#### Example 1.3.1.

Take the equation

\begin{equation*}
y' = xy .
\end{equation*}

Note that \(y=0\) is a solution. We will remember that fact and assume \(y \not =0\) from now on, so that we can divide by \(y\text{.}\) Write the equation as \(\frac{dy}{dx} = xy\) or \(\frac{dy}{y} = x \, dx\text{.}\) Then

\begin{equation*}
\int \frac{dy}{y} = \int x\,dx + C .
\end{equation*}

We compute the antiderivatives to get

\begin{equation*}
\ln \, \lvert y\rvert = \frac{x^2}{2} + C ,
\end{equation*}

or

\begin{equation*}
\lvert y \rvert = e^{\frac{x^2}{2} + C} = e^{\frac{x^2}{2}} e^C = D e^{\frac{x^2}{2}} ,
\end{equation*}

where \(D > 0\) is some constant. Because \(y=0\) is also a solution and because of the absolute value, we can write:

\begin{equation*}
y = D e^{\frac{x^2}{2}} ,
\end{equation*}

for any number \(D\) (including zero or negative).

We check:

\begin{equation*}
y' = D x e^{\frac{x^2}{2}} = x \left( D e^{\frac{x^2}{2}} \right) = xy .
\end{equation*}

Yay!