Let Y = Y ( X ) be a curve lying in the first quadrant such that the area enclosed by the line Y - y = Y ' ( x ) ( X - x ) and the co-ordinate axes, where ( x , y ) is any point on the curve, is always - y 2 2 Y ' ( x ) + 1 , Y ' x ≠ 0 . If Y ( 1 ) = 1 , then 12 Y ( 2 ) equals ________.

Step-by-step solution

Plotting the line Y - y = Y ' ( x ) ( X - x ) we get, Now, from the above diagram, The area of the triangle formed will be, A = 1 2 - y Y ' ( x ) + x ( y - x Y ' x ) Now, according to the given condition we get, 1 2 - y Y ' ( x ) + x ( y - x Y ' x ) = - y 2 2 Y ' ( x ) + 1 ⇒ - y + x Y ' ( x ) y - x Y ' ( x ) = - y 2 + 2 Y ' ( x ) ⇒ - y 2 + x y Y ' ( x ) + x y Y ' ( x ) - x 2 Y ' ( x ) 2 = - y 2 + 2 Y ' ( x ) ⇒ 2 x y - x 2 Y ' ( x ) = 2 ⇒ d y d x = 2 x y - 2 x 2 ⇒ d y d x - 2 x y = - 2 x 2 It is a linear differentia