Solutions Manual to Accompany Ordinary Differential Equations (eBook)

Michael D. Greenberg (Autor)

eBook Download: EPUB

2014 | 1. Auflage
208 Seiten
Sybex (Verlag)
9781118457566 (ISBN)

Infos

Features a balance between theory, proofs, and examples and provides applications across diverse fields of study Ordinary Differential Equations presents a thorough discussion of first-order differential equations and progresses to equations of higher order.

Autor

MICHAEL D. GREENBERG, PhD, is Professor Emeritus of Mechanical Engineering at the University of Delaware where he teaches courses on engineering mathematics and is a three-time recipient of the University of Delaware Excellence in Teaching Award. Greenberg's research has emphasized vortex methods in aerodynamics and hydrodynamics.

Inhalt

Chapter 1 First-Order Differential Equations 1

Chapter 2 Higher-Order Linear Equations 34

Chapter 3 Applications of Higher-Order Equations 56

Chapter 4 Systems of Linear Differential Equations 70

Chapter 5 Laplace Transform 107

Chapter 6 Series Solutions 127

Chapter 7 Systems of Nonlinear Differential Equations 148

CAS Tutorial 167

Chapter 1. Maple 167

Chapter 2. MATLAB 172

Chapter 3. Mathematica 176

Chapter 4. Maple for this Text, by Chapter 180

Textauszug

CHAPTER 2

Higher-Order Linear Equations

Section 2.2

The Answers to Exercises at the end of the text should suffice for Section 2.1, so we will omit it here and go right to Section 2.2. Sections 2.2 and 2.3 are particularly important, not only to understand, but also to retain beyond this course. Subsequent courses in engineering and science draw heavily upon this material, and it is expected that the student remembers it. Be familiar with the elementary functions that we encounter: exponentials (real exponentials in Section 2.2, complex exponentials in Section 2.3) and the circular and hyperbolic functions (sin, cos, sinh, and cosh).

It’s true that the difficulty in solving an ODE increases with the order of the equation; but as we begin with linear, homogeneous equations with constant coefficients, we find that we can solve them explicitly; that solution is given, for second-order equations, in Theorem 2.2.1.

EXAMPLES

Example 1. (General solution; distinct roots) Derive the general solution of

SOLUTION. Put into the latter and obtain cancel the and obtain the characteristic equation which gives the distinct roots Thus, we have been successful in finding two solutions of the assumed exponential form, Now, the DE is linear and homogeneous, so Theorem 2.1.1 applies and tells us that we can build up, from our results, a solution for arbitrary constants But we have no basis for claiming that the later is the general solution of the DE, except that Theorem 2.2.1 assures us that it is! Answer: General solution is How easy that was!

Suppose initial conditions were supplied, say Apply them to find and These two linear algebraic equations for give so the particular solution satisfying those conditions is Theorem 2.2.2 assures us that the interval of existence and uniqueness is

Example 2. (General solution; repeated roots) Derive the general solution of

SOLUTION. Put into the latter and obtain cancel the and obtain the characteristic equation that gives the repeated roots

Thus, our assumed exponential solution method has led us to only the one solution  We found, in our proof of Theorem 2.2.1, that when r is a repeated root, the “missing” solution is  By (11b) in Theorem 2.2.1, the general solution is 

Example 3. (Archaeology) If the characteristic roots are  then what was the (linear, homogeneous, constant-coefficient) DE?

SOLUTION. Then the characteristic equation is  so the DE was  

Example 4. (Inferring the equation) Another version of the preceding “archeology” problem occurs if we are given the general solution and asked for the DE. For instance, suppose we are told that the general solution of a certain DE is

(A)

That is, both  and  are solutions of the DE. To find  and  put the two solutions into the DE. That step gives  and  and these give  so the DE is 

Example 5. (A subtle point in the text Example 4) In text Example 4 we obtained the solution form

(A)

in which c1, c2 were arbtirary constants. We renamed

(B1)

(B2)

and said the solution is, equivalently,

(C)

in which  are also arbitrary. Does it really follow that if  are arbitrary in (A), then   are arbitrary in (C)? That is, if  are all arbitrary, does (A) imply (C) and (C) imply (A)?

SOLUTION. Certainly, (A) implies (C) for if we choose any values for  then we can compute   and pass from (A) to (C). But the reverse, passing from (C) to (A), requires us to be able, given any arbitrarily chosen values for  to compute  from (B1, B2). We can; adding and subracting them readily gives  and  Thus, our answer is: Yes, (A) and (C) are equivalent, with  all arbitrary.

Example 6. (Factoring the operator) We used factoring-the-operator as a method of deriving the general solution that is then given in Theorem 2.2. We do not favor it as a solution “method”; rather, we suggest seeking solutions in exponential form and proeeding as we did in Examples 1 and 2 above. But, to help you to understand the factorization solution steps, this example gives a concrete example: Use operator factorization to derive the general solution of

SOLUTION. Factor the DE  as  that is, If we define say, then we have these two first-order equations

that is,

which are readily solved by methods given in Section 1.2. First, the second gives  Then, put that in the first, obtaining  which is also solved as in Section 1.2; its general solution is given by (27) therein:

Then (27) gives 

Section 2.3

Theorem 2.2.1 gives the general solution even if the roots of the characteristic equation are complex, but how then are we to understand the complex exponentials that result? The key is Euler’s definition (2), which also gives us the connections (4a,b) and (5a,b) between the complex exponentials and the circular functions; those four formulas all come from the definition (2).

EXAMPLES

Example 1. (Evaluation of complex exponentials) Given the exponential on the left, evaluate it–that is, in standard Cartesian form 

SOLUTION.

 Do we get the same thing if we square  ?

Let’s see: But, the trigonometric identity   with  gives  so   as above.

Example 2. Evaluate 

SOLUTION.

Example 3. (General solution) Find the general solution of 

SOLUTION.  gives  so  

Example 4. (Archeology) If the characteristic roots are  what was the DE?

SOLUTION. The characteristic equation was

so the DE was 

Section 2.4

Linear dependence and independence are important here in connection with the general solution of linear differential equations, but are also fundamental concepts in mathematics.

EXAMPLES

Example 1. (LI or LD) Determine whether the given set is LI or LD on  which we will denote as I. If it is LD, give a nontrivial linear relationship on them–that is, of the form (2) with the aj’s not all zero.

(a) 

SOLUTION. It is LD because it contains the zero function. Thus,  for instance, in which the coefficients are not both zero. Remember, any set containing the zero function is LD.

(b)

SOLUTION. There is no obvious nontrivial relationship among them that comes to mind, so let us fall back on Theorem 2.4.1:

The latter is not just nonzero “somewhere” on the interval, it is nonzero everywhere on the interval; that’s even more than we need, but, in any case, the conclusion from Theorem 2.4.1 is that the set is LI on I. In fact, the set is LI on any  interval because W is nonzero for all .

(c)

SOLUTION. In this case, their Wronskian is 

for all x. However, in that case Theorem 2.4.1 gives no information; it is a “one way” result, telling us that if  anywhere in I then the set is LI; but if  everywhere on I, it simply gives no information. However, we can see by inspection that this set is LD, because we can write one member as a linear combination of the others. For instance,  that is,   which is of the form (2), with the coefficients not all zero.

Example 2. (LI and matching coefficients) One of the exercises shows that for

to hold on an interval I on which  is LI, it is necessary (and obviously sufficient) that the corresponding coefficients match: for each  Here is the problem: If possible, satisfy the given equation, on  by suitable choice of the capitalized constants. If it cannot be satisfied by any such choice, say so and give your reasoning.

(a) 

SOLUTION. Write it out,  Now,  are LI (on any interval), so set  If it is not clear that  are LI, note that their Wronskian is found to be

(b) 

SOLUTION. Write it out as What are the LI functions therein? They are  [One way to see that, quickly, is to remember (from the italicized comment following Definition 2.4.1) that if the set is LD then that means that at least one of the functions must be expressible as a linear combination of the others. Surely, 1 cannot be formed as a linear combination of  surely,  cannot be formed as a linear combination of  and so on. Alternatively, use the Wronskian test in Theorem 2.4.1.] Thus, matching coefficients of those LI functions on each side, we must set  These cannot be satisfied by any choice of A and B because 0 cannot equal 3. Thus, the relation is impossible.

(c) 

SOLUTION. Rewrite this as 

By the linear independence of  set  These are solvable, and give 

Example 3. (Basis) Consider (1) in the text. Let the  be continuous on I and suppose a is in I. Let  denote the solution satisfying initial conditions  and let  denote the solution satisfying initial conditions  Is  a basis for the DE (1) on I?

SOLUTION. We need only check that they are LI, and we can use the Wronskian test in Theorem 2.4.4(c) for that:  so  a basis (that is, a basis of solutions, or a fundamental set) for (1).

Example 4. (Basis) Show whether or not  and  comprise a basis for the DE   on the interval 

SOLUTION. First, verify that each is indeed a solution (on that interval I). That’s readily verified by substitution. Next, we must verify that they are LI on I. Since these are only two functions, the simplest test is whether or not one of them can be expressed as a multiple of the other (page 120 in the text); since  is not a scalar multiple of  nor vice versa, the set is LI....

Produkt-Details

• Erscheint lt. Verlag: 28.8.2014
• Sprache: englisch
• Themenwelt: Mathematik / Informatik ► Mathematik ► Analysis
• Themenwelt: Technik
• Schlagworte: across diverse • Applications • balance between • Differential • Differential Equations • Differentialgleichungen • discussion • Equations • Fields • Mathematics • Mathematik • Ordinary • Study • theory • thorough
• ISBN-13: 9781118457566 / 9781118457566