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(D) Given the differential equation $e^{dy/dx} = x$.Taking the natural logarithm on both sides,we get:$\frac{dy}{dx} = \log x$Integrating both sides w

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$y = x(\log x - 1) + 1$

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(D) Given the differential equation $e^{dy/dx} = x$.Taking the natural logarithm on both sides,we get:$\frac{dy}{dx} = \log x$Integrating both sides with respect to $x$:$\int dy = \int \log x \, dx$Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \log x$ and $dv = dx$:$y = x \log x - \int x \cdot \frac{1}{x} \, dx$$y = x \log x - \int 1 \, dx$$y = x \log x - x + C$$y = x(\log x - 1) + C$ (Equation $i$)Given the initial conditions $x = 1$ and $y = 0$,substitute these into Equation $i$:$0 = 1(\log 1 - 1) + C$$0 = 1(0 - 1) + C$$0 = -1 + C$$C = 1$Substituting $C = 1$ back into Equation $i$,we get:$y = x(\log x - 1) + 1$

The solution of $e^{dy/dx} = x$ with the initial conditions $x = 1$ and $y = 0$ is:

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