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If $x \cdot \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}$, then $f(x y)$ is equal to
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The correct answer is:
$k \cdot e^{\frac{x^{2}}{2}}$
$x \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}$
i.e., $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xy})=\mathrm{x} \frac{\mathrm{f}(\mathrm{x}, \mathrm{y})}{\mathrm{f}^{\prime}(\mathrm{x}, \mathrm{y})}$
$\Rightarrow \frac{\mathrm{f}^{\prime}(\mathrm{xy})}{\mathrm{f}(\mathrm{xy})} \mathrm{d}(\mathrm{xy})=\mathrm{x} \mathrm{dx}$
$\Rightarrow \int \frac{\mathrm{f}^{\prime}(\mathrm{xy})}{\mathrm{f}(\mathrm{xy})} \mathrm{d}(\mathrm{xy})=\int \mathrm{xdx}$
$\Rightarrow \log [\mathrm{f}(\mathrm{xy})]=\frac{\mathrm{x}^{2}}{2}+\mathrm{C}$
$\Rightarrow \mathrm{f}(\mathrm{xy})=\mathrm{e}^{\left(\mathrm{x}^{2} / 2+\mathrm{C}\right)}$
$=\mathrm{e}^{\frac{\mathrm{x}^{2}}{2} \cdot \mathrm{e}^{\mathrm{C}}=\mathrm{k} \cdot \mathrm{e}^{\frac{\mathrm{x}^{2}}{2}}}$
i.e., $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xy})=\mathrm{x} \frac{\mathrm{f}(\mathrm{x}, \mathrm{y})}{\mathrm{f}^{\prime}(\mathrm{x}, \mathrm{y})}$
$\Rightarrow \frac{\mathrm{f}^{\prime}(\mathrm{xy})}{\mathrm{f}(\mathrm{xy})} \mathrm{d}(\mathrm{xy})=\mathrm{x} \mathrm{dx}$
$\Rightarrow \int \frac{\mathrm{f}^{\prime}(\mathrm{xy})}{\mathrm{f}(\mathrm{xy})} \mathrm{d}(\mathrm{xy})=\int \mathrm{xdx}$
$\Rightarrow \log [\mathrm{f}(\mathrm{xy})]=\frac{\mathrm{x}^{2}}{2}+\mathrm{C}$
$\Rightarrow \mathrm{f}(\mathrm{xy})=\mathrm{e}^{\left(\mathrm{x}^{2} / 2+\mathrm{C}\right)}$
$=\mathrm{e}^{\frac{\mathrm{x}^{2}}{2} \cdot \mathrm{e}^{\mathrm{C}}=\mathrm{k} \cdot \mathrm{e}^{\frac{\mathrm{x}^{2}}{2}}}$
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