1.2.1 Example 1 \(y\left ( y^{\prime }\right ) ^{2}-4xy^{\prime }+y=0\)

Given

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

Let \(y^{\prime }=p\). The above becomes

\begin{align} yp^{2}-4xp+y & =0\nonumber \\ f\left ( x,y,p\right ) & =0 \tag {2}\end{align}

We have to either isolate \(x\) or \(y\) as both can be inside \(f\). Let us start by isolating \(x\). The above becomes

\[ x=\frac {yp^{2}+y}{4p}\]

Therefore \(f\left ( y,p\right ) =\frac {yp^{2}+y}{4p}\) now, We now apply (3A) (not 2A) and obtain

\[ \frac {dy}{dp}=\frac {p\frac {df}{dp}}{1-p\frac {\partial f}{\partial y}}\]

Hence

\begin{align*} \frac {dy}{dp} & =\frac {\frac {y\left ( p^{2}-1\right ) }{4p}}{1-\frac {p^{2}}{4}-\frac {1}{4}}\\ & =-\frac {y}{p}\frac {p^{2}-1}{p^{2}-3}\\ & =\frac {y}{p}\frac {\left ( p^{2}-1\right ) }{\left ( 3-p^{2}\right ) }\end{align*}

This is separable. Solving gives

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

Therefore the parametric solution is

\begin{align*} x & =\frac {yp^{2}+y}{4p}\\ y & =\frac {c_{1}}{\left ( p\left ( p^{2}-3\right ) \right ) ^{\frac {1}{3}}}\end{align*}

The above is the solution to (1) in parametric form where the dependency between \(y\) and \(x\) is via \(p\). We can stop here. But let see if we can get the solution as \(y\left ( x\right ) \) as the normal case is. Eliminating \(p\) between 4(1) and 4(2) results in the solution

\[ c^{6}-64c^{3}x^{3}+24c^{3}xy^{2}-48x^{2}y^{4}+16y^{6}=0 \]

And the above is the final nonparametric solution. It is an implicit solution.

Let try the other way. Let us start by isolating \(y.\) This gives

\begin{align*} yp^{2}-4xp+y & =0\\ f\left ( x,y,p\right ) & =0 \end{align*}

Isolating \(y\). The above becomes

\[ y=\frac {4xp}{1+p^{2}}\]

Therefore \(f\left ( y,p\right ) =\frac {4xp}{1+p^{2}}\) now, We now apply (2A) (not 3A) and obtain

\[ \frac {dx}{dp}=\frac {\frac {df}{dp}}{p-\frac {\partial f}{\partial x}}\]

Hence

\begin{align*} \frac {dx}{dp} & =\frac {\frac {-4xp^{2}+4x}{\left ( p^{2}+1\right ) ^{2}}}{p-\frac {4p}{p^{2}+1}}\\ & =4x\frac {p^{2}-1}{-p^{5}+2p^{3}+3p}\end{align*}

Solving gives

\[ x=\frac {\left ( p^{2}+1\right ) c_{1}}{\left ( p^{2}-3\right ) ^{\frac {1}{3}}p^{\frac {4}{3}}}\]

Thus the parametric solution is

\begin{align*} y & =\frac {4xp}{1+p^{2}}\\ x & =\frac {\left ( p^{2}+1\right ) c_{1}}{\left ( p^{2}-3\right ) ^{\frac {1}{3}}p^{\frac {4}{3}}}\end{align*}

Eliminating \(p\) gives the solution. But it was harder to eliminate \(p\) using this approach.

We might think this method is complicated, but it is actually much simpler than direct method. How would we solve original ode directly? We start by solving for \(y^{\prime }\) in (1) which gives two ode’s that we need to solve each on its own.

\begin{align} y^{\prime } & =\frac {2x+\sqrt {4x^{2}-y^{2}}}{y}\tag {5}\\ y^{\prime } & =\frac {2x-\sqrt {4x^{2}-y^{2}}}{y}\nonumber \end{align}

Starting with the first one above, we notice it is homogeneous ode. Let \(u=\frac {y}{x}\) and it becomes

\[ u^{\prime }=\frac {-u^{2}+\sqrt {-u^{2}+4}+2}{ux}\]

This is separable which results in

\[ \int \frac {u}{-u^{2}+\sqrt {-u^{2}+4}+2}du=\int \frac {1}{x}dx \]

The above integrals gives a very complicated antiderivative. After that we have to replace \(u\) back by \(\frac {y}{x}\) and simplify. We now do the same for the second ode in (5). It is clear here that the parametric method is simpler. But for the parametric method to work, we would have to be able isolate \(p\).