y = f(x) is a solution of y'' + e^{x}y = 0. Prove that f(x) is bounded.

**Solution**

We have 2 e^{-x} y' y'' + 2 y y' = 0. Integrating from 0 to k gives y(k)^{2} = y(0)^{2} - 2 ∫_{0}^{k} e^{-x} y' y'' dx. Integrating by parts gives 2 ∫_{0}^{k} e^{-x} y' y'' dx = e^{-x} (y')^{2}|_{0}^{k} + ∫_{0}^{k} (y')^{2} e^{-x} dx = e^{-k}y'(k)^{2} - y'(0)^{2} + ∫_{0}^{k} (y')^{2} e^{-x} dx = A - y'(0)^{2}, where A > 0.

Hence y(k)^{2} = y(0)^{2} + y'(0)^{2} - A < y(0)^{2} + y'(0)^{2}, which establishes that y is bounded.

© John Scholes

jscholes@kalva.demon.co.uk

25 Jan 2002