## Examples solving isobaric and homogeneous ﬁrst order odes

December 20, 2021   Compiled on December 20, 2021 at 2:38am

### 1 Introduction

These are examples showing how to solve isobaric ode’s showing 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.

An ode $$y^{\prime }=f\left (x,y\right )$$ is called isobaric of degree $$m$$ if $$f\left (tx,t^{m}y\right ) =t^{m-1}f\left (x,y\right )$$. On the other hand, It is called homogeneous ode if $$f\left (tx,ty\right ) =f\left (x,y\right )$$. 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. We need to be careful here. Homogeneous function is not the same as a homogeneous ode. A function $$M\left (x,y\right )$$ is homogeneous 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, we can 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 $$y^{\prime }=f\left (x,y\right ) =\frac {M\left ( x,y\right ) }{N\left (x,y\right ) }$$ where $$M,N$$ are both homogeneous functions but with diﬀerent degrees. Then the ode is not homogeneous ode in this case 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 setting

$f\left (tx,t^{m}y\right ) =f\left (x,y\right )$

Then $$m=1$$ is the only solution. Taking derivative of both sides w.r.t. $$t$$ and setting $$t=1$$ after that1 gives

$m=\frac {-x\frac {\partial f}{\partial x}}{y\frac {\partial f}{\partial y}}$

If the above is evaluated and results in $$m=1$$ then the ode is homogenous. If not, then the ode is not homogenous and we check if it is isobaric by using

$f\left (tx,t^{m}y\right ) =t^{m-1}f\left (x,y\right )$

Taking derivative of both sides w.r.t. $$t$$ and setting $$t=1$$ after that gives

$m=\frac {f+x\frac {\partial f}{\partial x}}{f-y\frac {\partial f}{\partial x}}$

If it is possible to simplify the rhs above to a 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.

### 2 Examples

#### 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+x\frac {\partial f}{\partial x}}{f-y\frac {\partial f}{\partial x}}\tag {2}\\ & =\frac {\frac {-\left (y^{2}+\frac {2}{x}\right ) }{2yx}+x\left (\frac {1}{x^{3}y}+\frac {y^{2}+\frac {2}{x}}{2x^{2}y}\right ) }{\frac {-\left ( y^{2}+\frac {2}{x}\right ) }{2yx}-y\left (\frac {-1}{x}+\frac {y^{2}+\frac {2}{x}}{2xy^{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 $$y=vx^{\frac {=1}{2}}$$.

#### 2.2 Example 2

\begin {equation} \frac {dy}{dx}=x\sqrt {x^{4}+4y}-x^{3} \tag {1} \end {equation}

We start by checking if it is isobaric or not. Using

\begin {align*} m & =\frac {f+x\frac {\partial f}{\partial x}}{f-y\frac {\partial f}{\partial x}}\\ & =\frac {\left (x\sqrt {x^{4}+4y}-x^{3}\right ) +x\left (\sqrt {x^{4}+4y}+\frac {2x^{4}}{\sqrt {x^{4}+4y}}-3x^{2}\right ) }{\left (x\sqrt {x^{4}+4y}-x^{3}\right ) -x^{3}-\frac {2xy}{\sqrt {x^{4}+4y}}}\\ & =\frac {4\frac {x}{\sqrt {x^{4}+4y}}\left (2y-x^{2}\sqrt {x^{4}+4y}+x^{4}\right ) }{\frac {x}{\sqrt {x^{4}+4y}}\left (2y-2x^{2}\sqrt {x^{4}+4y}+x^{4}\right ) }\\ & =\frac {4\frac {x}{\sqrt {x^{4}+4y}}}{\frac {x}{\sqrt {x^{4}+4y}}}\\ & =4 \end {align*}

Therefore this is isobaric of order $$4$$.  Substituting $$y=vx^{m}=vx^{4}$$ in (1) results in $v^{\prime }=\frac {-4v+\sqrt {1+4v}-1}{x}$ Which is separable. This is solved easily for $$v\relax (x)$$ and then $$y$$ is found from $$y=vx^{4}$$.

#### 2.3 Example 3

\begin {align} x\left (x-y^{3}\right ) \frac {dy}{dx} & =\left (3x+y^{3}\right ) y\nonumber \\ \frac {dy}{dx} & =\frac {\left (3x+y^{3}\right ) y}{x\left (x-y^{3}\right ) } \tag {1} \end {align}

We start by checking if it is isobaric or not. Using

\begin {align*} m & =\frac {f+x\frac {\partial f}{\partial x}}{f-y\frac {\partial f}{\partial x}}\\ & =\frac {\frac {\left (3x+y^{3}\right ) y}{x\left (x-y^{3}\right ) }+x\left (\frac {3y}{x\left (-y^{3}+x\right ) }-\frac {\left (y^{3}+3x\right ) y}{x^{2}\left (-y^{3}+x\right ) }-\frac {\left (y^{3}+3x\right ) y}{x\left (-y^{3}+x\right ) ^{2}}\right ) }{\frac {\left (3x+y^{3}\right ) y}{x\left (x-y^{3}\right ) }-y\left (\frac {3y^{3}}{x\left (-y^{3}+x\right ) }+\frac {y^{3}+3x}{x\left (-y^{3}+x\right ) }+\frac {3\left (y^{3}+3x\right ) y^{3}}{x\left (-y^{3}+x\right ) ^{2}}\right ) }\\ & =\frac {-4\frac {y^{4}}{\left (x-y^{3}\right ) ^{2}}}{-12\frac {y^{4}}{\left (x-y^{3}\right ) ^{2}}}\\ & =\frac {1}{3} \end {align*}

$$m=\frac {1}{3}$$ makes each term the same weight $$\frac {4}{3}$$. Hence the substitution $$y=vx^{\frac {1}{3}}$$ will make the ode separable. Substituting this in (1) results in $\frac {dv}{dx}=\frac {-4}{3x}\frac {v\left (v^{3}+2\right ) }{\left ( v^{3}-1\right ) }$ Which is separable. This is solved for $$v$$, and then $$y$$ is found from $$y=vx^{\frac {1}{3}}$$.

#### 2.4 Example 4

\begin {equation} y^{\prime }=\frac {y}{x}\ln \left (xy-1\right ) \tag {1} \end {equation}

We start by checking if it is isobaric or not. Using

\begin {align*} m & =\frac {f+x\frac {\partial f}{\partial x}}{f-y\frac {\partial f}{\partial x}}\\ & =\frac {\frac {y}{x}\ln \left (xy-1\right ) +x\left (\frac {-y\ln \left ( xy-1\right ) }{x^{2}}+\frac {y^{2}}{x\left (xy-1\right ) }\right ) }{\frac {y}{x}\ln \left (xy-1\right ) -y\left (\frac {\ln \left (xy-1\right ) }{x}+\frac {y}{xy-1}\right ) }\\ & =\frac {\frac {y^{2}}{xy-1}}{-\frac {y^{2}}{xy-1}}\\ & =-1 \end {align*}

Hence the substitution $$y=\frac {v}{x}$$ will make the ode separable. Substituting this in (1) results in $v^{\prime }=\frac {v\ln \relax (v) }{x}$ Which is separable. This is solved for $$v$$, and then $$y$$ is found from $$y=\frac {v}{x}$$.

#### 2.5 Example 5

\begin {equation} \left (y^{\prime }\right ) ^{2}=y\left (y-2y^{\prime }x\right ) ^{3} \tag {1} \end {equation}

One way to handle this is to ﬁrst solve for $$y^{\prime }$$ and then apply the above method. This will result in $$m=-1$$.

#### 2.6 Example 6

\begin {align} \left (x-y\right ) y^{\prime }-x-y & =0\nonumber \\ y^{\prime } & =\frac {x+y}{x-y} \tag {1} \end {align}

We start by checking if it homogenous or not. Using

\begin {align*} m & =-\frac {x\frac {\partial f}{\partial x}}{y\frac {\partial f}{\partial x}}\\ & =-\frac {x\left (\frac {1}{x-y}-\frac {x+y}{\left (x-y\right ) ^{2}}\right ) }{y\left (\frac {1}{x-y}+\frac {x+y}{\left (x-y\right ) ^{2}}\right ) }\\ & =1 \end {align*}

Since $$m=1$$ then this is homogeneous ode. Hence the substitution $$v=\frac {y}{x}$$ makes the ode (1) separable.

#### 2.7 Example 7

\begin {align} y^{\prime }x-y-2\sqrt {xy} & =0\nonumber \\ y^{\prime } & =\frac {y+2\sqrt {xy}}{x} \tag {1} \end {align}

We start by checking if it homogenous or not. Using

\begin {align*} m & =-\frac {x\frac {\partial f}{\partial x}}{y\frac {\partial f}{\partial x}}\\ & =-\frac {x\left (\frac {y}{x\sqrt {xy}}-\frac {y+2\sqrt {xy}}{x^{2}}\right ) }{y\left (\frac {1+\frac {x}{\sqrt {xy}}}{x}\right ) }\\ & =-\frac {-\frac {1}{x}\left (y+\sqrt {xy}\right ) }{\frac {1}{x}\left ( y+\sqrt {xy}\right ) }\\ & =1 \end {align*}

Since $$m=1$$ then this is homogeneous ode. Hence the substitution $$v=\frac {y}{x}$$ makes the ode (1) separable.

1Thanks to vv from the Maple forum for this trick.