Question

高等数学微分方程，求解～

Answer 1

f''(x) - f'(x) = cosx, 是二阶常系数线性微分方程，
特征方程 r^2 - r = 0, r = 0, 1
设特解 f(x) = Acosx + Bsinx,
代入微分方程得 - Acosx - Bsinx + Asinx - Bcosx = cosx
A - B = 0 , - A - B = 1, 解得 A = B = -1/2。
特解 f(x) = -(1/2)(cosx + sinx),
通解 f(x) = C1 + C2 e^x - (1/2)(cosx + sinx)
f(0) = 0, 得 C1 + C2 - 1/2 = 0
f'(x) = C2 e^x - (1/2)(-sinx+cosx)
= C2 e^x + (1/2)(sinx-cosx)
f'(0) = C2 - 1/2, f'(2π) = C2 e^(2π) - 1/2
最大值应是 f'(2π) = C2 e^(2π) - 1/2 = 2，
得 C2 = 5/[2e^(2π)]， 则 C1 = 1/2 - 5/[2e^(2π)]
f(x) = 1/2 - 5/[2e^(2π)] + 5/[2e^(2π)] e^x - (1/2)(cosx + sinx)

追问

为什么最大值在2pi处呢