全微分方程求同通解积分下限怎么取

2025-12-14 11:31:25
推荐回答(1个)
回答1:

∫(xy^2+y-1)dx=x^2y^2/2+xy-x+c1(y)
∫(x^2y+x+2)dy=x^2y^2/2+xy+2y+c2(x)
x^2y^2/2+xy-x+c1(y)=x^2y^2/2+xy+2y+c2(x)
c1(y)=2y+c,c2(x)=-x+c
原函数 f(x,y)=x^2y^2/2+xy-x+2y+c