40th Putnam 1979

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Problem B4

Find a non-trivial solution of the differential equation F(y) ≡ (3x2 + x - 1)y'' - (9x2 + 9x - 2)y' + (18x + 3)y = 0.
y = f(x) is the solution of F(y) = 6(6x + 1) such that f(0) = 1, and ( f(-1) - 2)( f(1) - 6) = 1. Find a relation of the form ( f(-2) - a)( f(2) - b) = c.

 

Solution

Answer: a = 6, b = 14, c = 1.

The differential equation looks fairly horrible; there is no obvious systematic way of solving it. So we try various types of solutions. Trying simple polynomials leads to x2 + x. Trying exponentials leads to e3x. Obviously, a particular solution to F(y) = 36x + 6 is y = 2, so the general solution is y = 2 + Ae3x + Bx + Bx2. f(0) = 1 implies A = -1. ( f(-1) - 2)( f(1) - 6) = (- e-3 )( -4 - e3 + 2B) = 1, so B = 2.

f(-2) = 2 - e-6 + 4, f(2) = 2 - e6 + 12, so we need a = 6, b = 14, c = 1 (and then the relation reduces to ( - e-6 )( - e 6) = 1, as required).

 


 

40th Putnam 1979

© John Scholes
jscholes@kalva.demon.co.uk
4 Dec 1999