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For the Equation (x216)3(x1)y2xy+y=0\left( x ^ { 2 } - 16 \right) ^ { 3 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0

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For the equation (x216) 3(x1) y2xy+y=0\left( x ^ { 2 } - 16 \right) ^ { 3 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0 , the point x=1x = 1 is


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