发布网友 发布时间:2022-04-27 10:32
共1个回答
热心网友 时间:2023-09-13 01:57
怎样下载
www.xjl222.com2(1) y' = dy/dx = (y-2x)/(x+2y) = (y/x-2)/(1+2y/x)
令 p = y/x, 则 y = xp,dy/dx = p+xdp/dx
则微分方程化为 p+xdp/dx = (p-2)/(1+2p)
xdp/dx = (p-2)/(1+2p)-p = -2(1+p^2)/(1+2p)
(1+2p)dp/(1+p^2) = -2dx/x
arctanp + ln(1+p^2) = -2lnx + lnC
(1+p^2)e^(arctanp) = C/x^2
(x^2+y^2)e^[arctan(y/x) = C
y(1) = 1 代入,得 C = 2e^(π/4)
则得 (x^2+y^2)e^[arctan(y/x) = 2e^(π/4)