### 17th Putnam 1957

Problem B6

y is the solution of the differential equation (x2 + 9) y'' + (x2 + 4)y = 0, y(0) = 0, y'(0) = 1. Show that y(x) = 0 for some x ∈ [√(63/53) π, 3π/2].

Solution

We compare the equation given with (1) z'' = -4/9 z, z(0) = 0, z'(0) = 1, and (2) w'' = -53/63 w, w(0) = 0, w'(0) = 1.

For all x > 0 we have (x2 + 4)/(x2 + 9) > 4/9, so y''z - yz'' = (4/9 - (x2 + 4)/(x2 + 9) ) yz. Integrating gives y'z - yz'|03π/2 = ∫ (4/9 - (x2 + 4)/(x2 + 9) ) yz dx (*). Now y(0) = z(0) = 0, z = sin 2x/3, so z(3π/2) = 0, z'(3π/2) = -2/3, hence y'z - yz'|03π/2 = 2/3 y(3π/2). Now if y has no zeros on the open interval (0, 3π/2), then y(3π/2) ≥ 0 and the integrand (4/9 - (x2 + 4)/(x2 + 9) ) yz is negative for the entire interval. Hence lhs (*) ≥ 0 and rhs (*) < 0. Contradiction. So y has at least on zero on (0, 3π/2).

Turning to (2), we note that π2 < 10, so (3π/2)2 < 22.5. Hence on the interval [0, 3π/2] we have (x2 + 4)/(x2 + 9) < 26.5/31.5 = 53/63. Now w = sin (√(53/63) x), so w(x) > 0 on the interval (0, k), where k = √(63/53) π.

Now suppose that y has a root in (0, k]. Let h be the smallest such root, so that y(x) is positive on (0, h) and y(h) = 0, y'(h) ≤ 0. Integrate w''y - w y'' = ( (x2 + 4)/(x2 + 9) - 53/63). The lhs is w'y - w y' |0h = w(h) y'(h) ≥ 0. The rhs is ∫0h ( (x2 + 4)/(x2 + 9) - 53/63) y w dx. But the integrand is negative on (0, h), so the rhs < 0. Contradiction. Hence y has no roots on (0, k]. So it must have a root on (k, 3π/2).

Comment. This is essentially the same question as 61/B6.