# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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7. y - y' + y - y = sec t, -n/2 < t < n/2

8. y - y = csc t, 0 < t < n

In each of Problems 9 through 12 find the solution of the given initial value problem. Then plot a graph of the solution.

9. y + y = sec t, y(0) = 2, y(0) = 1, y'(0) = —2

10. yv + 2y" + y = sint, y(0) = 2, y(0) = 0, y"(0) = -1, y"'(0) = 1

11. y — y'+ y — y = sec t, y(0) = 2, y(0) = — 1, y'(0) = 1

12. y — y = csc t, y(n/2) = 2, y (n/2) = 1, y'(n/2) = — 1

13. Given that x, x2, and 1 /x are solutions of the homogeneous equation corresponding to

x3y + x2/ - 2xy + 2y = 2x4, x > 0,

determine a particular solution.

230

Chapter 4. Higher Order Linear Equations

14. Find a formula involving integrals for a particular solution of the differential equation

/"- y" + y - y = g(t).

15. Find a formula involving integrals for a particular solution of the differential equation

yv - y = g(t).

Hint: The functions sin t, cos t, sinh t, and cosh t form a fundamental set of solutions of the homogeneous equation.

16. Find a formula involving integrals for a particular solution of the differential equation

/"- 3y" + 3y - y = g(t).

If g(t) = t-2e(, determine Y(t).

17. Find a formula involving integrals for a particular solution of the differential equation

v3 f - 3x2 f + 6xy - 6y = g(x), x > 0.

Hint: Verify that x, x2, and x3 are solutions of the homogeneous equation.

REFERENCES

Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ: Prentice Hall, 1961; New York: Dover, 1989).

Ince, E. L., Ordinary Differential Equations (London: Longmans, Green, 1927; New York: Dover, 1953).

CHAPTER

5

Series Solutions of Second Order Linear Equations

Finding the general solution of a linear differential equation rests on determining a fundamental set of solutions of the homogeneous equation. So far, we have given a systematic procedure for constructing fundamental solutions only if the equation has constant coefficients. To deal with the much larger class of equations having variable coefficients it is necessary to extend our search for solutions beyond the familiar elementary functions of calculus. The principal tool that we need is the representation of a given function by a power series. The basic idea is similar to that in the method of undetermined coefficients: We assume that the solutions of a given differential equation have power series expansions, and then we attempt to determine the coefficients so as to satisfy the differential equation.

5.1 Review of Power Series

In this chapter we discuss the use of power series to construct fundamental sets of solutions of second order linear differential equations whose coefficients are functions of the independent variable. We begin by summarizing very briefly the pertinent results about power series that we need. Readers who are familiar with power series may go on to Section 5.2. Those who need more details than are presented here should consult a book on calculus.

231

232

Chapter 5. Series Solutions of Second Order Linear Equations

1. A power series an (x — x0)” is said to converge at a point x if

n=0

lim Y] an (x — x0)n

m—»-to n u

n0

exists for that x. The series certainly converges for x = x0; it may converge for all x, or it may converge for some values of x and not for others.

TO

2. The series an (x — x0)” is said to converge absolutely at a point x if the series

n=0

2^K(x — x0)n 1 = Z^|an ||x — x0|n

n=0 n=0

converges. It can be shown that if the series converges absolutely, then the series also converges; however, the converse is not necessarily true.

3. One of the most useful tests for the absolute convergence of a power series is the ratio test. If an = 0, and if for a fixed value of x

lim

n^TO

an + 1(x — x0)n + 1

an (x — x0)n

= |x — x0| lim

n+1

= L |x — x01,

then the power series converges absolutely at that value of x if |x — x0| < 1/L, and diverges if |x — x0l > 1/L. If |x — x0l = 1/L, the test is inconclusive.

a

n

For which values of x does the power series

TO

J2(—1)n+1n(x — 2)n

n=1

converge?

To test for convergence we use the ratio test. We have

lim

n

(—1)n+2(n + 1)(x — 2)

n+1

(—1)n+1n(x — 2)n

= |x — 2| lim

= |x — 2|.

According to statement 3 the series converges absolutely for |x — 2| < 1, or 1 < x < 3, and diverges for |x — 2| > 1. The values of x corresponding to |x — 2| = 1 are x = 1 and x = 3. The series diverges for each of these values of x since the nth term of the series does not approach zero as n ^to.

1

n

n

TO

4. If the power series an (x — x0)n converges at x = x1, it converges absolutely

n=0

for |x — x0| < |x1 — x01; and if it diverges at x = x1, it diverges for |x — x0| >

|x1 — x01.

5. There is a nonnegative number p, called the radius of convergence, such that

TO

Y.an (x — x0)n converges absolutely for |x — x0| < p and diverges for |x — x01 >

n=0

p. For a series that converges only at x0, we define p to be zero; for a series that

5.1 Review of Power Series

233

converges for all x, we say that p is infinite. If p > 0, then the interval |x — x01 < p is called the interval of convergence; it is indicated by the hatched lines in Figure

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