求可分离变量微分方程的特解(4x-x^2)y'=yx=3时y=1

Question

求可分离变量微分方程的特解(4x-x^2)y'=y x=3时y=1

Answer 1

(4x-x^2)dy/dx=y
dy/y=dx/[x(4-x)]=1/4(1/x+1/(4-x))dx
两边积分：ln|y|=1/4∫dx/x+1/4∫dx/(4-x)=1/4ln|x|-1/4ln|4-x|+C=1/4ln|x/(4-x)|+C
y=C*|x/(4-x)|^(1/4) (C≠0)
代入x=3,y=1：1=C*3^(1/4),C=1/3^(1/4)
所以y=|x/(3(4-x))|^(1/4)