For x ∈ R , x ≠ 0 , if y x is a differentiable function such that x ∫ 1 x y t d t = x + 1 ∫ 1 x t y t d t , then y x equals (where C is a constant)

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Aicharya · Accepted Answer

x &#8747; 1 x y t d t = x &#8747; 1 x t y t d t + &#160; &#8747; 1 x t y t d t Differentiate w.r.t. x &#8747; 1 x y ( t ) d t + x y ( x ) = &#8747; 1 x t y ( t ) d t + x [ x y ( x ) ] + x y ( x ) &#8747; 1 x y ( t ) d t = &#8747; 1 x t y ( t ) d t + x 2 y ( x ) &#160; Differentiate again w.r.t. x y ( x ) = x y ( x ) + 2 x &#160; y ( x ) + x 2 y ' ( x ) &#160; 1 - 3 x y x = x 2 y ' x y ' x y x = 1 - 3 x x 2 1 y d y d x = 1 - 3 x x 2 Integrating on both sides &#8658; ln y = - 1 x - 3 ln x + C &#8658; ln ( y x 3 ) = - 1 x + C &#8658; y x 3 = &#160; e - 1 x + C &#8658; y = e - 1 x + C x 3 &#8658; y = C x 3 e - 1 x

For x ∈ R , x ≠ 0 , if y x is a differentiable function such that x ∫ 1 x y t d t = x + 1 ∫ 1 x t y t d t ,…

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