Examples

Example 1 Solve

\[ y^{\prime }=\frac {-y\left ( 2x^{2}y^{3}+3\right ) }{x\left ( x^{2}y^{3}+1\right ) }\]

The first step is to identify if this is class G and find \(F\). We start by multiplying the RHS by \(\frac {x}{y}\) (regardless of what is in the RHS) which gives

\begin{align*} y^{\prime } & =\frac {x}{y}\left ( \frac {-y\left ( 2x^{2}y^{3}+3\right ) }{x\left ( x^{2}y^{3}+1\right ) }\right ) \\ & =\frac {-2x^{2}y^{3}-3}{x^{2}y^{3}+1}\\ & =F\left ( x,y\right ) \end{align*}

Next we check if \(F\left ( x,y\right ) \) has \(y\) or not in it. If so, then let the RHS above be \(F\left ( x,y\right ) \) and now do

\begin{align*} f_{x} & =x\frac {\partial F}{\partial x}\\ & =x\left ( \frac {2y^{3}x}{\left ( x^{2}y^{3}+1\right ) ^{2}}\right ) \\ & =\frac {2y^{3}x^{2}}{\left ( x^{2}y^{3}+1\right ) ^{2}}\end{align*}

And let

\begin{align*} f_{y} & =y\frac {\partial F}{\partial y}\\ & =y\left ( \frac {3x^{2}y^{2}}{\left ( x^{2}y^{3}+1\right ) ^{2}}\right ) \\ & =\frac {3x^{2}y^{3}}{\left ( x^{2}y^{3}+1\right ) ^{2}}\end{align*}

Now we check, if \(f_{y}=0\) then this is not Homogeneous type G. Else we now need to determine value of \(\alpha \). This is done as follows.

\begin{align*} \alpha & =\frac {fx}{f_{y}}\\ & =\frac {2}{3}\end{align*}

If \(\alpha \) comes out not to have in it \(x\) nor \(y\) as in this case, then we are done. This ode is Homogeneous type G. But we have to do one more check.  We have to check that \(F\left ( x,y\right ) \) found above ends up with no \(x\) in it. Hence the solution is

\begin{equation} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \tau \right ) \right ) }d\tau =0 \tag {1}\end{equation}

Now let \(yx^{\alpha }=\tau \) or \(y=\frac {\tau }{x^{\alpha }}\ \)and substituting this into \(F\left ( x,y\right ) \) gives

\begin{align*} F\left ( \tau \right ) & =\frac {-2x^{2}\left ( \frac {\tau }{x^{\alpha }}\right ) ^{3}-3}{x^{2}\left ( \frac {\tau }{x^{\alpha }}\right ) ^{3}+1}\\ & =\frac {-2x^{2}\left ( \frac {\tau }{x^{\frac {2}{3}}}\right ) ^{3}-3}{x^{2}\left ( \frac {\tau }{x^{\frac {2}{3}}}\right ) ^{3}+1}\\ & =\frac {-2x^{2}\left ( \frac {\tau ^{3}}{x^{2}}\right ) -3}{x^{2}\left ( \frac {\tau ^{3}}{x^{3}}\right ) +1}\\ & =\frac {-2\tau ^{3}-3}{\tau ^{3}+1}\end{align*}

We see that \(F\left ( x,y\right ) \) ends up as \(F\left ( \tau \right ) =\frac {-2\tau ^{3}-3}{\tau ^{3}+1}\) after the transformation. It has no \(x\) left in it. If we end up with \(x\) then this method can not be used.

The solution (1) becomes

\begin{align*} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -\left ( \frac {-2\tau ^{3}-3}{\tau ^{3}+1}\right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{yx^{\frac {2}{3}}}\frac {3\tau ^{3}+3}{4\tau ^{4}+7\tau }d\tau & =0 \end{align*}

Solving the integral gives

\[ \ln x-c_{1}+\frac {3}{7}\ln \left ( yx^{\frac {2}{3}}\right ) +\frac {3}{28}\ln \left ( 4x^{2}y^{3}+7\right ) =0 \]

And this is the final answer. Now if earlier we have \(F\left ( x,y\right ) \) not have \(y\) in it. In this case we check if \(F\left ( x,y\right ) \) has \(x\). If not, then \(\alpha =0\) and we do the same as above. But if \(F\left ( x,y\right ) \) has \(x\) and not has \(y\) then it is not not Homogeneous type G.

Example 2 Solve

\[ y^{\prime }=\frac {2x\left ( -x^{4}-2x^{2}y+y^{2}\right ) }{y^{2}+2x^{2}y-x^{4}}\]

The first step is to identify if this is class G and find \(F\). We start by multiplying the RHS by \(\frac {x}{y}\) (regardless of what is in the RHS) which gives

\begin{align*} y^{\prime } & =\frac {x}{y}\left ( \frac {2x\left ( -x^{4}-2x^{2}y+y^{2}\right ) }{y^{2}+2x^{2}y-x^{4}}\right ) \\ & =\frac {2x^{2}\left ( x^{4}+2x^{2}y-y^{2}\right ) }{y\left ( x^{4}-2x^{2}y-y^{2}\right ) }\\ & =F\left ( x,y\right ) \end{align*}

Next we check if \(F\left ( x,y\right ) \) has \(y\) or not in it. If so, then let the RHS above be \(F\left ( x,y\right ) \) and now do

\begin{align*} f_{x} & =x\frac {\partial F}{\partial x}\\ & =x\left ( \frac {4x\left ( x^{8}-4x^{6}y-6x^{4}y^{2}-4x^{2}y^{3}+y^{4}\right ) }{y\left ( x^{4}-2x^{2}y-y^{2}\right ) ^{2}}\right ) \\ & =\frac {4x^{2}\left ( x^{8}-4x^{6}y-6x^{4}y^{2}-4x^{2}y^{3}+y^{4}\right ) }{y\left ( x^{4}-2x^{2}y-y^{2}\right ) ^{2}}\end{align*}

And let

\begin{align*} f_{y} & =y\frac {\partial F}{\partial y}\\ & =y\left ( \frac {-2x^{2}\left ( x^{8}-4x^{6}y-6x^{4}y^{2}-4x^{2}y^{2}-4x^{2}y^{3}+y^{4}\right ) }{y^{2}\left ( x^{4}-2x^{2}y-y^{2}\right ) ^{2}}\right ) \\ & =\frac {-2x^{2}\left ( x^{8}-4x^{6}y-6x^{4}y^{2}-4x^{2}y^{2}-4x^{2}y^{3}+y^{4}\right ) }{y\left ( x^{4}-2x^{2}y-y^{2}\right ) ^{2}}\end{align*}

Now we check, if \(f_{y}=0\) then this is not Homogeneous type G. Else we now need to determine value of \(\alpha \). This is done as follows.

\begin{align*} \alpha & =\frac {fx}{f_{y}}\\ & =-2 \end{align*}

If \(\alpha \) comes out not to have in it \(x\) nor \(y\) as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

\[ y^{\prime }=\frac {y}{x}F\left ( \frac {y}{x^{\alpha }}\right ) \]

Hence the solution is

\begin{equation} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \tau \right ) \right ) }d\tau =0 \tag {1}\end{equation}

Now let \(y=\frac {\tau }{x^{\alpha }}\) and substitute this into \(F\left ( x,y\right ) \) which results in

\begin{align*} F\left ( \tau \right ) & =\frac {2x^{2}\left ( x^{4}+2x^{2}\frac {\tau }{x^{\alpha }}-\left ( \frac {\tau }{x^{\alpha }}\right ) ^{2}\right ) }{\frac {\tau }{x^{\alpha }}\left ( x^{4}-2x^{2}\frac {\tau }{x^{\alpha }}-\left ( \frac {\tau }{x^{\alpha }}\right ) ^{2}\right ) }\\ & =\frac {2x^{2}\left ( x^{4}+2x^{2}\frac {\tau }{x^{-2}}-\left ( \frac {\tau }{x^{-2}}\right ) ^{2}\right ) }{\frac {\tau }{x^{-2}}\left ( x^{4}-2x^{2}\frac {\tau }{x^{-2}}-\left ( \frac {\tau }{x^{-2}}\right ) ^{2}\right ) }\\ & =\frac {2x^{2}\left ( x^{4}+2x^{4}\tau -x^{4}\tau ^{2}\right ) }{\tau x^{2}\left ( x^{4}-2x^{4}\tau -\tau ^{2}x^{4}\right ) }\\ & =\frac {2\left ( x^{4}+2x^{4}\tau -x^{4}\tau ^{2}\right ) }{\tau \left ( x^{4}-2x^{4}\tau -\tau ^{2}x^{4}\right ) }\\ & =\frac {2}{\tau }\frac {\left ( 1+2\tau -\tau ^{2}\right ) }{\left ( 1-2\tau -\tau ^{2}\right ) }\\ & =\frac {2}{\tau }\frac {\left ( \tau ^{2}-2\tau -1\right ) }{\left ( \tau ^{2}+2\tau -1\right ) }\end{align*}

The solution(1) becomes

\begin{align*} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \alpha \right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{x^{2}}}\frac {1}{\tau \left ( 2-\left ( \frac {2}{\tau }\frac {\left ( \tau ^{2}-2\tau -1\right ) }{\left ( \tau ^{2}+2\tau -1\right ) }\right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{x^{2}}}\frac {1}{2}\frac {\tau ^{2}+2\tau -1}{\tau ^{3}+\tau ^{2}+\tau +1}d\tau & =0 \end{align*}

Solving the integral gives

\[ \ln x-c_{1}-\frac {1}{2}\ln \left ( \frac {x^{2}+y}{x^{2}}\right ) +\frac {1}{2}\ln \left ( \frac {x^{4}+y^{2}}{x^{4}}\right ) =0 \]

Example 3 Solve

\[ y^{\prime }=-\frac {1}{2}\frac {3y^{2}-x}{y\left ( y^{2}-3x\right ) }\]

The first step is to identify if this is class G and find \(F\). We start by multiplying the RHS by \(\frac {x}{y}\) (regardless of what is in the RHS) which gives

\begin{align*} y^{\prime } & =\frac {x}{y}\left ( -\frac {1}{2}\frac {3y^{2}-x}{y\left ( y^{2}-3x\right ) }\right ) \\ & =-\frac {1}{2}\frac {3xy^{2}-x^{2}}{y^{4}-3xy^{2}}\\ & =F\left ( x,y\right ) \end{align*}

Next we check if \(F\left ( x,y\right ) \) has \(y\) or not in it. If so, then let the RHS above be \(F\left ( x,y\right ) \) and now do

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

And let

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

Now we check, if \(f_{y}=0\) then this is not Homogeneous type G. Else we now need to determine value of \(\alpha \). This is done as follows.

\begin{align*} \alpha & =\frac {fx}{f_{y}}\\ & =-\frac {1}{2}\end{align*}

If \(\alpha \) comes out not to have in it \(x\) nor \(y\) as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

\[ y^{\prime }=\frac {y}{x}F\left ( \frac {y}{x^{\alpha }}\right ) \]

Hence the solution is

\begin{equation} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \tau \right ) \right ) }d\tau =0 \tag {1}\end{equation}

Now let \(y=\frac {\tau }{x^{\alpha }}\) and substitute this into \(F\left ( x,y\right ) \) which results in

\begin{align*} F\left ( \tau \right ) & =-\frac {1}{2}\frac {3xy^{2}-x^{2}}{y^{4}-3xy^{2}}\\ & =-\frac {1}{2}\frac {3x\left ( \frac {\tau }{x^{\alpha }}\right ) ^{2}-x^{2}}{\left ( \frac {\tau }{x^{\alpha }}\right ) ^{4}-3x\left ( \frac {\tau }{x^{\alpha }}\right ) ^{2}}\\ & =-\frac {1}{2}\frac {3x\left ( \frac {\tau }{x^{-\frac {1}{2}}}\right ) ^{2}-x^{2}}{\left ( \frac {\tau }{x^{-\frac {1}{2}}}\right ) ^{4}-3x\left ( \frac {\tau }{x^{-\frac {1}{2}}}\right ) ^{2}}\\ & =-\frac {1}{2}\frac {3x^{2}\tau ^{2}-x^{2}}{\tau ^{4}x^{2}-3x\tau ^{2}x}\\ & =-\frac {1}{2}\frac {3\tau ^{2}-1}{\tau ^{4}-3\tau ^{2}}\end{align*}

The solution(1) becomes

\begin{align*} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \alpha \right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{\sqrt {x}}}\frac {1}{\tau \left ( \frac {1}{2}-\left ( -\frac {1}{2}\frac {3\tau ^{2}-1}{\tau ^{4}-3\tau ^{2}}\right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{\sqrt {x}}}2\tau \frac {\tau ^{2}-3}{\tau ^{4}-1}d\tau & =0 \end{align*}

Solving the integral gives

\[ \ln x-c_{1}-\frac {1}{2}\ln \left ( \frac {y}{\sqrt {x}}-1\right ) -\ln \left ( \frac {y}{\sqrt {x}}+1\right ) +2\ln \left ( \frac {y^{2}}{x}+1\right ) =0 \]

Example 4 Solve

\[ y^{\prime }=-\frac {1}{2}\frac {y\left ( 1+\sqrt {x^{2}y^{4}+1}\right ) }{x}\]

The first step is to identify if this is class G and find \(F\). We start by multiplying the RHS by \(\frac {x}{y}\) (regardless of what is in the RHS) which gives

\begin{align*} y^{\prime } & =\frac {x}{y}\left ( -\frac {1}{2}\frac {y\left ( 1+\sqrt {x^{2}y^{4}+1}\right ) }{x}\right ) \\ & =-\frac {1}{2}\left ( 1+\sqrt {x^{2}y^{4}+1}\right ) \\ & =F\left ( x,y\right ) \end{align*}

Next we check if \(F\left ( x,y\right ) \) has \(y\) or not in it. If so, then let the RHS above be \(F\left ( x,y\right ) \) and now do

\begin{align*} f_{x} & =x\frac {\partial F}{\partial x}\\ & =-\frac {1}{2}\frac {x^{2}y^{4}}{\sqrt {x^{2}y^{4}+1}}\end{align*}

And let

\begin{align*} f_{y} & =y\frac {\partial F}{\partial y}\\ & =\frac {-x^{2}y^{4}}{\sqrt {x^{2}y^{4}+1}}\end{align*}

Now we check, if \(f_{y}=0\) then this is not Homogeneous type G. Else we now need to determine value of \(\alpha \). This is done as follows.

\begin{align*} \alpha & =\frac {fx}{f_{y}}\\ & =\frac {1}{2}\end{align*}

If \(\alpha \) comes out not to have in it \(x\) nor \(y\) as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

\[ y^{\prime }=\frac {y}{x}F\left ( \frac {y}{x^{\alpha }}\right ) \]

Hence the solution is

\begin{equation} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \tau \right ) \right ) }d\tau =0 \tag {1}\end{equation}

Now let \(y=\frac {\tau }{x^{\alpha }}\) and substitute this into \(F\left ( x,y\right ) \) which results in

\begin{align*} F\left ( \tau \right ) & =-\frac {1}{2}\left ( 1+\sqrt {x^{2}y^{4}+1}\right ) \\ & =-\frac {1}{2}\left ( 1+\sqrt {x^{2}\left ( \frac {\tau }{x^{\alpha }}\right ) ^{4}+1}\right ) \\ & =-\frac {1}{2}\left ( 1+\sqrt {x^{2}\left ( \frac {\tau }{x^{\frac {1}{2}}}\right ) ^{4}+1}\right ) \\ & =-\frac {1}{2}\left ( 1+\sqrt {x^{2}\frac {\tau ^{4}}{x^{2}}+1}\right ) \\ & =-\frac {1}{2}\left ( 1+\sqrt {\tau ^{4}+1}\right ) \end{align*}

The solution(1) becomes

\begin{align*} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \alpha \right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{y\sqrt {x}}\frac {1}{\tau \left ( -\frac {1}{2}-\left ( -\frac {1}{2}\left ( 1+\sqrt {\tau ^{4}+1}\right ) \right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{y\sqrt {x}}\frac {2}{\tau \sqrt {\tau ^{4}+1}}d\tau & =0 \end{align*}

Solving the integral gives

\[ \ln x-c_{1}-\operatorname {arctanh}\left ( \frac {1}{\sqrt {x^{2}y^{4}+1}}\right ) =0 \]

Example 5 Solve

\[ y^{\prime }=x\left ( 1+\frac {2y}{x}+\frac {y^{2}}{x^{4}}\right ) \]

The first step is to identify if this is class G and find \(F\). We start by multiplying the RHS by \(\frac {x}{y}\) (regardless of what is in the RHS) which gives

\begin{align*} y^{\prime } & =\frac {x}{y}\left ( x\left ( 1+\frac {2y}{x}+\frac {y^{2}}{x^{4}}\right ) \right ) \\ & =\frac {x^{2}}{y}+2x+\frac {y}{x^{2}}\\ & =\frac {\left ( x^{2}+y\right ) ^{2}}{x^{2}y}\\ & =F\left ( x,y\right ) \end{align*}

Next we check if \(F\left ( x,y\right ) \) has \(y\) or not in it. If so, then let the RHS above be \(F\left ( x,y\right ) \) and now do

\begin{align*} f_{x} & =x\frac {\partial F}{\partial x}\\ & =\frac {2x^{4}-2y^{2}}{x^{2}y}\end{align*}

And let

\begin{align*} f_{y} & =y\frac {\partial F}{\partial y}\\ & =\frac {-x^{4}+y^{2}}{x^{2}y}\end{align*}

Now we check, if \(f_{y}=0\) then this is not Homogeneous type G. Else we now need to determine value of \(\alpha \). This is done as follows.

\begin{align*} \alpha & =\frac {fx}{f_{y}}\\ & =-2 \end{align*}

If \(\alpha \) comes out not to have in it \(x\) nor \(y\) as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

\[ y^{\prime }=\frac {y}{x}F\left ( \frac {y}{x^{\alpha }}\right ) \]

Hence the solution is

\begin{equation} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \tau \right ) \right ) }d\tau =0 \tag {1}\end{equation}

Now let \(y=\frac {\tau }{x^{\alpha }}\) and substitute this into \(F\left ( x,y\right ) \) which results in

\begin{align*} F\left ( \tau \right ) & =\frac {\left ( x^{2}+y\right ) ^{2}}{x^{2}y}\\ & =\frac {\left ( x^{2}+\frac {\tau }{x^{\alpha }}\right ) ^{2}}{x^{2}\frac {\tau }{x^{\alpha }}}\\ & =\frac {\left ( x^{2}+\frac {\tau }{x^{-2}}\right ) ^{2}}{x^{2}\frac {\tau }{x^{-2}}}\\ & =\frac {\left ( x^{2}+\tau x^{2}\right ) ^{2}}{x^{4}\tau }\\ & =\frac {x^{4}+\tau ^{2}x^{4}+2\tau x^{4}}{x^{4}\tau }\\ & =\frac {1+\tau ^{2}+2\tau }{\tau }\end{align*}

The solution(1) becomes

\begin{align*} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \alpha \right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{x^{2}}}\frac {1}{\tau \left ( 2-\left ( \frac {1+\tau ^{2}+2\tau }{\tau }\right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{x^{2}}}-\frac {1}{\tau ^{2}+1}d\tau & =0\\ \ln x-c_{1}-\int ^{\frac {y}{x^{2}}}\frac {1}{\tau ^{2}+1}d\tau & =0 \end{align*}

Solving the integral gives

\begin{align*} \ln x-c_{1}-\arctan \left ( \frac {y}{x^{2}}\right ) & =0\\ y & =-\tan \left ( c_{1}-\ln x\right ) x^{2}\end{align*}

Example 6 Solve

\[ \left ( y^{\prime }\right ) ^{2}=4y-x^{2}\]

Hence

\[ y^{\prime }=\pm \sqrt {4y-x^{2}}\]

For the first ode, the first step is to identify if this is class G and find \(F\). We start by multiplying the RHS by \(\frac {x}{y}\) (regardless of what is in the RHS) which gives

\begin{align*} y^{\prime } & =\frac {x}{y}\sqrt {4y-x^{2}}\\ & =F\left ( x,y\right ) \end{align*}

Next we check if \(F\left ( x,y\right ) \) has \(y\) or not in it. If so, then let the RHS above be \(F\left ( x,y\right ) \) and now do

\begin{align*} f_{x} & =x\frac {\partial F}{\partial x}\\ & =-\frac {2x\left ( x^{2}-2y\right ) }{y\sqrt {4y-x^{2}}}\end{align*}

And let

\begin{align*} f_{y} & =y\frac {\partial F}{\partial y}\\ & =\frac {x\left ( x^{2}-2y\right ) }{y\sqrt {4y-x^{2}}}\end{align*}

Now we check, if \(f_{y}=0\) then this is not Homogeneous type G. Else we now need to determine value of \(\alpha \). This is done as follows.

\begin{align*} \alpha & =\frac {fx}{f_{y}}\\ & =-2 \end{align*}

If \(\alpha \) comes out not to have in it \(x\) nor \(y\) as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

\[ y^{\prime }=\frac {y}{x}F\left ( \frac {y}{x^{\alpha }}\right ) \]

Hence the solution is

\begin{equation} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \tau \right ) \right ) }d\tau =0 \tag {1}\end{equation}

Now let \(y=\frac {\tau }{x^{\alpha }}\) and substitute this into \(F\left ( x,y\right ) \) which results in

\begin{align*} F\left ( \tau \right ) & =\frac {x}{y}\sqrt {4y-x^{2}}\\ & =\frac {x}{\frac {\tau }{x^{\alpha }}}\sqrt {4\frac {\tau }{x^{\alpha }}-x^{2}}\\ & =\frac {x}{\tau x^{2}}\sqrt {4\tau x^{2}-x^{2}}\end{align*}

Since the requirement is that \(F\left ( \tau \right ) \) ends up free of \(x\), then the only way to use this method and simplify the above to eliminate \(x\) is to assume \(x>0\). Now the above simplifies to

\[ F\left ( \tau \right ) =\frac {1}{\tau }\sqrt {4\tau -1}\]

The solution(1) becomes

\begin{align*} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \alpha \right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{x^{2}}}\frac {1}{\tau \left ( 2-\frac {1}{\tau }\sqrt {4\tau -1}\right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{x^{2}}}\frac {1}{2\tau -\sqrt {4\tau -1}}d\tau & =0 \end{align*}

Solving the integral gives long complicated expression which is verified correct. So better to keep the solution implicit as the above. Now we solve the second ode \(y^{\prime }=-\sqrt {4y-x^{2}}\) in similar way.