dx比dt常微分方程组的求解

问题描述

精选答案

dx/dt=5x+4y,dy/dt=4x+5y

∴dy/dx=(4x+5y)/(5x+4y)

令z=y/x,则dy/dx=z+xdz/dx

∴z+xdz/dx=(4+5z)/(5+4z)

==>(5+4z)dz/(1-z²)=4dx/x

==>[9/(1-z)+1/(1+z)]dz=8dx/x

==>ln|1+z|-9ln|1-z|=8ln|x|+ln|C| (C是积分常数)

==>(1+z)/(1-z)^9=Cx^8

==>(1+y/x)/(1-y/x)^9=Cx^8

==>(x+y)/(x-y)^9=C

==>x+y=C(x-y)^9 (C是积分常数)

故原常微分方程组的通解是：x+y=C(x-y)^9 (C是积分常数).