3.5.3.23 Example 23

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

Let \(y^{\prime }=p\) and rearranging gives

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

Hence

\begin{align*} f & =-\frac {1}{p}\\ g & =\frac {1}{2}p \end{align*}

Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative of (1) w.r.t. \(x\) gives

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

But \(f\left ( p\right ) =-\frac {1}{p},f^{\prime }\left ( p\right ) =\frac {1}{p^{2}},g=\frac {1}{2}p,g^{\prime }=\frac {1}{2}\) and the above becomes

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

The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p^{2}+1=0\). Hence \(p=\pm i\) But these do not verify the ode. Hence no singular solutions exist.

The general solution is when \(\frac {dp}{dx}\neq 0\) in (2). This gives the ode

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

But this is non-linear in \(p\). Hence inversion is needed. This becomes

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

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

\begin{equation} x=\frac {\left ( \frac {1}{2}\operatorname {arcsinh}\left ( p\right ) +c_{1}\right ) p}{\sqrt {p^{2}-1}} \tag {3}\end{equation}

We now need to eliminate \(p\). We have two equations to do that, (1) and (3). Here they are side by side

\begin{align} y & =-x\frac {1}{p}+\frac {1}{2}p\tag {1}\\ x & =\frac {\left ( \frac {1}{2}\operatorname {arcsinh}\left ( p\right ) +c_{1}\right ) p}{\sqrt {p^{2}-1}} \tag {3}\end{align}

We can either solve for \(p\) from (1) and plugin in the value found into (3). Or we can solve for \(p\) from (3) and plugin the value found in (1). In this case it is easier to solve for \(p\) from (1) which gives

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

Substituting each of these into (3) gives these two general solutions

\begin{align*} x & =\frac {\left ( \frac {1}{2}\operatorname {arcsinh}\left ( y+\sqrt {2x+y^{2}}\right ) +c_{1}\right ) \left ( y+\sqrt {2x+y^{2}}\right ) }{\sqrt {\left ( y+\sqrt {2x+y^{2}}\right ) ^{2}-1}}\\ x & =\frac {\left ( \frac {1}{2}\operatorname {arcsinh}\left ( y-\sqrt {2x+y^{2}}\right ) +c_{1}\right ) \left ( y-\sqrt {2x+y^{2}}\right ) }{\sqrt {\left ( y-\sqrt {2x+y^{2}}\right ) ^{2}-1}}\end{align*}