3.5.3.7 Example 7

\(yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives

\begin{align} y & =\frac {x+p^{2}}{p}\tag {1}\\ & =\frac {1}{p}x+p\nonumber \\ & =xf+g\nonumber \end{align}

Hence \(f=\frac {1}{p},g\left ( p\right ) =p\). Taking derivative w.r.t. \(x\) gives

\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}

Using \(f=\frac {1}{p},g=p\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. the above simplifies to

\begin{equation} p-\frac {1}{p}=\left ( -\frac {x}{p^{2}}+1\right ) \frac {dp}{dx} \tag {2A}\end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in (2) which results in \(Q\left ( p\right ) =0\) or \(p-1=0\) or \(p=1\). Substituting these values in (1) gives the solutions

\begin{equation} y_{1}\left ( x\right ) =x+1 \tag {3}\end{equation}

The general solution is found by finding \(p\) from (2A). Since (2A) is not linear and not separable in \(p\), then inversion is needed. Writing (2) as

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

Hence

\[ \frac {dx}{dp}+\frac {x}{p\left ( p^{2}-1\right ) }=\frac {p^{2}}{p\left ( p^{2}-1\right ) }\]

This is now linear ODE in \(x\left ( p\right ) \). The solution is

\begin{align} x & =\frac {p\sqrt {\left ( p-1\right ) \left ( 1+p\right ) }\ln \left ( p+\sqrt {p^{2}-1}\right ) }{\left ( 1+p\right ) \left ( p-1\right ) }+c_{1}\frac {p}{\sqrt {\left ( 1+p\right ) \left ( p-1\right ) }}\nonumber \\ & =\frac {p\sqrt {p^{2}-1}\ln \left ( p+\sqrt {p^{2}-1}\right ) }{p^{2}-1}+c_{1}\frac {p}{\sqrt {p^{2}-1}} \tag {4}\end{align}

Now we need to eliminate \(p\) from (1,4). From (1) since \(y=\frac {1}{p}x+p\) then solving for \(p\) gives

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

Substituting each \(p_{i}\) in (4) gives the general solution (implicit) in \(y\left ( x\right ) \). First solution is

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

And second solution is

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