1.2.6 Example 6 \(x\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }+4x=0\)

This can also be solved as dAlembert ode.

\begin{align} x\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }+4x & =0\nonumber \\ xp^{2}-2yp+4x & =0 \tag {1}\end{align}

We start by trying to isolate \(x\)

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

Eq (3A) gives

\begin{align*} \frac {dy}{dp} & =\frac {p\frac {df}{dp}}{1-p\frac {\partial f}{\partial y}}\\ & =\frac {2yp}{p^{2}+4}\end{align*}

This is separable (and linear). solving gives

\begin{equation} y=cp^{2}+4c \tag {3}\end{equation}

Eliminating \(p\) from  (2,3) gives

\[ p=\frac {x}{2c}\]

And implicit solution for \(y\)

\[ y\left ( 16c^{2}-4cy+x^{2}\right ) =0 \]

Which implies

\begin{align*} y & =0\\ 16c^{2}-4cy+x^{2} & =0 \end{align*}

Or

\begin{align*} y & =0\\ y & =\frac {x^{2}+16c^{2}}{4c}\\ & =\frac {x^{2}+c_{2}^{2}}{c_{2}}\end{align*}

But the solution \(y=0\) does not satisfy the ode. So it is removed. So the solution is

\[ y=\frac {x^{2}+c_{2}^{2}}{c_{2}}\]

This is the same general solution obtained using dAlembert, but with dAlembert method, we also obtain singular solutions \(y=\pm 2x\). This method does not find these, only the general solution. See my dAlembert showing how this problem is solved using that method.

Let us now solving this problem by isolating \(y\) instead of \(x\) and see if we get same solution. From original ode

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

Using (2A) gives

\begin{align*} \frac {dx}{dp} & =\frac {\frac {df}{dp}}{p-\frac {\partial f}{\partial x}}\\ & =\frac {x}{p}\end{align*}

This is easilt solved giving

\begin{equation} x=c_{1}p \tag {5}\end{equation}

Eliminating \(p\) between (4,5) gives

\begin{align*} y & =\frac {4c_{1}^{2}+x^{2}}{2c_{1}}\\ & =\frac {c^{2}+x^{2}}{c}\end{align*}

Which is the same solution found earlier.