Question
Solve the following differential equations:$e^{d y / d x}=x$

Answer

$
\begin{gathered}
e^{d y / d x}=x \quad \therefore \frac{d y}{d x}=\log x \\
\therefore d y=\log x d x \quad \therefore \int 1 d y=\int \log x d x \\
\text { Now } \int \log x d x=\int(\log x)(1) d x \quad \ldots \ldots \ldots \text { (1) } \\
=(\log x) \int 1 d x-\int\left[\frac{d}{d x}(\log x) \cdot \int 1 d x\right] d x \\
\\
=(\log x)(x)-\int \frac{1}{x} \cdot x d x=x \log x-\int 1 d x \\
=x \log x-x
\end{gathered}
$
$\therefore$ from (1), the general solution is
$
y=x \log x-x+c, \text { i.e. } y=x(\log x-1)+c
$

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