2.18 \(y=xy^{\prime }+ay^{\prime }\left ( 1-y^{\prime }\right ) \) \begin{align*} y & =xy^{\prime }+ay^{\prime }\left ( 1-y^{\prime }\right ) \\ y & =xp+ap\left ( 1-p\right ) \\ F & =xp+ap-ap^{2}-y\\ & =-ap^{2}+p\left ( a+x\right ) -y\\ & =0 \end{align*}
We first check that \(\frac {\partial F}{\partial y}=-1\neq 0\) . This is quadratic in \(p\) .
\begin{align*} b^{2}-4ac & =0\\ \left ( a+x\right ) ^{2}-4\left ( -a\right ) \left ( -y\right ) & =0\\ \left ( a+x\right ) ^{2}-4ay & =0\\ y & =\frac {\left ( x+a\right ) ^{2}}{4a}\qquad a\neq 0 \end{align*}
Now we check if this satisfies the ode itself. We see it does. The general solution can be
found to be
\begin{align*} \Psi \left ( x,y,c\right ) & =y-cx-ac\left ( 1-c\right ) \\ & =y-cx-ac+ac^{2}\\ & =y-c\left ( x+a\right ) +ac^{2}\\ & =0 \end{align*}
This is quadratic in \(c\) .
\begin{align*} b^{2}-4aC & =0\\ \left ( x+a\right ) ^{2}-4\left ( a\right ) \left ( y\right ) & =0\\ y & =\frac {\left ( x+a\right ) ^{2}}{4a}\qquad a\neq 0 \end{align*}
Which is the same obtained using p-discriminant. Hence this is the singular solution. The
following plot shows the singular solution as the envelope of the family of general solution
plotted using different values of \(c\) . For this, \(a=1\) was used.