HMMT 二月 2010 · CALC 赛 · 第 9 题

HMMT February 2010 — CALC Round — Problem 9

[ 7 ] Let x ( t ) be a solution to the differential equation 2 ′ ′′ ( x + x ) + x · x = cos t √ ( ) ′ 2 π with x (0) = x (0) = . Compute x . 5 4 n ∑ 1

解析

[ 7 ] Let x ( t ) be a solution to the differential equation 2 ′ ′′ ( x + x ) + x · x = cos t √ ( ) 2 π ′ with x (0) = x (0) = . Compute x . 5 4 √ 4 450 2 ′ ′ ′ 2 ′ ′ Answer: Rewrite the equation as x + 2 xx + ( xx ) = cos t . Let y = x , so y = 2 xx and 5 ′ 1 ′′ the equation becomes y + y + y = cos t . The term cos t suggests that the particular solution should 2 4 2 be in the form A sin t + B cos t . By substitution and coefficient comparison, we get A = and B = . 5 5 4 2 2 2 Since the function y ( t ) = sin t + cos t already satisfies the initial conditions y (0) = x (0) = and 5 5 5 4 ′ ′ y (0) = 2 x (0) x (0) = , the function y also solves the initial value problem. Note that since x is 5 ( ) π π 2 positive at t = 0 and y = x never reaches zero before t reaches , the value of x must be positive. 4 4 √ √ √ √ ( ) ( ) 4 π π 6 2 450 Therefore, x = + y = · = . 4 4 5 2 5 n ∑ 1