2.10.1.1 Example 1

The first step is to see if we can write the above as

\begin{equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1}\end{equation}

Hence

\begin{equation} y^{\prime }=\frac {y}{x}-\frac {2}{x}e^{\frac {-y}{x}} \tag {2}\end{equation}

Comparing (2) to (1) shows that

\begin{align*} n & =1\\ m & =1\\ g\left ( x\right ) & =-\frac {2}{x}\\ b & =-1\\ f\left ( b\frac {y}{x}\right ) & =e^{-\frac {y}{x}}\end{align*}

Hence the solution is

\begin{equation} y=ux \tag {A}\end{equation}

Where \(u\) is the solution to

\begin{equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) \tag {3}\end{equation}

Therefore \(f\left ( bu\right ) =e^{-u}\) and \(\left ( 3\right ) \) becomes

\[ u^{\prime }=-\frac {2}{x^{2}}e^{-u}\]

This is separable.

\begin{align*} e^{u}du & =-\frac {2}{x^{2}}dx\\ \int e^{u}du & =-2\int \frac {1}{x^{2}}dx\\ e^{u} & =\frac {2}{x}+c_{1}\\ u & =\ln \left ( \frac {2}{x}+c_{1}\right ) \end{align*}

Hence (A) becomes

\[ y=x\ln \left ( \frac {2}{x}+c_{1}\right ) \]