# MC GRAW, Differential Equations (Schaum's Easy Outlines)

(Parte 1 de 3)

SCHAUM’S Easy OUTLINES

Based on Schaum’s

Outline of Theory and Problems of

Differential Equations, Second Edition by Richard Bronson, Ph.D.

Abridgement Editor Erin J. Bredensteiner, Ph.D.

SCHAUM’S OUTLINE SERIES McGRA W-HILL

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Contents

Chapter 1 Basic Concepts and Classifying

Differential Equations 1

Chapter 2Solutions of First-Order

Differential Equations 8

Chapter 3Applications of First-Order

Differential Equations 20

Chapter 4Linear Differential Equations:

Theory of Solutions29

Chapter 5 Solutions of Linear Homogeneous

Differential Equations with Constant Coefﬁcients 3

Chapter 6Solutions of Linear

Nonhomogeneous Equations and Initial-Value Problems39

Chapter 7Applications of Second-Order

Linear Differential Equations47

Chapter 8Laplace Transforms and Inverse

Laplace Transforms 5

Chapter 9Solutions by Laplace Transforms65 Chapter 10Matrices and the Matrix

Exponential 69

Chapter 11Solutions of Linear Differential

Equations with Constant Coefﬁcients by Matrix Methods78

Chapter 12Power Series Solutions85 Chapter 13Gamma and Bessel Functions98 Chapter 14Numerical Methods104 Chapter 15Boundary-Value Problems and Fourier Series115

Appendix Laplace Transforms 124 Index 133 vi DIFFERENTIALEQUATIONS

Chapter 1

Basic Concepts and Classifying

Differential Equations

In This Chapter:

✔ Differential Equations ✔ Notation

✔ Solutions

✔Initial-Value and Boundary-Value

Problems ✔Standard and Differential Forms

✔Linear Equations

✔ Bernoulli Equations

✔ Homogeneous Equations

✔ Separable Equations

✔Exact Equations

Differential Equations

Adifferential equationis an equation involving an unknown function and its derivatives.

Example 1.1:The following are differential equations involving the unknown function y.

Adifferential equation is an ordinary differential equation if the unknown function depends on only one independent variable. If the unknown function depends on two or more independent variables, the differential equation is a partial differential equation. In this book we will be concerned solely with ordinary differential equations.

Example 1.2:Equations 1.1 through 1.4 are examples of ordinary differential equations, since the unknown function ydepends solely on the variable x. Equation 1.5 is a partial differential equation, since ydepends on both the independent variables tand x.

dydx y dydx y dy x dydx xy++ =(sin ) e dydx dydx

2 DIFFERENTIALEQUATIONS

Note!

The orderof a differential equation is the order of the highest derivative appearing in the equation.

Example 1.3:Equation 1.1 is a ﬁrst-order differential equation; 1.2, 1.4, and 1.5 are second-order differential equations. (Note in 1.4 that the order of the highest derivative appearing in the equation is two.) Equation 1.3 is a third-order differential equation.

Notation

 spectively, the ﬁrst, second, third, fourth, , nth derivatives of ywith re-

The expressions are often used to represent, respect to the independent variable under consideration. Thus, repre-

sents if the independent variable is x, but represents if the independent variable is p. Observe that parenthesis are used in y(n) to distinguish it from the nth power, yn. If the independent variable is time, usually denoted by t, primes are often replaced by dots. Thus, represent, respectively.

Solutions

Asolutionof a differential equation in the unknown function yand the independent variable xon the interval is a function y(x)that satisﬁes the differential equation identically for all xin .

CHAPTER 1: Basic Concepts and Classiﬁcation3

Thus, satisﬁes the differential equation for all values of xand is a solution on the interval .

Example 1.5:Determine whether is a solution of .

Note that the left side of the differential equation must be nonnegative for every real function y(x)and any x, since it is the sum of terms raised to the second and fourth powers, while the right side of the equation is negative. Since no function y(x)will satisfy this equation, the given differential equation has no solutions.

We see that some differential equations have inﬁnitely many solutions (Example 1.4), whereas other differential equations have no solutions (Example 1.5). It is also possible that a differential equation has exactly one solution. Consider , which for reasons identical to those given in Example 1.5 has only one solution .

You Need to Know

Aparticular solutionof a differential equation is any one solution.Thegeneral solutionof a differential equation is the set of all solutions.

Example 1.6:The general solution to the differential equation in Example 1.4 can be shown to be (see Chapters Four and Five)

. That is, every particular solution of the differential equation has this general form. Afew particular solutions are: (a) (choose and ), (b) (choose

4 DIFFERENTIALEQUATIONS

The general solution of a differential equation cannot always be expressed by a single formula. As an example consider the differential equa- tion , which has two particular solutions and .

Initial-Value and Boundary-Value Problems

Adifferential equation along with subsidiary conditions on the unknown function and its derivatives, all given at the same value of the independent variable, constitutes an initial-value problem. The subsidiary conditions are initial conditions. If the subsidiary conditions are given at more than one value of the independent variable, the problem is a boundary-value problemand the conditions are boundary conditions.

Example 1.7:The problem is an initial value problem, because the two subsidiary conditions are both given at

. The problem is a boundary-value problem, because the two subsidiary conditions are given at x=0 and x = 1.

Asolution to an initial-value or boundary-value problem is a function y(x) that both solves the differential equation and satisﬁes all given subsidiary conditions.

Standard and Differential Forms

Standard form for a ﬁrst-order differential equation in the unknown function y(x) is where the derivative appears only on the left side of 1.6. Many, but not all, ﬁrst-order differential equations can be written in standard form by algebraically solving for and then setting f(x,y) equal to the rightside of the resulting equation. ′y

The right side of 1.6 can always be written as a quotient of two other functions M(x,y) and −N(x,y). Then 1.6 becomes which is equivalent to the differential form

Linear Equations

Consider a differential equation in standard form 1.6. If f(x,y) can be written as (that is, as a function of xtimes y, plus another function of x), the differential equation is linear. First-order linear differential equations can always be expressed as

Linear equations are solved in Chapter Two.

Bernoulli Equations ABernoulli differential equation is an equation of the form where ndenotes a real number. When n = 1 or n = 0, a Bernoulli equation reduces to a linear equation. Bernoulli equations are solved in Chapter Two.

Homogeneous Equations Adifferential equation in standard form (1.6) is homogeneousif for every real number t. Homogeneous equations are solved in Chapter Two.

6 DIFFERENTIALEQUATIONS

Note!

In the general framework of differential equations, the word “homogeneous” has an entirely different meaning (see Chapter Four). Only in the context of ﬁrst-order differential equations does “homogeneous” have the meaning deﬁned above.

Separable Equations

Consider a differential equation in differential form (1.7). If M(x,y) =A(x) (a function only of x) and N(x,y) =B(y) (a function only of y), the differential equation is separable, or has its variables separated. Separable equations are solved in Chapter Two.

Exact Equations Adifferential equation in differential form (1.7) is exact if

Exact equations are solved in Chapter Two (where a more precise deﬁnition of exactness is given).

Mx y Nx yx

CHAPTER 1: Basic Concepts and Classiﬁcation7

Chapter 2

Solutions of

First-Order

Differential

Equations

In This Chapter:

✔ Separable Equations ✔ Homogeneous Equations

✔Exact Equations

✔Linear Equations

✔ Bernoulli Equations

✔Solved Problems

Separable Equations

General Solution

The solution to the ﬁrst-order separable differential equation (see Chapter One).

is (2.2) where crepresents an arbitrary constant.

(See Problem 2.1) The integrals obtained in Equation 2.2 may be, for all practical purposes, impossible to evaluate. In such case, numerical techniques (see Chapter 14) are used to obtain an approximate solution. Even if the indicated integrations in 2.2 can be performed, it may not be algebraically possible to solve for y explicitly in terms ofx. In that case, the solution is left in implicit form.

Solutions to the Initial-Value Problem The solution to the initial-value problem (2.3) can be obtained, as usual, by ﬁrst using Equation 2.2 to solve the differential equation and then applying the initial condition directly to evaluate c. Alternatively, the solution to Equation 2.3 can be obtained from

(2.4) where sand tare variables of integration.

Homogeneous Equations The homogeneous differential equation (2.5)

having the property f(tx,ty)=f(x, y) (see Chapter One) can be transformed into a separable equation by making the substitution

A s ds B t dt x y

The resulting equation in the variables v andx is solved as a separable differential equation; the required solution to Equation 2.5 is obtained by back substitution.

Alternatively, the solution to 2.5 can be obtained by rewriting the differential equation as

into Equation 2.8. After simplifying, the resulting differential equation will be one with variables (this time, uand y) separable.

Ordinarily, it is immaterial which method of solution is used. Occasionally, however, one of the substitutions 2.6 or 2.9 is deﬁnitely superior to the other one. In such cases, the better substitution is usually apparent from the form of the differential equation itself. (See Problem 2.2)

Exact Equations dxdy uy du dy =+ vx dv dx =+

10 DIFFERENTIALEQUATIONS

Note!

Test for exactness: If M(x,y) and N(x,y) are continuous functions and have continuous ﬁrst partial derivatives on some rectangle of the xy-plane, then Equation 2.1 is exact if and only if

Method of Solution To solve Equation 2.1, assuming that it is exact, ﬁrst solve the equations (2.14)

(2.15) for g(x, y). The solution to 2.1 is then given implicitly by g(x, y) =c(2.16) where crepresents an arbitrary constant.

Equation 2.16 is immediate from Equations 2.1 and 2.12. If 2.12 is substituted into 2.1, we obtain dg(x, y(x)) =0. Integrating this equation

(note that we can write 0 as 0dx), we have , which, in turn, implies 2.16.

Integrating Factors

In general, Equation 2.1 is not exact. Occasionally, it is possible to transform 2.1 into an exact differential equation by a judicious multiplication. Afunction I(x, y) is an integrating factorfor 2.1 if the equation

CHAPTER 2: Solutions of First-Order Differential Equations11 is exact. Asolution to 2.1 is obtained by solving the exact differential equation deﬁned by 2.17. Some of the more common integrating factors are displayed in Table 2.1 and the conditions that follow:

If ,a function of xalone, then

(2.18) If , a function of yalone, then

If M=yf(xy) and N=xg(xy), then (2.20)

In general, integrating factors are difﬁcult to uncover. If a differential equation does not have one of the forms given above, then a search for an integrating factor likely will not be successful, and other methods of solution are recommended. (See Problems 2.3–2.6)

Linear Equations

Method of Solution Aﬁrst-order lineardifferential equation has the form (see Chapter One) (2.21)

An integrating factor for Equation 2.21 is

12 DIFFERENTIALEQUATIONS

which depends only on xand is independent of y. When both sides of 2.21 are multiplied by I(x), the resulting equation is exact. This equation can be solved by the method described previously. Asimpler procedure is to rewrite 2.23 as

CHAPTER 2: Solutions of First-Order Differential Equations13 Table 2.1

integrate both sides of this last equation with respect to x, and then solve the resulting equation for y. The general solution for Equation 2.21 is

where cis the constant of integration. (See Problem 2.7)

Bernoulli Equations ABernoulli differential equation has the form (2.24) where nis a real number. The substitution (2.25) transforms 2.24 into a linear differential equation in the unknown function z(x). (See Problem 2.8)

Solved Problems Solved Problem 2.1Solve This equation may be rewritten in the differential form dydx x y y I x q x dx c

14 DIFFERENTIALEQUATIONS

Solving for y,we obtain the solution in implicit form as with k= −2c. Solving fory explicitly, we obtain the two solutions

Solved Problem 2.2Solve

This differential equation is not separable. Instead it has the form with where

so it is homogeneous. Substituting Equations 2.6 and 2.7 into the equation, we obtain which can be algebraically simpliﬁed to

This last equation is separable; its solution is

which, when evaluated, yields or (2.26)

where we have set and have noted that Finally, substituting v=y/x back into 2.26, we obtain the solution to the given differential equation as yxkx=ln||.

x dv vx dvdx xv x ft x ty ty txtx ty xtx yxx fx y yxx

CHAPTER 2: Solutions of First-Order Differential Equations15

Solved Problem 2.3Solve .

This equation has the form of Equation 2.1 with M(x, y) =2xyand N(x, y) =1 +x2. Since the differential equation is exact. Because this equation is exact, we now determine a function g(x, y) that satisﬁes Equations 2.14 and 2.15. Substituting M(x, y) =2xy into 2.14, we obtain Integrating both sides of this equation with respect to x, we ﬁnd

or (2.27)

Note that when integrating with respect to x, the constant (with respect to x) of integration can depend on y.

We now determine h(y). Differentiating 2.27 with respect to y, we obtain (y) Substituting this equation along with N(x, y) = 1 +x2 into 2.15, we have

Integrating this last equation with respect to y, we obtain h(y) =y+c1 (c1=constant). Substituting this expression into 2.27 yields

The solution to the differential equation, which is given implicitly by 2.16 as g(x, y) =c,is

 x2y+y= c2 (c2=c − c1)

Solved Problem 2.4Determine whether the differential equation ydx− xdy=0 is exact.

This equation has the form of Equation 2.1 with M(x, y) =yand N(x, y) =−x. Here

which are not equal, so the differential equation is not exact.

Solved Problem 2.5Determine whether −1/x2is an integrating factor for the differential equation ydx−xdy=0.

It was shown in Problem 2.4 that the differential equation is not exact. Multiplying it by −1/x2,we obtain

Equation 2.28 has the form of Equation 2.1 with M(x, y) =−y/x2and N(x, y) =1/x.Now

so 2.28 is exact, which implies that −1/x2 is an integrating factor for the original differential equation.

Solved Problem 2.6Solve ydx−xdy=0.

Using the results of Problem 2.5, we can rewrite the given differential equation as which is exact. Equation 2.28 can be solved using the steps described in Equations 2.14 through 2.16.

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