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(D) Given the equation: $x \int_{1}^{x} y(t) dt = (x + 1) \int_{1}^{x} t y(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz ru

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

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(D) Given the equation: $x \int_{1}^{x} y(t) dt = (x + 1) \int_{1}^{x} t y(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $\int_{1}^{x} y(t) dt + x y(x) = \int_{1}^{x} t y(t) dt + (x + 1) x y(x)$. Rearranging the terms: $\int_{1}^{x} y(t) dt - \int_{1}^{x} t y(t) dt = (x^2 + x - x) y(x) = x^2 y(x)$. Differentiating again with respect to $x$: $y(x) - x y(x) = 2x y(x) + x^2 y'(x)$. Simplifying: $y(x) (1 - x - 2x) = x^2 y'(x) \implies y(x) (1 - 3x) = x^2 y'(x)$. Separating variables: $\frac{y'(x)}{y(x)} = \frac{1 - 3x}{x^2} = \frac{1}{x^2} - \frac{3}{x}$. Integrating both sides: $\ln|y(x)| = -\frac{1}{x} - 3 \ln|x| + K$. $\ln|y(x)| + \ln|x^3| = -\frac{1}{x} + K$. $\ln|y(x) x^3| = -\frac{1}{x} + K$. $y(x) x^3 = C e^{-\frac{1}{x}}$. Thus,$y(x) = \frac{C e^{-\frac{1}{x}}}{x^3}$.

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

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