PROBLEMS AND SOLUTIONS 261

and E 1, we obtain

G(p, q) (p + q + a + z- 1) 3F2[q z,

from which the symmetry is obvious.

p-z 1"

p+q+a+x, p+q+a+y;

REFERENCE

[1] W. N. BAILEY, Generalized Hypergeometric Series, Stechert-Hafner, New York and London, 1964.

Also solved by O. P. LOSSERS (Technological University, Eindhoven, theNetherlands).

Problem 73-12, A Nonlinear Differential Equation, by OTTO G. RUEHR (MichiganTechnological University).

Determine the general solution of

x2 + 2 f= -x + 2x -d-x + x

The problem arose in modeling the Helmholtz equation in two dimensions.

Solution by the proposer.

Assuming that f’ + 2x 0, we have

f"+2f’+2x

Integration yields

and hence

f’ X

log If’ + 2xl log Ifl + log Icl + dx,

f’+ =cexp dx

Letting x/f= u"/u’ and integrating, we obtain

f(u’)z c2 exp {cu}.Substitutingf= xu’/u" and noting the identity

(d2u/ dx2)/(du/dx)3 d2x/du2,

we obtain the linear differential equation

d2x exp {-cu}du2 - c 2

x--O.

Let

=t+c,Dow

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262 PROBLEMS AND SOLUTIONS

where c is a constant. Then

f c2 exp {clu} (dx/du)2 (dx/dt),and the equation for x becomes

d2x dx-t- 4-x =0.dt2 + c dt

Hence, the general solution in parametric form is

x AJo(t + c) + B Yo(t + c),

f [Ad,(t + c) + BY,(t + c)] 2,

where A and B are arbitrary constants. Note that the limit c --, o corresponds tothe solution f(x)= C- x2.

Editorial note. It is instructive to consider the initial value problem for thisdifferential equation. Letting 0 correspond to the initial point (xo,f(xo),f’(xo)), one finds that c is given by

2x//f(xo)-- f’(Xo) + 2Xo’and then A and B are determined by the equations

AJo(c) + BYo(c) Xo,

AJI(c) + B Y,(c) w/f(xo).The solution holds in an interval [x,x2] containing Xo, where x and x2 aredetermined by the condition dx/dt 0. [C.C.R.]

Also solved by H. J. GAWLIK (Royal Armament Research and DevelopmentEstablishment, Fort Halstead, Sevenoaks, Kent, England).

Problem 73-13, An Integral Inequality, by RICHARD ASKEY (University of Wis-consin).

Heisenberg’s inequality can be stated as

IF(x)] 2 dx,Mo x21F(x)l2 dx, t2lF(t)l 2 dt >-

where

and

By Schl6milch’s inequality,

Mo{a, b} x

F(t) F(x) e2ixt dx.

(1) Mr x21F(x)l2 dx, t21F(t)l 2 dt >= IF(x)I 2 dxDow

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