# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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and

c2 + y/X c3co^VXL - "\/Xc4si^V"XL = 0. Thus c1 = c4 = 0 and for nontrivial solutions to exist X must satisfy the equation yfXLcosV"XL - sin a/X L = 0. In this case c2 = (-ä/Ã cos\/~X L)(c3) and it follows that the eigenfunction ô1 is given by

ô1(x) = sin( 1/4 x) - VX1 xco^X1 L where X1 is the

smallest positive solution of the equation

\pk L = ta^V"XL. A graphical or numerical estimate of X1

2

reveals that X1 = (4.4934)2/L .

25c. Assuming X Ô 0 the eigenvalue equation is

2(1-cosx) = xsinx, where x = yfX L. Graphing

f(x) = 2(1-cosx) and g(x) = xsinx we see that there is an

intersection for 6 < x < 7. Since both f(x) and g(x) are

zero for x = 2n, this then is the precise root and thus 22

X = (2n) /L . In addition, it appears there might be an

intersection for 0 < x < 1. Using a Taylor series representation for f(x) and g(x) about x = 0, however, shows there is no intersection for 0 < x < 1. Of course x = 0 is also an intersection, which yields X = 0, which gives the trivial solution and hence X = 0 is not an eigenvalue.

Section 11.3, Page 651

1. We must first find the eigenvalues and normalized

eigenfunctions of the associated homogeneous problem y" + Xy = 0, y(0) = 0, y(1) = 0. This problem has the solutions ô (x) = k sinnnx, for X = n2n2, n = 1,2,... .

n n n

Section 11.3

243

Choosing k so that ^ô2dx = 1 we find k = \j~2 . Hence

n J0 Tn n v

the solution of the original nonhomogeneous problem is given by y = Ü.ô .(x), where the coefficients Ü. are

n=1

found from Eq.(12), b = c /(A -2) where c is given by

n n n n

cn = ^ I xsinnnxdx (Eq.9). [Note that the original

problem can be written as -y" = 2y + x and therefore comparison with Eq.(1) yields r(x) = 1 and f(x) = x]. Integrating the expression for c. by parts yields c. =

_ ¦ ² ë/2 (-1)n+1 _

y/~2 (-1)n+1/nn and thus y = Ó——-—------ ^/"2sinnnx.

(n2n2-2)nn

n=1

2. From Problem 1 of Section 11.2 we have

ô = \l~2 sin[(2n-1)nx/2] for A = (2n-1)2n2/4 and from

nn

Problem 7 of that section we have

c. = 4ä/"2 (-1)n+1/(2n-1)2n2. Substituting these values into b = c / (A -2) and y = Ó b ô yields the desired

n n n nn

n=1

result.

3. Referring to Problem 3 of Section 11.2 we have

y = b + Ó b(y2 cosnnx), where b = c / (A -2) for

0 n ’ n n n

n=1

n = 0,1,2,... . The rest of the calculations follow

those of Problem 1.

5. Note that the associated eigenvalue problem is the same as for Problem 1 and that |1-2x| = 1-2x for 0 < x < 1/2 while |1-2x| = 2x-1 for 1/2 < x < 1.

8. Writing the D.E. in the form Eq.(1), we have -y" = p.y + f(x),

so r(x) = 1. The associated eigenvalue problem is y" + Ay = 0, y'(0) = 0, y'(1) = 0 which has the eigenvalues

An = n2n2, n = 0,1,2... and the normalized eigenfunctions

ô0 = 1, ôn(x) = äÄcosnnx, n = 1,2..., as found in Problem 3,

Section 11.2. Now, we assume y(x) = b0 + Ó ü.ôn(x) = y

n=1 n=0

cn 1

ü.ô n(x), Eq.(5), and thus bn = -------------, where cn = f(x^ n(x)dx,

An-Ö j0

244

Section 11.3

10.

11.

12.

14.

18.

n = 0,1,2..., Eq.(9). Thus

X"1 cn ,— x"1 cncosnnx

y(x) = I -------ôn(x) = -^/ö + V 2 I --- -----, where we

ÏÒ0 Õï-Ö Ï=1 Õï-Ö

have assumed ö Ô Xn, n = 0,1,2... .

Since ö = n2 is an eigenvalue of the corresponding homogeneous equation, Theorem 11.3.1 tells us that a solution will exist only if -(a+x) is orthogonal to the corresponding eigenfunction '\/~2sinTCx. Thus we require

J1(a+x)sinnxdx = 0, which yields a = -1/2. With a = -1/2,

0

we find that Y = (x-1/2)/n2 and yc = csinnx + dcosnx by

methods of Chapter 3. Setting y = yc + Y and choosing d

to satisfy the B.C. we obtain the desired family of solutions.

Note that in this case ö = 4n2 and ô2 = \pl sin2nx are the

eigenvalue and eigenfunction respectively of the corresponding homogeneous equation. However, there is no value of a for which -\J~2 J1(a+x)sin2nxdx = 0, and thus

there is no solution.

In this case a solution will exist only if -a is

orthogonal to V~2cosnx., that is if 11acosnxdx = 0.

0

Since this condition is valid for all a, a family of solutions exists.

Since c^ n(x) converges to zero we have c^ n(x) = 0.

n=1 n=1

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