muyanli
幼苗
共回答了18个问题采纳率:88.9% 举报
(1+2y)xdx+(1+x^2)dy=0
分离变量:
x/(1+x^2) dx +1/(1+2y) dy=0
积分:
∫x/(1+x^2)dx+∫1/(1+2y)dy=0
1/2∫2x/(1+x^2)dx+1/2∫1/(1+2y)d2y=0
1/2∫1/(1+x^2)dx^2+1/2∫1/(1+2y)d2y=0
∫1/(1+x^2)d(1+x^2)+∫1/(1+2y)d(1+2y)=0
ln(1+x^2)+ln(1+2y)=C1
(1+x^2)(1+2y)=e^C1 =C
so :(1+x^2)(1+2y)=C
1年前
追问
8