Question

微分方程（1+x2）dy-(1+y2)dx=0的通解是______________什么是通解？

Answer 1

-ArcTan[x] + ArcTan[y]=c 

In[1]:= DSolve[D[y[x], x] == (1 + (y[x])^2)/(1 + x^2), y, x]

Out[1]= {{y -> Function[{x}, Tan[ArcTan[x] + C[1]]]}}

Answer 2

(1+x²)dy-(1+y²)dx=0

(1+x²)dy=(1+y²)dx
dy/(1+y²)=dx/(1+x²)
∫dy/(1+y²)=∫dx/(1+x²)
arctany=arctanx+C