题目

求解一阶微分方程
(t-2y-4)dt+(2t-y-5)dy=0
求y(t)=?

提问时间：2021-04-22

答案

(t-2y-4)dt+(2t-y-5)dy=0
dy/dt=(t-2y-4)/(2t-y-5)
令Y=y+1 T=t-2代入：
dY/dT=(T-2Y)/(2T-Y)=(1-2Y/T)/(2-Y/T)
令u=Y/T Y=Tu dY/dT=u+Tdu/dT
u+Tdu/dT=(1-2u)/(2-u)
Tdu/dT=(1-2u)/(2-u)-u=(1-4u+u^2)/(2-u)
(2-u)du/(1-4u+u^2)=dT/T
-2(2-u)du/(1-4u+u^2)=-2dT/T 积分得通
ln((1-4u+u^2)=-2lnT+lnC
(1-4u+u^2)T^2=C
(1-4(y+1)/(t-2)+(y+1)^2/(t-2)^2)(t-2)^2=C
(t-2)^2-4(y+1)(t-2)+(y+1)^2=C