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+xf_{x}}{f-yf_{y}}\\ & =\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 \(v\left ( x\right ) \) and then \(y\) is found from \(y=vx^{4}\).