3.2.2.8 Example 9 \(y^{\prime }=-\frac {x}{2}-1+\frac {1}{2}\sqrt {x^{2}+4x+4y}\)

This is clairaut ode.

\begin{align} p & =-\frac {x}{2}-1+\frac {1}{2}\sqrt {x^{2}+4x+4y}\nonumber \\ \frac {1}{2}\sqrt {x^{2}+4x+4y} & =p+1+\frac {x}{2}\nonumber \\ \sqrt {x^{2}+4x+4y} & =2p+2+x\nonumber \\ x^{2}+4x+4y & =\left ( 2p+2+x\right ) ^{2}\nonumber \\ x^{2}+4x+4y & =4p^{2}+4px+8p+x^{2}+4x+4\nonumber \\ 4y-8p-4px-4p^{2}-4 & =0 \tag {1}\end{align}

Isolating \(x\) gives

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

Using (3A)

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

But \(1-p\frac {\partial f}{\partial y}=0\), so this did not work. Let us try to isolate \(y\) to see if we get better luck. Isolating \(y\) from (1) gives

\begin{align*} y & =p^{2}+xp+2p+1\\ & =f\left ( x,p\right ) \end{align*}

Using (2A)

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

But here we also get \(p-\frac {\partial f}{\partial x}=0\). So it is not possible to use parametric method on this ode. Does this happen always on clairaut ode? No, we solved clairaut using this method in example 2 above. It is just by chance we get denominator zero for this specific ode. I think it is because \(p\) is linear in the ode. This method should be used for ode which has non linear \(y^{\prime }\left ( x\right ) \) in it. But I need to double check more on this. For now, I check that \(p\) is nonlinear before using this method.