3.5.3.20 Example 20

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

Solving for \(y\) gives

\begin{align} y & =xp+\left ( 1+2p+p^{2}\right ) \tag {1}\\ y & =xf+g\nonumber \end{align}

Hence \(f=p,g=\left ( 1+2p+p^{2}\right ) \). Since \(f=p\) then this is Clairaut.  Taking derivative of the above w.r.t. \(x\) gives

\begin{align*} y^{\prime } & =f+x\frac {df}{dp}\frac {dp}{dx}+\frac {dg}{dp}\frac {dp}{dx}\\ p & =f+\frac {dp}{dx}\left ( x\frac {df}{dp}+\frac {dg}{dp}\right ) \end{align*}

But \(\frac {df}{dp}=1,\frac {dg}{dp}=2+2p\). The above becomes

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

But \(f=p\). The above simplifies to

\begin{equation} 0=\frac {dp}{dx}\left ( x+2+2p\right ) \tag {2}\end{equation}

The general solution is when \(\frac {dp}{dx}=0\). Hence \(p=c_{1}\). Substituting this into (1) gives

\[ y=xc_{1}+\left ( 1+2c_{1}+c_{1}^{2}\right ) \]

The singular solution is when \(\frac {dp}{dx}\neq 0\) in (2) which gives

\begin{align*} x+2+2p & =0\\ p & =\frac {-x-2}{2}\end{align*}

Substituting this in (1) gives

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

Checking this solution against the ode shows it is verifies the ode. Hence there are two solutions, one general and one singular

\[ y=\left \{ \begin {array} [c]{c}xc_{1}+1+2c_{1}+c_{1}^{2}\\ -\frac {1}{4}x^{2}-x \end {array} \right . \]