Question

微分方程dx/dy=e^(x-y)满足y(0)=1的解,

Answer 1

e^(-y) dy = e^(-x) ex
积分：-e^(-y) = -e^(-x) + C
y = - ln[ e^(-x) - C]
y(0)=1 => 1 = -ln(1-C1) => C=1-1/e
y = - ln[ e^(-x) -1 + 1/e] 为所求特解.