3.1.12.19 Example \(y^{\prime }=\frac {y}{x}+\frac {1}{x}e^{-\frac {y}{x}}\)

This is homogeneous class D  \(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left ( \frac {y}{x}\right ) \). Hence from lookup table

\begin{align*} \xi & =x^{2}\\ \eta & =xy \end{align*}

From above we found the solution to be

\[ Y=\int \frac {1}{F\left ( X\right ) }dX+c \]

In this case \(F\left ( X\right ) =e^{-X}\). Hence

\begin{align*} Y & =\int e^{X}dX+c\\ Y & =e^{X}+c \end{align*}

Now we just need to find canonical coordinates \(\left ( X,Y\right ) \) since \(\xi ,\eta \) are known. From above

\begin{align*} X & =\frac {y}{x}\\ Y & =-\frac {1}{x}\end{align*}

Hence the solution becomes

\begin{align*} -\frac {1}{x} & =e^{\frac {y}{x}}+c\\ e^{\frac {y}{x}} & =c_{2}-\frac {1}{x}\\ \frac {y}{x} & =\ln \left ( c_{2}-\frac {1}{x}\right ) \\ y & =x\ln \left ( c_{2}-\frac {1}{x}\right ) \end{align*}

The nice thing about this method is that once we solve for one pattern of an ode, then the same solution in canonical coordinates is used, the only change need is to plug-in in the RHS of the original ode in the solution and integrate.