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First find y'y' = -x/y lets name it eq1Then find y'' y'' = (-y + xy')/y^2Substitute x in terms of y and y' from eq 1 we gety'' = -(1 + y'^2)/y lets name it eq2Substitute y' = -x/yWe gety'' = -(x^2 + y^2)/y^3Substitute x^2 + y^2 = a^2y'' = -a^2/y^3a^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''Thereforea = |-(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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