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