2.13.2.1 Example 1

\begin{equation} \frac {dy}{dx}=\frac {-\left ( y^{2}+\frac {2}{x}\right ) }{2yx} \tag {1}\end{equation}

Here \(f\left ( x,y\right ) =\frac {-\left ( y^{2}+\frac {2}{x}\right ) }{2yx}\). We start by checking if it is isobaric or not. To find \(m\) such that \(f\left ( tx,t^{m}y\right ) =t^{m-1}f\left ( x,y\right ) \) we do (as given in the introduction)

\begin{align} m & =\frac {f+xf_{x}}{f-yf_{y}}\tag {2}\\ & =\frac {\frac {-\left ( y^{2}+\frac {2}{x}\right ) }{2yx}+x\left ( \frac {xy^{2}+4}{2x^{3}y}\right ) }{\frac {-\left ( y^{2}+\frac {2}{x}\right ) }{2yx}-y\left ( -\frac {xy^{2}-2}{2x^{2}y^{2}}\right ) }\nonumber \\ & =\frac {\frac {1}{x^{2}y}}{-\frac {2}{x^{2}y}}\nonumber \\ & =-\frac {1}{2}\nonumber \end{align}

Hence this is isobaric of index \(m=-\frac {1}{2}\) because it has a numerical solution as a result.

To verify this result, here \(M\left ( x,y\right ) =\left ( -y^{2}-\frac {2}{x}\right ) ,N\left ( x,y\right ) =2yx\). Let us start by checking for isobaric (since homogeneous is special case).

\begin{align*} M\left ( tx,t^{m}y\right ) & =\left ( -t^{2m}y^{2}+\frac {2}{tx}\right ) \\ & =\frac {1}{t}\left ( -t^{2m+1}y^{2}+\frac {2}{x}\right ) \\ & =t^{-1}\left ( -t^{2m+1}y^{2}+\frac {2}{x}\right ) \end{align*}

The above is same as \(\left ( -y^{2}-\frac {2}{x}\right ) \) when \(2m+1=0\) or \(m=-\frac {1}{2}\). From the above we also see that \(r=-1\). This is by comparing the last result above to \(t^{r}M\left ( x,y\right ) \). Now that we found candidate \(m\) and \(r\), then all what we have to do is check \(N\left ( tx,t^{m}y\right ) =t^{r-m-1}N\left ( x,y\right ) \) or not. If it is, then we are done and the ode is isobaric of degree \(m\)

\begin{align*} N\left ( tx,t^{m}y\right ) & =2t^{m}ytx\\ & =2t^{\frac {-1}{2}}ytx\\ & =t^{\frac {1}{2}}\left ( 2yx\right ) \\ & =t^{\frac {1}{2}}N\left ( x,y\right ) \end{align*}

Now we check if \(\frac {1}{2}=r-m+1\). Which it is. Since \(r-m+1=-1-\left ( -\frac {1}{2}\right ) +1=\frac {1}{2}\). Hence this ode is isobaric. From now on Eq (2) will be used to find \(m\).

Hence the substitution \(y=vx^{m}\) will make the ode separable. This is the whole point of isobaric ode’s. The hardest part is to find \(m\). Substituting \(y=vx^{\frac {=1}{2}}\) in (1) results in

\[ v\frac {dv}{dx}=-\frac {1}{x}\]

This is solved for \(v\) easily since separable, and then \(y\) is found from \(y=vx^{\frac {=1}{2}}\).