SOLUTION MANUAL
All Chapters Included
ADVANCED ENGINEERING
MATHEMATICS
VOLUME 1
,Chap. 1 First-Order ODEs
Sec. 1.1 Basic Concepts. Modeling
To get a good start into this chapter and this section, quickly review your basic calculus.
Take a look at the front matter of the textbook and see a review of the main differentiation
and integration formulas. Also, Appendix 3, pp. A63–A66, has useful formulas for such
functions as exponential function, logarithm, sine and cosine, etc. The beauty of ordinary
differential equations is that the subject is quite systematic and has different methods for
different types of ordinary differential equations, as you shall learn. Let us discuss some
Examples of Sec. 1.1, pp. 4–7.
Example 2, p. 5. Solution by Calculus. Solution Curves. To solve the first-
order ordinary differential equation (ODE)
y· = cos x
means that we are looking for a function whose derivative is cos x. Your first answer
might be that the desired function is sin x, because (sin x)· = cos x. But your answer
would be incomplete because also (sin x + 2)· = cos x, since the derivative of 2 and of
any constant is 0. Hence the complete answer is y = cos x + c, where c is an arbitrary
constant. As you vary the constants you get an infinite family of solutions. Some of
these solutions are shown in Fig. 3. The lesson here is that you should never
forget your constants!
Example 4, pp. 6–7. Initial Value Problem. In an initial value problem (IVP) for a
first-order ODE we are given an ODE, here y· = 3y, and an initial value condition y(0)
= 5.7. For such a problem, the first step is to solve the ODE. Here we obtain y(x) = ce3x
as shown in Example 3, p. 5. Since we also have an initial condition, we must
substitute that condition into our solution and get y(0) = ce3·0 = ce0 = c · 1 = c = 5.7.
Hence the complete solution is y(x) = 5.7e3x. The lesson here is that for an initial
value problem you get a unique solution, also known as a particular solution.
,
, 2 Ordinary Differential Equations (ODEs) Part A
Modeling means that you interpret a physical problem, set up an appropriate
mathematical model, and then try to solve the mathematical formula. Finally, you
have to interpret your answer.
Examples 3 (exponential growth, exponential decay) and 5 (radioactivity) are examples of
modeling problems. Take a close look at Example 5, p. 7, because it outlines all the steps
of modeling.
Problem Set 1.1. Page 8
3. Calculus. From Example 3, replacing the independent variable t by x we know that
y· = 0.2y has a solution y = 0.2ce0.2x. Thus by analogy, y· = y has a solution
1 · ce1·x = cex,
where c is an arbitrary constant.
Another approach (to be discussed in details in Sec. 1.3) is to write the ODE as
dy
= y,
dx
and then by algebra
obtain 1
dy = y dx, so that dy = dx.
y
Integrate both sides, and then apply exponential functions on both sides to obtain
the same solution as above
r r
1
dy = dx, ln |y|= x + c, eln |y| = ex+c, y = ex · ec = c∗ ex,
y
(where c∗ = ec is a constant).
The technique used is called separation of variables because we separated the variables
so that y appeared on one side of the equation and x on the other side before we integrate
f
7. Solve by integration. Integrating y· = cosh 5.13x we obtain (chain rule!) y = cosh 5.13x d
1
= 5.13 (sinh 5.13x) + c. Check: Differentiate your answer:
( )·
1
5.13 (sinh 5.13x) + c =5.131 (cosh 5.13x) · 5.13 = cosh 5.13x, which is correct.
11. Initial value problem (IVP). (a) Differentiation of y = (x + c)ex by product rule and definition
y gives
y· = ex + (x + c)ex = ex + y.
But this looks precisely like the given ODE y· = ex + y. Hence we have shown that indeed
y = (x + c)ex is a solution of the given ODE. (b) Substitute the initial value condition into
the solution to give y(0) = (0 + c)e0 = c · 1 = 1 . Hence c = 1 so that the answer to the IVP is
2 2
y = (x + 21 )ex.
(c) The graph intersects the x-axis at x = 0.5 and shoots exponentially upward.
All Chapters Included
ADVANCED ENGINEERING
MATHEMATICS
VOLUME 1
,Chap. 1 First-Order ODEs
Sec. 1.1 Basic Concepts. Modeling
To get a good start into this chapter and this section, quickly review your basic calculus.
Take a look at the front matter of the textbook and see a review of the main differentiation
and integration formulas. Also, Appendix 3, pp. A63–A66, has useful formulas for such
functions as exponential function, logarithm, sine and cosine, etc. The beauty of ordinary
differential equations is that the subject is quite systematic and has different methods for
different types of ordinary differential equations, as you shall learn. Let us discuss some
Examples of Sec. 1.1, pp. 4–7.
Example 2, p. 5. Solution by Calculus. Solution Curves. To solve the first-
order ordinary differential equation (ODE)
y· = cos x
means that we are looking for a function whose derivative is cos x. Your first answer
might be that the desired function is sin x, because (sin x)· = cos x. But your answer
would be incomplete because also (sin x + 2)· = cos x, since the derivative of 2 and of
any constant is 0. Hence the complete answer is y = cos x + c, where c is an arbitrary
constant. As you vary the constants you get an infinite family of solutions. Some of
these solutions are shown in Fig. 3. The lesson here is that you should never
forget your constants!
Example 4, pp. 6–7. Initial Value Problem. In an initial value problem (IVP) for a
first-order ODE we are given an ODE, here y· = 3y, and an initial value condition y(0)
= 5.7. For such a problem, the first step is to solve the ODE. Here we obtain y(x) = ce3x
as shown in Example 3, p. 5. Since we also have an initial condition, we must
substitute that condition into our solution and get y(0) = ce3·0 = ce0 = c · 1 = c = 5.7.
Hence the complete solution is y(x) = 5.7e3x. The lesson here is that for an initial
value problem you get a unique solution, also known as a particular solution.
,
, 2 Ordinary Differential Equations (ODEs) Part A
Modeling means that you interpret a physical problem, set up an appropriate
mathematical model, and then try to solve the mathematical formula. Finally, you
have to interpret your answer.
Examples 3 (exponential growth, exponential decay) and 5 (radioactivity) are examples of
modeling problems. Take a close look at Example 5, p. 7, because it outlines all the steps
of modeling.
Problem Set 1.1. Page 8
3. Calculus. From Example 3, replacing the independent variable t by x we know that
y· = 0.2y has a solution y = 0.2ce0.2x. Thus by analogy, y· = y has a solution
1 · ce1·x = cex,
where c is an arbitrary constant.
Another approach (to be discussed in details in Sec. 1.3) is to write the ODE as
dy
= y,
dx
and then by algebra
obtain 1
dy = y dx, so that dy = dx.
y
Integrate both sides, and then apply exponential functions on both sides to obtain
the same solution as above
r r
1
dy = dx, ln |y|= x + c, eln |y| = ex+c, y = ex · ec = c∗ ex,
y
(where c∗ = ec is a constant).
The technique used is called separation of variables because we separated the variables
so that y appeared on one side of the equation and x on the other side before we integrate
f
7. Solve by integration. Integrating y· = cosh 5.13x we obtain (chain rule!) y = cosh 5.13x d
1
= 5.13 (sinh 5.13x) + c. Check: Differentiate your answer:
( )·
1
5.13 (sinh 5.13x) + c =5.131 (cosh 5.13x) · 5.13 = cosh 5.13x, which is correct.
11. Initial value problem (IVP). (a) Differentiation of y = (x + c)ex by product rule and definition
y gives
y· = ex + (x + c)ex = ex + y.
But this looks precisely like the given ODE y· = ex + y. Hence we have shown that indeed
y = (x + c)ex is a solution of the given ODE. (b) Substitute the initial value condition into
the solution to give y(0) = (0 + c)e0 = c · 1 = 1 . Hence c = 1 so that the answer to the IVP is
2 2
y = (x + 21 )ex.
(c) The graph intersects the x-axis at x = 0.5 and shoots exponentially upward.
