**Example – 9**

Solve the DE \(xdy - ydx = \sqrt {{x^2} + {y^2}} dx.\)

**Solution:** Upon rearrangement, we have

\[\begin{align} & \frac{{dy}}{{dx}} = \frac{{y + \sqrt {{x^2} + {y^2}} }}{x}\\ &\quad \;\; = \left( {\frac{y}{x}} \right) + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}}\end{align}\]

This is obviously a first degree homogeneous DE. We substitute \(y = vx\) to obtain:

\[\begin{align} & \qquad \;\;v + x\frac{{dv}}{{dx}} = v + \sqrt {1 + {v^2}} \\& \Rightarrow \quad \frac{{dv}}{{\sqrt {1 + {v^2}} }} = \frac{{dx}}{x}\end{align}\]

Integrating both sides, we have

\[\begin{align}&\qquad \ln \left| {v + \sqrt {1 + {v^2}} } \right| = \ln x + \ln C\\ &\qquad\qquad\qquad\qquad \;\;\;= \ln Cx\\&\Rightarrow \quad \frac{y}{x} + \sqrt {1 + \frac{{{y^2}}}{{{x^2}}}} = Cx\end{align}\]

**Example – 10**

Solve the DE \(\left( {1 + {e^{x/y}}} \right)dx + {e^{x/y}}\left( {1 - \frac{x}{y}} \right)dy = 0.\)

**Solution: **This DE can be rearranged as

\[\frac{{dx}}{{dy}} = \frac{{{e^{x/y}}\left( {\frac{x}{y} - 1} \right)}}{{{e^{x/y}} + 1}}\]

Using the substituting \(x = vy\) (note : not \(y = vx)\) can reduce this DE to a VS form. (We did not use \(y = vx\) since that would’ve led to an expression involving complicated exponentials).

We now have

\[\begin{align} & \qquad v + y\frac{{dv}}{{dy}} = \frac{{{e^v}\left( {v - 1} \right)}}{{{e^v} + 1}}\\&\Rightarrow \quad \frac{{dy}}{y} \quad\;\;= - \frac{{{e^v} + 1}}{{{e^v} + v}}dv\end{align}\]

Integrating both sides, we have

\[\begin{align} & \qquad \;\;\ln\,y = - \ln \left| {{e^v} + v} \right| + \ln C\\ &\Rightarrow \quad y({e^v} + v) = C\\&\Rightarrow \quad {e^{x/y}} + \frac{x}{y} = \frac{C}{y} \end{align}\]

This example should serve the show that \(y = vx\) will not always be the most appropriate substitution to solve a homogeneous DE; \(x = vy\) could be more appropriate in such a scenario, as in the example above.