∵∫dx/√(1-x²)=arcsinx+C (C是积分常数)∴由(1-x²)^1/2*y'-(1-y²)^1/2=0==>dy/√(1-y²)=dx/√(1-x²)==>arcsiny=arcsinx+C故 原方程的通解是arcsiny=arcsinx+C (C是积分常数)....