证明:函数y=√(2x-x^2 满足关系式 y^3y'+1=0

y'=(2-2x)/[2*√(2x-x^2)]
y^3=(2x-x^2) √(2x-x^2)
y^3y'=(2-2x)/2 * (2x-x^2)=(1-x)/(2x-x^2)怎么证明:函数y=√(2x-x^2 满足关系式 y^3y“+1=0y'=(2-2x)/[2*√(2x-x^2)]=(1-x)/√(2x-x^2)y"=-1/√(2x-x^2) - (x-1)^2/[2(2x-x^2)√(2x-x^2)]y^3y"=(2x-x^2) √(2x-x^2) * {-1/√(2x-x^2 + 1/[2(x-1)(2x-x^2)√(2x-x^2)])}=x^2-2x-(1-x)^2=-1y^3y"+1=0