∵y''²=(100-x)²/4-[y'(100-x)/2]²
==>y''²=(100-x)²/4-y'²(100-x)²/4
==>4y''²=(1-y'²)(100-x)²
==>2y''=±(100-x)√(1-y'²)
==>2dy'/√(1-y'²)=±(100-x)dx
==>2arcsin(y')=2C1±(100x-x²/2) (C1是积分常数)
==>y'=sin[C1±(50x-x²/4)]
==>y=∫sin[C1±(50x-x²/4)]dx+C2 (C1是积分常数)
∴原方程的通解是y=∫sin[C1±(50x-x²/4)]dx+C2 (C1,C2是积分常数).