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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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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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