For x∈R, x≠0, if yx is differentiable function such that x∫1xy(t)dt=(x+1)∫1xty(t)dt, then yx equals: (where C is a constant)

x ∫1xy(t) dt =x+1∫1xt y(t)dt differentiate w.r.t x by Newtoris Leibintz Rale x y+∫1xy(t) dt=(x+1) x y+(1)∫1xt y(t) dt ∫1xy(t) dt=x2y+∫1x t y(t) dt differentiate again y=x2dydx+y(2x)+xy x2dydx=y(1-3x ) dyy=(1-3x)x2dx ∫dyy=∫1x2-3x dx log y=-1x -3 log x+log c log y x3c=-1x yx3c=e-1/x ⇒ y=c e-1/xx3

For x∈R, x≠0, if yx is differentiable function such that x∫1xy(t)dt=(x+1)∫1xty(t)dt, then yx equals: (where C is a constant)

x ∫1xy(t) dt =x+1∫1xt y(t)dt differentiate w.r.t x by Newtoris Leibintz Rale x y+∫1xy(t) dt=(x+1) x y+(1)∫1xt y(t) dt ∫1xy(t) dt=x2y+∫1x t y(t) dt differentiate again y=x2dydx+y(2x)+xy x2dydx=y(1-3x ) dyy=(1-3x)x2dx ∫dyy=∫1x2-3x dx log y=-1x -3 log x+log c log y x3c=-1x yx3c=e-1/x ⇒ y=c e-1/xx3

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For $x \in R, x$ $\neq 0,$ if $y (x)$ is differentiable function such that $x \int_{1}^{x} y (t) d t = (x + 1) \int_{1}^{x} t y (t) d t,$ then $y (x)$ equals: (where C is a constant)

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$\frac{C}{x^{3}} e^{- \frac{1}{x}}$

$\frac{C}{x^{2}} e^{\frac{1}{x}}$

$\frac{C}{x} e^{- \frac{1}{x}}$

$C x^{3} e^{\frac{1}{x}}$

answer is D.

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Detailed Solution

$x \int_{1}^{x} y (t) dt = (x + 1) \int_{1}^{x} t y (t) dt differentiate w . r . t x^{} by Newtoris Leibintz Rale x y + \int_{1}^{x} y (t) dt = (x + 1) x y + (1) \int_{1}^{x} t y (t) dt \int_{1}^{x} y (t) dt = x^{2} y + \int_{1}^{x} t y (t) dt differentiate again y = x^{2} \frac{dy}{dx} + y (2 x) + xy x^{2} \frac{dy}{dx} = y (1 - 3 x^{}) \frac{dy}{y} = \frac{(1 - 3 x)}{x^{2}} dx \int \frac{dy}{y} = \int (\frac{1}{x^{2}} - \frac{3}{x}) dx \log y = \frac{- 1}{x} - 3 \log x + \log c \log \frac{y x^{3}}{c} = \frac{- 1}{x} \frac{{yx}^{3}}{c} = e^{- 1 / x} \Rightarrow y = \frac{c e^{- 1 / x}}{x^{3}}$