求下列微分方程的通解1.y'=x/(y+siny)2.y'=e^(x-y)

Question

求下列微分方程的通解1.y'=x/(y+siny) 2.y'=e^(x-y)

Answer 1

dy/dx=x/(y+siny)
(y+siny)dy=xdx
两端同时积分得
∫(y+siny)dy=∫xdx
y²/2+cosy=x²/2+C 即为微分方程的通解
dy/dx=e^(x-y)
dy/dx=(e^x)/(e^y)
e^ydy=e^xdx
两端同时积分得
∫e^ydy=∫e^xdx
e^y=e^x+C