If y=e^x(log x), then x y_2+(x-1) y= | vedclass

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(A) Given, $y = e^x \log x$. Differentiating with respect to $x$, we get: $y_1 = \frac{d y}{d x} = e^x \log x + e^x \cdot \frac{1}{x} = y + \frac{e^x}

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$(2 x-1) y_1$

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(A) Given, $y = e^x \log x$. Differentiating with respect to $x$, we get: $y_1 = \frac{d y}{d x} = e^x \log x + e^x \cdot \frac{1}{x} = y + \frac{e^x}{x}$. This implies $x y_1 = x y + e^x$. Differentiating again with respect to $x$: $x y_2 + y_1 = x y_1 + y + e^x$. Substituting $e^x = x y_1 - x y$ from the first derivative equation: $x y_2 + y_1 = x y_1 + y + (x y_1 - x y)$. $x y_2 = x y_1 + y + x y_1 - x y - y_1$. $x y_2 = (2 x - 1) y_1 + (1 - x) y$. Rearranging the terms: $x y_2 + (x - 1) y = (2 x - 1) y_1$.

If $y=e^x(\log x)$, then $x y_2+(x-1) y=$

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