First find y' y' = -x/y lets name it eq1 Then find y'' y'' = (-y + xy')/y^2 Substitute x in terms of y and y' from eq 1 we get y'' = -(1 + y'^2)/y lets name it eq2 Substitute y' = -x/y We get y'' = -(x^2 + y^2)/y^3 Substitute x^2 + y^2 = a^2 y'' = -a^2/y^3 a^2 = -y^3 × y'' a = (-y^3 × y'')^(1/2) a = |y| × (-yy'')^(1/2) From eq 2 1 + y'^2 = -yy'' and y = -(1 + y'^2)/y'' Therefore a = |-(1 + y^2)/y''| (1 + y'^2)^(1/3) We can write |-(1 + y^2)| = {(1 + y^3)^2}^(1/2) Therefore a = {(1 + y'^2)^3}^(1/2)/|y''|
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