7 Converting first order ODE which is homogeneous to separable ODE

(Added July, 2017).

If the ODE \(M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\) has both \(M\) and \(N\) homogeneous functions of same power, then this ODE can be converted to separable. Here is an example. We want to solve

\begin{equation} \left ( x^{3}+8x^{2}y\right ) +\left ( 4xy^{2}-y^{3}\right ) y^{\prime }=0 \tag {1}\end{equation}

The above is homogeneous in \(M,N\), since the total powers of each term in them is \(3\).

\[ \left ( \overset {3}{\overbrace {x^{3}}}+8\overset {3}{\overbrace {x^{2}y}}\right ) +\left ( 4\overset {3}{\overbrace {xy^{2}}}-\overset {3}{\overbrace {y^{3}}}\right ) y^{\prime }=0 \]

So we look at each term in \(N\) and \(M\) and add all the powers on each \(x,y\) in them. All powers should add to same value, which is \(3\) in this case. Of course \(N,M\) should be polynomials for this to work. So one should check that they are polynomials in \(x,y\) before starting this process. Once we check \(M,N\) are homogeneous, then we let

\[ y=xv \]

Therefore now

\begin{align} M & =x^{3}+8x^{2}\left ( xv\right ) \nonumber \\ & =x^{3}+8x^{3}v \tag {2}\end{align}

And

\begin{align} N & =4x\left ( xv\right ) ^{2}-\left ( xv\right ) ^{3}\nonumber \\ & =4x^{3}v^{2}-x^{3}v^{3} \tag {3}\end{align}

And

\begin{equation} y^{\prime }=v+xv^{\prime } \tag {4}\end{equation}

Substituting (3,4,5) into (1) gives

\begin{align*} \left ( x^{3}+8x^{3}v\right ) +\left ( 4x^{3}v^{2}-x^{3}v^{3}\right ) \left ( v+xv^{\prime }\right ) & =0\\ \left ( x^{3}+8x^{3}v\right ) +\left ( 4x^{3}v^{3}-x^{3}v^{4}\right ) +\left ( 4x^{4}v^{2}-x^{4}v^{3}\right ) v^{\prime } & =0 \end{align*}

Dividing by \(x^{3}\neq 0\) it simplifies to

\[ \left ( 1+8v\right ) +\left ( 4v^{3}-v^{4}\right ) +x\left ( 4v^{2}-v^{3}\right ) v^{\prime }=0 \]

Which can be written as

\begin{align*} x\left ( 4v^{2}-v^{3}\right ) v^{\prime } & =-\left ( \left ( 1+8v\right ) +\left ( 4v^{3}-v^{4}\right ) \right ) \\ v^{\prime } & =\frac {-\left ( \left ( 1+8v\right ) +\left ( 4v^{3}-v^{4}\right ) \right ) }{\left ( 4v^{2}-v^{3}\right ) }\left ( \frac {1}{x}\right ) \end{align*}

We see that it is now separable. We now solve this for \(v\left ( x\right ) \) by direct integration of both sides And then using \(y=xv\) find \(y\left ( x\right ) \).