Solution of \(e^{d y / d x}=x\) when \(x=1\) and \(y=0\) is

Given, \(e^{d y / d x}=x\)
Taking logarithm on both sides, we get
\[
\begin{aligned}
\log e^{d y} / d x &=\log x \\
\frac{d y}{d x} &=\log x \\
d y &=\log x d x
\end{aligned}
\]
On integrating, we get
\[
\begin{aligned}
&\int \begin{aligned}
\int \mathrm{dy} &=\int \log \mathrm{xdx} \\
&=\log \mathrm{x} \int 1 \mathrm{dx}-\int\left[\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x} \int 1 \mathrm{dx}\right] \mathrm{dx} \\
&=\mathrm{x} \log \mathrm{x}-\int \frac{1}{\mathrm{x}} \times \mathrm{x} \mathrm{dx} \\
&=\mathrm{x} \log \mathrm{x}-\int \mathrm{dx} \\
&=\mathrm{x} \log \mathrm{x}-\mathrm{x} \\
\mathrm{y} &=\mathrm{x}(\log \mathrm{x}-1)+\mathrm{C} \\
\text { when } \mathrm{x} &=1 \text { and } \mathrm{y}=0 \\
\Rightarrow \quad 0=1(\log 1-1)+\mathrm{C}
\end{aligned}
\end{aligned}
\]
\[
\begin{array}{ll}
\Rightarrow & 0=(0-1)+C \\
\Rightarrow & C=1
\end{array}
\]
\(\therefore \mathrm{Eq}\). (i) becomes
\[
y=x(\log x-1)+1
\]