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

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which give c1 = c2 = 0 and thus X = 0 is not an

eigenvalue.

If X > 0, the general solution of the D.E. is y = c^in^/X x + c2cosV"X x. The B.C. require that

c2 - \[X c1 = 0, and

(sin \fx + \fx cos \[X ) c1 + (cosV"X - \fX sin \[X )c2 = 0. In order to have nontrivial solutions X must satisfy (X-1)sinV~X - 2\[X co^\[X = 0. In this case c2 = "\[X c1

and thus ôn = sin^x + -\jXn co^Xn x. If X Ô 1, the

eigenvalue equation is equivalent to tan^/X = 2\[X /(X-1) and thus by graphing f^\[X ) = tan^/~X and

g(\J~X ) = 2\J~X / (X-1 ) we can estimate the eigenvalues. Since g(V"X) has a vertical asymptote at X = 1 and f^\[X ) has a vertical asymptote at \[X = n/2, we see that 1 < X1 < n/2.

By interating numerically, we find ó/X1 = 1.30655 and thus X1 = 1.7071. The second eigenvalue will lie to the right of n, the second zero of tan^/X . Again iterating numerically, we find \!X2 = 3.6732 and thus X2 = 13.4924. For large values

of n, we see from the graph that ^X~ = (n-1)n, which are the zeros of tan^/X . Thus Xn = (n-1)2n2 for large n.

For X < 0, the discussion follows the pattern of Example 1 yielding y(x) = c^inh^/^T x + c2cosh^/^ x. The B.C. then yield c2 - ä/"ö c1 = 0 and

(sinhV"^ + cos^V"^ )c1 + (coshV"^ + sinhV"^ )c2 =0, which

236

Section 11.1

have a non_zero solution if and only if

(p.+ 1)sinh\/"^ + 2ä/"ö coshV"^ = 0. By plotting y = tanh^"^ and

y = _^V"^ /(ö+1) we see that they intersect only at ö = 0, and

thus there are no negative eigenvalues.

10. If X = 0, the general solution of the D.E. is

y = c1 + c2x. The B.C. y(0) + y'(0) = 0 requires

c + c = 0 and the B.C. y(1) = 0 requires c + c = 0

1 2 1 2

and thus X = 0 is an eigenvalue with corresponding

eigenfunction ô0(x) = 1_x.

If X < 0, set _ X = ö2 to obtain y = c1cosp.x + c2sinp,x.

In this case the B.C. require c1 + ö^ = 0 and

c1cosp. + c2sinp, = 0 which yields nontrivial solutions for c1

and c2 (i.e., c1 = _M-c2) if and only if tanp, = ö. By

plotting on the same graph f(p.) = ö and g(p.) = tanp,, we see that they intersect at Ö0 = 0 (ö = 0 ^ X = 0, which has

already been discussed), ö1 = 4.49341 (which is just to the

left of the vertical asymptote of tanö at ö = 3n/2, ö2 = 7.72525 (which is just to the left of the vertical asymptote of tanö at ö = 5n/2) and for larger values ö = (2n+1)n/2. Since X = _ö2, we have X = _20.1907,

r n n r n 1

X2 = _59.6795, X = _(2n+1)2n2/4 and ô = siïö x _ ö cosö x.

n Tn ãïãïãï

If X > 0, the general solution of the D.E. is

y(x) = c^osh^/X x + c2sinh^/X x. The B.C. respectively

require that c1 + \[X c2 = 0 and c^oshVX + c2sinhV"X = 0 and

thus X must satisfy tanh^/X = '\/X in order to have nontrivial solutions. The only solution of this equation is X = 0 and thus there are no positive eigenvalues.

11a. From Eq.(i) the coefficient of y' is öÎ and from Eq.(ii) the

coefficient of y' is (öÐ)'. Thus (öÐ)' = öÎ, which gives

Eq.(iii).

11b. Eq.(iii) is both linear and separable. Using the latter

approach we have dö/ö = [(Q_P')/P]dx and thus

lnP. Taking the exponential of

both sides yields Eq.(iv). The choice of x0 simply alters the constant of integration, which is immaterial here.

237

13. Since P(x) = x2 and Q(x) = x, we find that

p.(x) = (1/x2)exp[^x (s/s2)ds] = k/x, where k is an

0

arbitrary constant which may be set equal to 1. It follows that Bessel's equation takes the form (xy')' + (x2-u2/x)y = 0.

18a.Assuming y = s(x)u, we have y' = s'u + su' and

y" = s"u + 2s'u' + su" and thus the D.E. becomes su" + (2s'+4s)u' + [s" + 4s' + (4 + 9X)s]u = 0. Setting 2s' + 4s = 0 we find s(x) = e-2x and the D.E. becomes u" + 9Xu = 0. The B.C. y(0) = 0 yields s(0)u(0) = 0, or u(0) = 0 since s(0) Ô 0. The B.C. at L

is y'(L) = s' (L)u(L) + s(L)u'(L) = e-2L (-2u(L) + u'(L)) = 0

and thus u'(L) - 2u(L) = 0. Thus the B.V.P. satisfied by

u(x) is u" + 9Xu = 0, u(0) = 0, u'(L) - 2u(L) = 0.

If X < 0, the general solution of the D.E. u" + 9Xu = 0

is u = c1sinh3p.x + c2cosh3p.x where -X = ö2. The B.C.

require that c2 = 0, c1(3p.cosh3p.L - 2sinh3p.L) = 0. In order to have nontrivial solutions ö must satisfy the equation 3ö/2 = tanh3p.L. A graphical analysis reveals that for L < 1/2 this equation has no solutions for ö Ô 0 so there are no negative eigenvalues for L < 1/2. If L > 1/2 there is one solution and hence one negative eigenvalue with eigenfuction ô-1(x) = e_2xsinh3p.x.

If X = 0, the general solution of the D.E. u" + 9Xu = 0

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