Question

求积分通解要详细过程

图片最好，谢谢

Answer 1

解：∵ylnydx+(x-lny)dy=0
==>(lnydx+xdy/y)-lnydy/y=0 (等式两端同除y)
==>∫(lnydx+xdy/y)-∫lnydy/y=0
==>xlny-(lny)^2/2=C/2 (C是积分常数)
==>2xlny-(lny)^2=C
∴此方程的通解是2xlny-(lny)^2=C。

Answer 2

∫ y e^(-y) dy
=-∫ y de^(-y)
=-y.e^(-y) +∫ e^(-y) dy
=-y.e^(-y) - e^(-y) +C1
e^y .(∫ y e^(-y) dy + C)
=e^y .(-y.e^(-y) - e^(-y) +C1+ C)
=e^y .(-y.e^(-y) - e^(-y) +C2 )
= -y - 1 + C2e^y