this,
Ek·18000 = _ 1, y(36000) = y ek·2·18000 = y (ek·18000 ) 2 = y (_ 1)2 = _ 1y .
0 2 0 0 0 2 4
22. For 1 year and annual, daily, and continuous compounding we obtain the values
Ya(1) = 1060.00, y d (1) = 1000(1 + 0.06/365)365 = 1061.83,
Yc(1) = 1000e0.06 = 1061.84,
Respectively. Similarly for 5 years,
Ya(5) = 1000 · 1.06 5 = 1338.23, y d (5) = 1000(1 + 0.06/365)365·5 = 1349.83,
Yc(5) = 1000e0.06·5 = 1349.86.
We see that the difference between daily compounding and continuous compounding
is very small.
The ode for continuous compounding is y c' = ry c.
Section 1.2. Geometric meaning of y ' = ƒ(x, y ). Direction fields, page 9
Purpose. To give the student a feel for the nature of odes and the general behavior of
fields of solutions. This amounts to a conceptual clarification before entering into formal
manipulations of solution methods, the latter being restricted to relatively small—albeit
important—classes of odes. This approach is becoming increasingly important, especially
because of the graphical power of computer software. It is the analog of conceptual
studies of the derivative and integral in calculus as opposed to formal techniques of
differentiation and integration.
Comment on isoclines
These could be omitted because students sometimes confuse them with solutions. In the
computer approach to direction fields they no longer play a role.
, Instructor’s manual 3
Comment on order of sections
This section could equally well be presented later in chap. 1, perhaps after one or two
formal methods of solution have been studied.
Solutions to problem set 1.2, page 11
2. Semi-ellipse x 2 /4 + y 2 /9 = 13/9, y > 0. To graph it, choose the y-interval large
enough, at least 0 % y % 4.
4. Logistic equation (verhulst equation; sec. 1.5). Constant solutions y = 0 and y = _ 12.
_1
for these, y ' = 0. Increasing solutions for 0 < < _21, decreasing for > 2.
y(0) y(0)
6. The solution (not of interest for doing the problem) is obtained by using
Dy/dx = 1/(dx/dy) and solving dx/dy = 1/(1 + sin y) by integration,
x + c = —2/(tan 2_ 1 y + 1); thus y = —2 arctan ((x + 2 + c) /(x + c)).
8. Linear ode. The solution involves the error function.
12. By integration, y = c — 1/x.
16. The solution (not needed for doing the problem) of y ' = 1/y can be obtained by
separating variables and using the initial condition; y 2 /2 = t + c, y = √2¯
t — 1.
18. The solution of this initial value problem involving the linear ode y ' + y = t2 is
Y = 4e—t + t2 — 2t + 2.
20. Cas project. (a) verify by substitution that the general solution is y = 1 + ce—x .
Limit y = 1 (y(x) = 1 for all x), increasing for y(0) < 1, decreasing
for y(0) > 1.
(b) Verify by substitution that the general solution is x 4 + y 4 = c. More “square-
shaped,” isoclines y = kx. Without the minus on the right you get “hyperbola -
like” curves y 4 — x 4 = const as solutions (verify!). The direction fields should turn
out in perfect shape.
(c) The computer m ay be better if the isoclines are complicated; but the computer
m a y give you nonsense even in simpler cases, for instance when y(x) becomes
imaginary. Much will depend on the choice of x- and y-intervals, a method of trial
and error. Isoclines m a y be preferable if the explicit form of the ode contains roots
on the right.
Section 1.3. Separable odes. Modeling, page 12
Purpose. To familiarize the student with the first “big” method of solving odes, the
separation of variables, and an extension of it, the reduction to separable form by a
transformation of the ode, namely, by introducing a new unknown function.
The section includes standard applications that lead to separable odes, namely,
1. The ode giving tan x as solution
2. The ode of the exponential function, having various applications, such as in
radiocarbon dating
3. A mixing problem for a single tank
4. Newton’s law of cooling
5. Torricelli’s law of outflow.
, 4 instructor’s m a n u a l
In reducing to separability we consider
6. The transformation u = y/x, giving perhaps the most important reducible class
of odes.
Ince’s classical book [a 11] contains m a n y further reductions as well as a systematic
theory of reduction for certain classes of odes.
Comment on problem 5
From the implicit solution we can get two explicit solutions
Y = +√ c¯
— ( 6 x¯
)2
representing semi-ellipses in the upper half-plane, and
Y = —√ c¯
— ( 6 x¯
)2
Representing semi-ellipses in the lower half-plane. [similarly, we can get two explicit
solutions x(y) representing semi-ellipses in the left and right half-planes, respectively.]
On the x-axis, the tangents to the ellipses are vertical, so that y '(x) does not exist. Similarly
for x '(y) on the y-axis.
This also illustrates that it is natural to consider solutions of odes on open rather than
on closed intervals.
Comment on separability
An analytic function ƒ(x, y) in a domain d of the xy-plane can be factored in d,
ƒ(x, y) = g(x)h(y), if and only if in d,
Ƒxyƒ = ƒxƒy
[d. Scott, american math. Monthly 92 (1985), 422–423]. Simple cases are easy to decide,
but this m ay save time in cases of more complicated odes, some of which m ay perhaps be
of practical interest. You m ay perhaps ask your students to derive such a criterion.
Comments on application
Each of those examples can be modified in various ways, for example, by changing the
application or by taking another form of the tank, so that each example characterizes a
whole class of applications.
The m any odes in the problem set, much more than one would ordinarily be willing
and have the time to consider, should serve to convince the student of the practical
importance of odes; so these are odes to choose from, depending on the students’ interest
and background.
Comment on footnote 3
Newton conceived his method of fluxions (calculus) in 1665–1666, at the age of 22.
Philosophiae naturalis principia mathematica was his most influential work.
Leibniz invented calculus independently in 1675 and introduced notations that were
essential to the rapid development in this field. His first publication on differentialcalculus
appeared in 1684.
Solutions to problem set 1.3, page 18
2. Dy/y 2 = —(x + 2) dx. The variables are now separated. Integration on both sides gives
1 _1 2 2
— — = —2x — 2x + c*. Hence y = — — .
2
Y X + 4x + c
