导数题。y'=2y(y+5) f(t)=y 当t=2 y=3 求当t=1 y=?
发布网友
发布时间：2024-09-27 08:11
共3个回答
热心网友
时间：2024-10-06 06:48
解
热心网友
时间：2024-10-06 06:53
分离变量
dy/[y(y+5)]=-2dt
两边积分
∫dy/[y(y+5)]=-2∫dt
而
1/y(y+5)
={[(y+5)-y]/y(y+5)}/5
=(1/5)[1/y-1/(y+5)]
∫dy/[y(y+5)]
=(1/5)[∫dy/y-∫dy/(y+5)]
=(1/5)[lny-ln(y+5)]
-2∫dt=-2t+C
lny-ln(y+5)=10t+C
t=2,y=3代入
ln3-ln8=20+C
C=ln(3/8)-20
t=1
ln(y/(y+5))=10*1+ln(3/8)-20
y/(y+5)=exp(ln(3/8)-10)=(3/8)exp(-10)
y=(3/8)exp(-10)(y+5)
y=(15/8)exp(-10)/[1-(3/8)exp(-10)]
=15/[8exp(10)-3]
热心网友
时间：2024-10-06 06:56
dy/dt=-2y(y+5)
dy/(y(y+5))=-2dt
∫ dy/5(1/y-1/(y+5))=∫-2dt
(lny-ln(y+5))/5=-2t+C
lny-ln(y+5)=-10t+C
当t=2 时 y=3 ln(1/5)=-20+C lny/(y+5)=-10+C ln[y/(y+5)]-ln(1/5)=10
ln[5y/(y+5)]=10 e^10=5y/(y+5)
y=5e^10/(5-e^10)
