Question

解初值问题。数学。微积分。

Answer 1

Answer 2

dy/(y^2-y) = e^x dx
dy[1/y - 1/(y-1)] = e^x dx
lny -ln(y-1) = e^x+C
ln(y/(y-1)) =e^x+C
x=0,y=2代入得到
ln2 = 1+C, C=ln2 -1
所以ln(y/(y-1)) = e^x+ln2-1