ode internal name "isobaric"
This is a generalization of the homogeneous ODE, where the substitution \(y=v\left ( x\right ) x^{m}\) makes the ODE separable. The weight \(m\) needs to be found first.
These are examples showing how to solve isobaric ode’s step by step method. The same method is also used to solve homogeneous ode, which is special case of isobaric.
The hardest part is to determine if the ode is isobaric or homogeneous and to find the degree of the isobaric. If the weight (or degree) \(m\) is one then it is just homogeneous ode. If the weight is not \(1\) then it is isobaric ode. An ode \(y^{\prime }=f\left ( x,y\right ) \) is called isobaric of degree \(m\) if \[ f\left ( tx,t^{m}y\right ) =t^{m-1}f\left ( x,y\right ) \] It is called homogeneous ode if \(m=1\) \[ f\left ( tx,ty\right ) =f\left ( x,y\right ) \] So homogeneous ode is special case of isobaric ode when \(m=1\). Another common definition of a homogeneous ode is that when writing the ode as \begin {align*} y^{\prime } & =f\left ( x,y\right ) \\ & =\frac {M\left ( x,y\right ) }{N\left ( x,y\right ) } \end {align*}
Then \(M,N\) must both be homogeneous functions of same degree. Care is needed here, Homogeneous function is not the same as a homogeneous ode. A function \(M\left ( x,y\right ) \) is homogeneous function of degree \(n\) if \(M\left ( tx,ty\right ) =t^{n}M\left ( x,y\right ) \) where \(n\) here do not have to be zero.
Using this second definition of homogeneous ode of \(y^{\prime }=\frac {M\left ( x,y\right ) }{N\left ( x,y\right ) }\), we can now check if \(M\left ( x,y\right ) \) and \(N\left ( x,y\right ) \) are both homogeneous functions and also have same degree (whatever this degree happened to be). If this is the case, then we say the ode itself is homogeneous ode.
It is possible to have an ode \(y^{\prime }=\frac {M\left ( x,y\right ) }{N\left ( x,y\right ) }\) where \(M,N\) are both homogeneous functions but with different degrees. In this case the ode is not homogeneous ode even though both \(M,N\) are each homogeneous functions.
We can use similar way to view isobaric ode. By saying that an isobaric ode is one when it is written as\begin {align*} y^{\prime } & =f\left ( x,y\right ) \\ & =\frac {M\left ( x,y\right ) }{N\left ( x,y\right ) } \end {align*}
Then given \(M\left ( tx,t^{m}y\right ) =t^{r}M\left ( x,y\right ) \) is homogeneous function of degree \(r\) and \(N\left ( tx,t^{m}y\right ) =t^{r-m+1}N\left ( x,y\right ) \) is homogeneous function of degree \(r-m+1\). In this case we say that the ode itself is isobaric of degree \(m\), since \begin {align*} f\left ( tx,t^{m}y\right ) & =\frac {t^{r}M\left ( x,y\right ) }{t^{r-m+1}N\left ( x,y\right ) }\\ & =t^{m-1}\frac {M\left ( x,y\right ) }{N\left ( x,y\right ) }\\ & =t^{m-1}f\left ( x,y\right ) \end {align*}
The above gives us another method to determine if an ode is homogeneous ode or isobaric ode. We start by writing the ode as \(y^{\prime }=\frac {M\left ( x,y\right ) }{N\left ( x,y\right ) }\). If \(M,N\) are both homogeneous functions of same degree, then the ode is homogeneous ode and we stop.
If however \(M\) satisfies \(M\left ( tx,t^{m}y\right ) =t^{r}M\left ( x,y\right ) \) and \(N\) satisfies \(N\left ( tx,t^{m}y\right ) =t^{r-m+1}N\left ( x,y\right ) \) where \(r\) is positive integer, then we say the ode is isobaric of degree \(m\).
Why is it important to know if an ode is homogeneous or isobaric? This is because if an ode is isobaric of degree \(m\) then the substitution \(y=ux^{m}\) or \(u=\frac {y}{x^{m}}\) converts to separable ode in \(u\). If an ode is homogeneous then the substitution \(y=ux\) or \(u=\frac {y}{x}\) converts to separable ode in \(u\).
This is why it is very useful to determine if an ode is isobaric or homogeneous ode. Because it allows us to use this substitution to convert it to separable. Separable ode’s are easy to solve, since they involve only integration. Of course the integrals can be very difficult to solve, but this is another issue.
How to determine if an ode is homogeneous or isobaric in practice? To check if an ode is homogeneous, we start with the definition that ode \(y^{\prime }=f\left ( x,y\right ) \) is homogeneous ode if in\begin {equation} f\left ( tx,t^{m}y\right ) =t^{m-1}f\left ( x,y\right ) \tag {A} \end {equation} then if \(m=1\) then the ode is homogenous. If not, then the ode is not homogenous and we check if it is isobaric by solving for \(m\). How to find \(m\)?
This is done by taking derivative of both sides of equation (A) w.r.t. \(t\) and setting \(t=1\) after that. This results in\begin {align*} xf_{x}+myf_{y} & =\left ( m-1\right ) f\\ xf_{x}+myf_{y} & =mf-f\\ xf_{x}+f & =m\left ( f-yf_{y}\right ) \end {align*}
Hence\[ m=\frac {f+xf_{x}}{f-yf_{y}}\] Here is the important point. If it is possible to simplify the RHS above to an actual numerical value, then \(m\) is the degree of isobaric and the ode is indeed isobaric. If it is not possible to obtain a numerical \(m\) value, then the ode is not isobaric. The best way to learn how to do this is by examples. Note in the above \(f_{x}\) is partial derivative. Which means taking derivative of \(f\) w.r.t \(x\) while keeping \(y\) fixed.