If y = ax + b/x^2 + c then (2xy1 + y)y3 =

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(a) 3(xy2 + y1)y2 Here, y = ax + b/x^2 + c \Rightarrow \left( x^2 + c ight)y = ax + b 	ext { Diffferentiating w . r . t . x, we get } 2xy + \left( x^2 + c ight)dy/dx = a 	ext { Diffferentiating w . r . t . x, we get } 2y + 2x y_1 + 2x y_1 + \left( x^2 + c ight) y_2 = 0 \Rightarrow 2y + 4x y_1 + \left( x^2 + c ight) y_2 = 0 	ext { Diffferentiating again w . r . t . x, we get } 2 y_1 + 4 y_1 + 4x y_2 + \left( x^2 + c ight) y_3 + 2x y_2 = 0 \Rightarrow 6 y_1 + 6x y_2 + \left( x^2 + c ight) y_3 = 0 \Rightarrow 6 y_1 + 6x y_2 + \left( - 2y - 4x y_1/y_2 ight) y_3 = 0 \left[ \because 2y + 4x y_1 + \left( x^2 + c ight) y_2 = 0 ight] \Rightarrow 6 y_1 y_2 + 6x \left( y_2 ight)^2 - 2y - 4x y_1 y_3 = 0 \Rightarrow 3 y_1 y_2 + 3x \left( y_2 ight)^2 - y - 2x y_1 y_3 = 0 \Rightarrow \left( y_1 + x y_2 ight)3 y_2 = \left( 2x y_1 + y ight) y_3

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