题目

二元微分方程
x''(t) = a - ky(t)'
y(t)'' = kx(t)'
初值 x(0) = 0,y(0) = 0,x'(0) = 0,y'(0) = v.｜t = 0
求 x(t) = y(t) =

提问时间：2021-01-12

答案

x[t] = ( 2 (a - k v0) Sin[(k t)/2]^2 )/k^2,
y[t] = ( a k t + (-a + k v0) Sin(k t) )/k^2