3.3.13.0 Examples

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

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

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

\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

Which is separable. This is solved easily for \(v\left ( x\right ) \) and then \(y\) is found from \(y=vx^{4}\).

\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}

\begin{align*} m & =\frac {f+xf_{x}}{f-yf_{y}}\\ & =\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

Which is separable. This is solved for \(v\), and then \(y\) is found from \(y=vx^{\frac {1}{3}}\).

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

\begin{align*} m & =\frac {f+xf_{x}}{f-yf_{y}}\\ & =\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

Which is separable. This is solved for \(v\), and then \(y\) is found from \(y=\frac {v}{x}\).

\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 \(m=-1\).

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

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

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

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

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

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

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

\begin{align*} m & =\frac {f+xf_{x}}{f-yf_{y}}\\ & =\frac {\frac {-y\left ( y^{2}+3x^{2}+2x\right ) }{x^{2}+y^{2}}+x\frac {d}{dx}\left ( \frac {-y\left ( y^{2}+3x^{2}+2x\right ) }{x^{2}+y^{2}}\right ) }{\frac {-y\left ( y^{2}+3x^{2}+2x\right ) }{x^{2}+y^{2}}-y\frac {d}{dy}\left ( \frac {-y\left ( y^{2}+3x^{2}+2x\right ) }{x^{2}+y^{2}}\right ) }\\ & =\frac {\frac {-y\left ( y^{2}+3x^{2}+2x\right ) }{x^{2}+y^{2}}+x\left ( -2\frac {y\left ( -x^{2}+2xy^{2}+y^{2}\right ) }{\left ( x^{2}+y^{2}\right ) ^{2}}\right ) }{\frac {-y\left ( y^{2}+3x^{2}+2x\right ) }{x^{2}+y^{2}}-y\left ( -\frac {3x^{4}+2x^{3}-2xy^{2}+y^{4}}{\left ( x^{2}+y^{2}\right ) ^{2}}\right ) }\\ & =\frac {3x^{4}+8x^{2}y^{2}+4xy^{2}+y^{4}}{4x^{2}y^{2}+4xy^{2}}\end{align*}

Since this does not simplify to numerical value, it is not homogenous ode. This turns out to be homogenous type D. See earlier note on this. There is a slight diﬀerence in deﬁnition between homogenous ode and homogenous type D. In Maple terms, homogenous ode is called homogenous ode type A. A homogenous type D is one in which the substitution \(y=ux\) makes the ode separable or quadrature.

\begin{equation} y^{\prime }=\frac {\left ( -108y^{2}+12\sqrt {-108y^{3}x^{3}+81y^{4}}\right ) ^{\frac {2}{3}}+12xy}{6\left ( -108y^{2}+12\sqrt {-108y^{3}x^{3}+81y^{4}}\right ) ^{\frac {1}{3}}} \tag {1}\end{equation}

Hence the substitution \(y=vx^{m}\) will make the ode separable. Substituting \(y=vx^{3}\) in (1) results in separable ode. But for this case, we have to assume \(x>0\) in order to simplify it. The resulting ode is too long to write now, but veriﬁed to be separable using the computer.