2.22 Example 22 \(\left ( y^{\prime }\right ) ^{2}+2xy^{\prime }-y=0\)

\begin{align*} \left ( y^{\prime }\right ) ^{2}+2xy^{\prime }-y & =0\\ p^{2}+2xp-y & =0\\ F & =p^{2}+2xp-y\\ & =0 \end{align*}

Checking \(\frac {\partial F}{\partial y}=-1\neq 0\). Since quadratic in \(p\) then

\begin{align*} b^{2}-4ac & =0\\ \left ( 2x\right ) ^{2}-4\left ( 1\right ) \left ( -y\right ) & =0\\ 4x^{2}+4y & =0\\ y & =-x^{2}\end{align*}

Now we check that this satisfies the ode itself.  We see it does not. Now we try the c-discriminant method. The general solution is too complicated to write here. But Mathematica and Maple claim there is no singular solution. So will leave it there for now. The paper I took this example from is wrong. It claimed \(y=x^{2}\) is the envelope. It is not.