1.2.27 \(p^{2}\left ( 1-x^{2}\right ) =1-y^{2}\)

Problem (27)

\[ p^{2}\left ( 1-x^{2}\right ) =1-y^{2}\]

Hence

\[ p^{2}\left ( 1-x^{2}\right ) -\left ( 1-y^{2}\right ) =0 \]

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

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

Hence we have \(y=\pm 1\) and also \(x=\pm 1\). These lines are the signular solution. This diagram below shows few solution curves and these 4 lines. The solution can be found to be

\[ x^{2}+y^{2}-2c_{1}xy=1-c_{1}^{2}\]

Lets now see what the general solution gives using C-discriminant. Since the solution is already quadratic in \(c_{1}\) then we can use the C-discriminant directly from the quadratic equation, and no need to use elimination. Writing the solution as

\[ c_{1}^{2}-2c_{1}xy+x^{2}+y^{2}-1=0 \]

Then \(b^{2}-4ac=0\) gives

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

Which gives same result as p-discriminant. The plot shows solution curves (in blue) for different values of \(c_{1}\) with the 4 singular solution lines in dashed red style.

Figure 21: Singular solution envelope with general solution curves