Let us consider a curve, y = f x passing through the point - 2 , 2 and the slope of the tangent to the curve at any point ( x , f ( x ) ) is given by f ( x ) + x f ' ( x ) = x 2 . Then

Question

Let us consider a curve, y = f x passing through the point - 2 , 2 and the slope of the tangent to the curve at any point ( x , f ( x ) ) is given by f ( x ) + x f ' ( x ) = x 2 . Then, x 3 - 3 x f ( x ) - 4 = 0, x 2 + 2 x f ( x ) - 12 = 0, x 3 + x f ( x ) + 12 = 0, x 2 + 2 x f ( x ) + 4 = 0

MARKS App · Accepted Answer

y + x d y d x = x 2 d y d x + y x = x I . F . = e ∫ 1 x d x = x Solution of differential equation, y · x = ∫ x · x d x x y = x 3 3 + C Passes through - 2 , 2 , So, - 12 = - 8 + 3 C C = - 4 3 ∴ 3 x y = x 3 - 4 i.e 3 x · f x = x 3 - 4 x 3 - 3 x f x - 4 = 0

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