Eliminating the arbitrary constants B and C in the expression y = (2/3C)(Cx – 1)^(3/2) + B, we get
x [1 + (dy/dx)²] = d²y/dx²
2x(dy/dx)(d²y/dx²) = 1 + (dy/dx)²
(dy/dx)(d²y/dx²) = 1
(dy/dx)² + 1 = d²y/dx²
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