2.28 \(y^{2}\left ( 1+p^{2}\right ) =4\)

Problem (28)

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

The ode is already quadratic in \(p\), therefore the p-discriminant can be used directly to find singular solutions and there is no need to use elimination. The p-discriminant is given by \(b^{2}-4ac=0\) or

\begin{align*} 0-4\left ( y^{2}\right ) \left ( y^{2}-4\right ) & =0\\ y^{2}\left ( y^{2}-4\right ) & =0 \end{align*}

The solution \(y=0\) does not satisfy the ode, hence it is not singular solution. Using \(ET^{2}C\) (see introduction), shows \(y=0\) is the \(T\) solution, or Tac locus. The solutions \(y=\pm 2\) satisfy the ode, hence these are the envelope \(E\).  The general solution to the ode can be found to be

\[ y^{2}+\left ( x+c_{1}\right ) ^{2}=4 \]

The general solution are circles with radius \(2\) with centers shifted by \(c_{1}\). The plot shows solution curves (in blue) for different values of \(c_{1}\) with the 2 envelopes and also the Tac locus.

Figure 22: Singular solution envelope with general solution curves