Question

[(x*根号下1-y的平方)]dx+[y*根号下1-x的平方]dy=0

Answer 1

[x√(1-y²)]dx+[y√(1-x²)]dy=0

[y√(1-x²)]dy＝-[x√(1-y²)]dx

分离变量得 ydy/√(1-y²)=-xdx/√(1-x²)

取积分得：-(1/2)∫d(1-y²)/√(1-y²)=(1/2)∫d(1-x²)/√(1-x²)

积分之得：-√(1-y²)=√(1-x²)+C

即通解为：√(1-x²)+√(1-y²)+C=0