2 Examples

2.1 \(9p^{2}\left ( 2-y\right ) ^{2}=4\left ( 3-y\right ) \)

\begin{align*} 9p^{2}\left ( 2-y\right ) ^{2} & =4\left ( 3-y\right ) \\ 9p^{2}\left ( 2-y\right ) ^{2}-4\left ( 3-y\right ) & =0 \end{align*}

Since this is quadratic in \(p\), we do not have to use elimination and can just use the quadratic discriminant

\begin{align*} b^{2}-4ac & =0\\ 0-4\left ( 9\left ( 2-y\right ) ^{2}\right ) \left ( -4\left ( 3-y\right ) \right ) & =0\\ 9\left ( 2-y\right ) ^{2}\left ( 3-y\right ) & =0\\ \left ( 2-y\right ) ^{2}\left ( 3-y\right ) & =0 \end{align*}

Comparing this to the form \(ET^{2}C=0\) we see that \(y=3\) is \(E\) (the envelop) and \(y=2\) is \(T\) (the Tac locus). This does not satisfy the ode. Let us see what happens if we use elimination method.

\begin{align*} F & =9p^{2}\left ( 2-y\right ) ^{2}-4\left ( 3-y\right ) =0\\ \frac {\partial F}{\partial p} & =18p\left ( 2-y\right ) ^{2}=0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}=-18p^{2}\left ( 2-y\right ) +4\neq 0\).  Eliminating \(p\). Second equation gives \(p=0\) and \(y=2\). Substituting \(p=0\) into the first equation gives

\[ y=3 \]

We now have to check if this solution satisfies the ode. We see it does. Hence it is the envelope. The second solution is \(y=2\) which does not satisfy the ode. This happens to be the Tac locus. We see that we obtain the same result using elimination as when using the quadratic discriminant directly.

To do the same thing use the solution, we have to first solve the ode. The general solution (also called the primitive) can be found to be

\begin{align*} \Psi \left ( x,y,c\right ) & =\left ( x+c\right ) ^{2}-y^{2}\left ( 3-y\right ) \\ & =x^{2}+c^{2}+2xc-y^{2}\left ( 3-y\right ) \\ & =0 \end{align*}

Since this is already quadratic in \(c\), we can use the the quadratic discriminant directly. (will use \(C\) instead of \(c\) so not to confuse it with the constant of integration \(c\) in the solution)

\begin{align*} b^{2}-4aC & =0\\ \left ( 2x\right ) ^{2}-4\left ( 1\right ) \left ( -y^{2}\left ( 3-y\right ) +x^{2}\right ) & =0\\ 4x^{2}+4y^{2}\left ( 3-y\right ) -4x^{2} & =0\\ 4y^{2}\left ( 3-y\right ) & =0\\ y^{2}\left ( 3-y\right ) & =0 \end{align*}

Comparing this to the form \(EN^{2}C^{3}=0\) shows that \(E=3\) which is same as found earlier using p-discriminant and \(y=0\) is \(N\) (nodal locus).

Hence in summary, we see that \(E=3,T=2,N=0\). Only the \(E\) (envelope solution \(y=3\)) satisfies the ode. The others do not.

The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\) and also shows the \(N\) and \(T\) curves.

Figure 2: Singular solution envelope with general solution curves

2.2 \(p^{2}=xy\)

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

Since this is quadratic in \(p\), we do not have to use elimination and can just use the quadratic discriminant

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

This has form \(ET^{2}C=0\). We see that \(y=0\) satisfies the ode, hence it is \(E\) (because \(C\) do not satisfy the ode). We found no \(T\) nor \(C\).

To apply c-discriminant we have to find the general solution. It will be

\begin{align*} y_{1} & =\frac {1}{36}\left ( 4x^{3}-12x^{\frac {3}{2}}c+9c^{2}\right ) \\ y_{2} & =\frac {1}{36}\left ( 4x^{3}+12x^{\frac {3}{2}}c+9c^{2}\right ) \end{align*}

Hence we have two general solutions. These can be written as

\begin{align*} \Psi _{1}\left ( x,y,c\right ) & =y-\frac {1}{36}\left ( 4x^{3}-12x^{\frac {3}{2}}c+9c^{2}\right ) =0\\ & =y+\frac {1}{3}cx^{\frac {3}{2}}-\frac {1}{4}c^{2}-\frac {1}{9}x^{3}=0\\ \Psi _{2}\left ( x,y,c\right ) & =y-\frac {1}{36}\left ( 4x^{3}+12x^{\frac {3}{2}}c+9c^{2}\right ) =0\\ & =y-\frac {1}{3}cx^{\frac {3}{2}}-\frac {1}{4}c^{2}-\frac {1}{9}x^{3}=0 \end{align*}

Since these are quadratic in \(c\), we can use the quadratic discriminant. First equation above gives

\begin{align*} b^{2}-4aC & =0\\ \left ( \frac {1}{3}x^{\frac {3}{2}}\right ) ^{2}-4\left ( -\frac {1}{4}\right ) \left ( y-\frac {1}{9}x^{3}\right ) & =0\\ \frac {1}{9}x^{3}+y-\frac {1}{9}x^{3} & =0\\ y & =0 \end{align*}

Comparing to \(EN^{2}C^{3}=0\) shows this is \(E\). Same as before. Second solution will give same result.

Hence \(y=0\) is the singular solution. No \(T,N,C\,\) were found. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

Figure 3: Singular solution envelope with general solution curves

2.3 \(27y-8p^{3}=0\)

\begin{align*} 27y-8\left ( y^{\prime }\right ) ^{3} & =0\\ 27y-8p^{3} & =0 \end{align*}

Since this is not quadratic in \(p\), we can not use the discriminant dirctly and have to use elimination.

\begin{align*} F & =27y-8p^{3}=0\\ \frac {\partial F}{\partial p} & =-24p^{2}=0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}=27\neq 0\).  Second equation gives \(p=0\). Substituting this into first equation gives \(27y=0\) or \(y=0\). We see this also satisfies the ode. Hence it is \(E\) (the envelope). The general solution can be found as

\[ \Psi \left ( x,y,c\right ) =y^{2}-\left ( x+c\right ) ^{3}=0 \]

Applying c-discriminant

\begin{align*} \Psi \left ( x,y,c\right ) & =y^{2}-\left ( x+c\right ) ^{3}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =-3\left ( x+c\right ) ^{2}=0 \end{align*}

Second equation gives \(\left ( x+c\right ) ^{2}=0\) or \(c=-x\). From first equation this gives \(y^{2}=0\) or \(y=0\). This is the same as \(y_{s}\) found from p-discriminant, hence

\[ y=0 \]

The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

Figure 4: Singular solution envelope with general solution curves

2.4 \(y-2xp-\ln p=0\)

\begin{align*} y-2xy^{\prime }-\ln y^{\prime } & =0\\ y-2xp-\ln p & =0 \end{align*}

Applying p-discriminant method via elimination gives

\begin{align*} F & =y-2xp-\ln p=0=0\\ \frac {\partial F}{\partial p} & =-2x-\frac {1}{p}=0 \end{align*}

Second equation gives \(p=-\frac {1}{2x}\). Substituting in the first equation gives

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

This does not satisfy the ode. Hence no singular solution exist.

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

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

We first check that \(\frac {\partial F}{\partial y}=1\neq 0\). Since this is quadratic in \(p\), we do not have to use elimination and can just use the quadratic discriminant

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

This does not satisfy the ode. Hence no singular solution exist.

2.6 \(y-2xp-\sin \left ( p\right ) =0\)

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

Applying p-discriminant method gives

\begin{align*} F & =y-2xp-\sin \left ( p\right ) =0\\ \frac {\partial F}{\partial y^{\prime }} & =-2x-\cos \left ( p\right ) =0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}=1\neq 0\).  Now we apply p-discriminant.  Second equation gives \(-2x-\cos \left ( p\right ) =0\) or \(p=\arccos \left ( -2x\right ) .\) Substituting in the first equation gives \(y-2x\arccos \left ( -2x\right ) -\sin \left ( \arccos \left ( -2x\right ) \right ) =0.\) I need to look at this more. This should give \(y_{s}=0\) but now it does not.

2.7 \(y-p^{2}x+\frac {1}{p}=0\)

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

Applying p-discriminant method gives

\begin{align*} F & =y-p^{2}x+\frac {1}{p}=0\\ \frac {\partial F}{\partial y^{\prime }} & =-2xp-\frac {1}{p^{2}}=0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}=1\neq 0\).  Now we apply p-discriminant.  Second equation gives 3 solutions for \(p.\)

\begin{align*} p_{1} & =\frac {\left ( -\frac {1}{2}\right ) ^{\frac {1}{3}}}{x^{\frac {1}{3}}}\\ p_{2} & =\frac {1}{2^{\frac {1}{3}}x^{\frac {1}{3}}}\\ p_{3} & =-\frac {\left ( -1\right ) ^{\frac {2}{3}}}{2^{\frac {1}{3}}x^{\frac {1}{3}}}\end{align*}

Using the first solution, then the first equation gives

\begin{align*} y-\left ( \frac {\left ( -\frac {1}{2}\right ) ^{\frac {1}{3}}}{x^{\frac {1}{3}}}\right ) ^{2}x+\frac {1}{\left ( \frac {\left ( -\frac {1}{2}\right ) ^{\frac {1}{3}}}{x^{\frac {1}{3}}}\right ) } & =0\\ y_{s} & =\frac {3}{2}\left ( -1\right ) ^{\frac {2}{3}}\sqrt [3]{2}\sqrt [3]{x}\end{align*}

Now we check if this satisfies the ode \(F=0\). It does not. Trying the second solution \(p_{2}=\frac {1}{2^{\frac {1}{3}}x^{\frac {1}{3}}}\). Substituting into \(F=0\) gives

\begin{align*} y-\left ( \frac {1}{2^{\frac {1}{3}}x^{\frac {1}{3}}}\right ) ^{2}x+\frac {1}{\left ( \frac {1}{2^{\frac {1}{3}}x^{\frac {1}{3}}}\right ) } & =0\\ y & =-\frac {1}{2}\sqrt [3]{2}\sqrt [3]{x}\end{align*}

Now we check if this satisfies the ode \(F=0\). It does not. Trying the third solution \(p_{3}=-\frac {\left ( -1\right ) ^{\frac {2}{3}}}{2^{\frac {1}{3}}x^{\frac {1}{3}}}\). Substituting into \(F=0\) gives

\begin{align*} y-\left ( -\frac {\left ( -1\right ) ^{\frac {2}{3}}}{2^{\frac {1}{3}}x^{\frac {1}{3}}}\right ) ^{2}x+\frac {1}{\left ( -\frac {\left ( -1\right ) ^{\frac {2}{3}}}{2^{\frac {1}{3}}x^{\frac {1}{3}}}\right ) } & =0\\ y & =-\frac {3}{2}\sqrt [3]{-1}\sqrt [3]{2}\sqrt [3]{x}\end{align*}

Now we check if this satisfies the ode \(F=0\). It does not. Hence no singular exist.

2.8 \(xp+p-p^{2}-y=0\)

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

This is quadratic in \(p\).

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

Now we verify this satisfies the ode. We see it does. Now we have to find the general solution. It will be

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

This is quadratic in \(c\).

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

We see this is the same as \(y\) from the p-discriminant method. Hence it is a singular solution. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

Figure 5: Singular solution envelope with general solution curves

2.9 \(y=px+\sqrt {4+p^{2}}\)

\begin{align*} y & =y^{\prime }x+\sqrt {4+y^{\prime 2}}\\ y-px-\sqrt {4+p^{2}} & =0 \end{align*}

Applying p-discriminant method gives

\begin{align*} F & =y-px-\sqrt {4+p^{2}}=0\\ \frac {\partial F}{\partial p} & =-x-\frac {p}{\sqrt {4+p^{2}}}=0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}=1\neq 0\).  Now we apply p-discriminant.  Second equation gives \(x\sqrt {4+p^{2}}+p=0\) which gives

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

Trying the negative root and substituting it in \(F=0\) gives

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

Which satisfies the ode.  The general solution can be found to be

\[ \Psi \left ( x,y,c\right ) =y-xc-\sqrt {4+c^{2}}=0 \]

Now we have to eliminate \(c\) using the c-discriminant method

\begin{align*} \Psi \left ( x,y,c\right ) & =y-xc-\sqrt {4+c^{2}}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =-x-\frac {2c}{2\sqrt {4+c^{2}}}=0 \end{align*}

Second equation gives

\[ c=\pm \frac {2x}{\sqrt {1-x^{2}}}\]

Taking the negative root, and substituting into the first equation gives

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

Which is the same obtained using the p-discriminant. Hence

\[ y=2\sqrt {1-x^{2}}\]

Is singular solution. We have to try the other root also. But graphically, the above seems to be the only valid singular solution.  The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

Figure 6: Singular solution envelope with general solution curves

2.10 \(p^{2}-xp+y=0\)

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

This also satisfies the ode. Hence it is \(E\). Now we check using p-discriminant method. General solution can be found to be

\[ \Psi \left ( x,y,c\right ) =y-xc+c^{2}=0 \]

This is quadratic in \(c\).

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

This is the same as \(y\) obtained using p-discriminant method then it is singular solution. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

Figure 7: Singular solution envelope with general solution curves

2.11 \(p=y\left ( 1-y\right ) \)

\begin{align*} y^{\prime } & =y\left ( 1-y\right ) \\ F & =p-y\left ( 1-y\right ) \\ & =0 \end{align*}

Since this is linear in \(p\), then there is no singular solution in the sense of envelope. We can not use the p-discriminant method. Mathematica gives singular solutions, but these are not the envelope. It calls them equilibrium solutions. Lets find them. By inspection we see this is separable. Hence the candidate singular solutions are obtain when \(y\left ( 1-y\right ) =0\). This is because this is what we have to divide both sides by to integrate. Therefore

\begin{align*} y_{s} & =0\\ y_{s} & =1 \end{align*}

The general solution is found from

\begin{align*} \int \frac {dy}{y\left ( 1-y\right ) } & =\int dx\\ -\ln \left ( y-1\right ) +\ln y & =x+c\\ \ln \frac {y}{y-1} & =x+c\\ \frac {y}{y-1} & =c_{1}e^{x}\\ y & =\left ( y-1\right ) c_{1}e^{x}\\ y-yc_{1}e^{x} & =-c_{1}e^{x}\\ y\left ( 1-c_{1}e^{x}\right ) & =-c_{1}e^{x}\\ y & =\frac {c_{1}e^{x}}{c_{1}e^{x}-1}\\ y & =\frac {c_{1}}{c_{1}-e^{-x}}\\ y & =\frac {1}{1-c_{2}e^{-x}}\end{align*}

Now we ask, can the singular solutions \(y_{s}=0,y_{s}=1\) be obtained from the above general solution for any value of \(c_{2}\)? We see when \(c_{2}=0\) then \(y=1\). Also when \(c_{2}=\infty \) then \(y=0\). So these are not really singular solutions. Mathematica call these equilibrium solutions. But these should not be called singular solutions. Mathematica generates these when using the option IncludeSingularSolutions. But Maple does not give these when using the option singsol=all.

The following plot shows these equilibrium solutions with the general solution plotted using different values of \(c\).

Figure 8: Equilibrium solution envelope with general solution curves

2.12 \(p^{2}x+py\ln y-y^{2}\left ( \ln y\right ) ^{4}=0\)

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

Applying p-discriminant method gives

\begin{align*} F & =p^{2}x+py\ln y-y^{2}\left ( \ln y\right ) ^{4}=0\\ \frac {\partial F}{\partial p} & =2px+y\ln y=0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}\neq 0\).  Now we apply p-discriminant. Eliminating \(p\). Second equation gives \(p=-\frac {y}{2x}\ln y\). Substituting into the first equation gives

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

Hence we obtain the solutions

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

Or

\begin{align} y & =0\tag {1}\\ y & =1\tag {2}\\ 1+4x\left ( \ln y\right ) ^{2} & =0 \tag {3}\end{align}

The solution \(y=0\) does not satisfy the ode. But \(y=1\) does. The solution \(1+4x\left ( \ln y\right ) ^{2}=0\) gives \(y=\left \{ \begin {array} [c]{c}e^{\frac {-i}{2\sqrt {x}}}\\ e^{\frac {i}{2\sqrt {x}}}\end {array} \right . \) But these do not satisfy the ode.

The primitive can be found to be

\[ \Psi \left ( x,y,c\right ) =y-e^{\frac {c}{c^{2}-x}}=0 \]

Now we have to eliminate \(c\) using the c-discriminant method

\begin{align*} \Psi \left ( x,y,c\right ) & =y-e^{\frac {c}{c^{2}-x}}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =\left ( \frac {1}{c^{2}-x}-\frac {2c^{2}}{\left ( c^{2}-x\right ) ^{2}}\right ) e^{\frac {c}{c^{2}-x}}=0 \end{align*}

Second equation gives \(\frac {1}{c^{2}-x}-\frac {2c^{2}}{\left ( c^{2}-x\right ) ^{2}}=0\) or \(e^{\frac {c}{c^{2}-x}}=0\). For the first one, \(c=\pm \sqrt {-x}\). Substituting \(\sqrt {-x}\) in first equation gives

\begin{align} y-e^{\frac {\sqrt {-x}}{-x-x}} & =0\nonumber \\ y & =e^{\frac {\sqrt {-x}}{-2x}}\nonumber \\ \ln y & =\frac {\sqrt {-x}}{-2x}\nonumber \\ \left ( \ln y\right ) ^{2} & =\frac {-x}{4x^{2}}\nonumber \\ 4x\left ( \ln y\right ) ^{2}+1 & =0\nonumber \\ y_{s} & =\left \{ \begin {array} [c]{c}e^{\frac {-i}{2\sqrt {x}}}\\ e^{\frac {i}{2\sqrt {x}}}\end {array} \right . \tag {4}\end{align}

These do not satisfy the ode. Can not obtain \(y=1\) solution using c-discriminant.

See paper by C.N. SRINIVASINGAR, example 4. The following plot shows the singular solution above as the envelope of the family of general solution plotted using different values of \(c\). Added also \(y_{s}=1\).

Figure 9: Singular solution envelope with general solution curves

2.13 \(p^{2}-4y=0\)

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

This is quadratic in \(p\).

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

We see this also satisfies the ode. Hence it is the envelope. The primitive can be found to be

\begin{align*} \Psi \left ( x,y,c\right ) & =y-\left ( x+c\right ) ^{2}=0\\ & =y-\left ( x^{2}+c^{2}+2xc\right ) =0\\ & =y-x^{2}-c^{2}-2xc=0 \end{align*}

This is quadratic in \(c\).

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

This is the same as found by p-discriminant method then 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 \(c\).

Figure 10: Singular solution envelope with general solution curves

2.14 \(1+p^{2}-\frac {1}{y^{2}}=0\)

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

We first check that \(\frac {\partial F}{\partial y}=2\frac {1}{y^{3}}\neq 0\). This is quadratic in \(p\).

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

We see both solutions also satisfy the ode. The primitive can be found to be

\begin{align*} \Psi \left ( x,y,c\right ) & =y^{2}+\left ( x+c\right ) ^{2}-1\\ & =y^{2}+x^{2}+c^{2}+2xc-1\\ & =0 \end{align*}

This is quadratic in \(c\).

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

Which agrees with the p-discriminant. Hence these are the singular solutions. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

Figure 11: Singular solution envelope with general solution curves

2.15 \(y-p^{2}+3xp-3x^{2}=0\)

\begin{align*} y-\left ( y^{\prime }\right ) ^{2}+3xy^{\prime }-3x^{2} & =0\\ y-p^{2}+3xp-3x^{2} & =0\\ F & =y-p^{2}+3xp-3x^{2}\\ & =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 ( 3x\right ) ^{2}-4\left ( -1\right ) \left ( y-3x^{2}\right ) & =0\\ 9x^{2}+4y-12x^{2} & =0\\ y & =\frac {3}{4}x^{2}\end{align*}

Which satisfies the ode. Hence it is the envelope. The primitive can be found to be

\[ \Psi \left ( x,y,c\right ) =y-cx-c^{2}-x^{2}=0 \]

This is quadratic in \(c\).

\begin{align*} b^{2}-4aC & =0\\ \left ( -x\right ) ^{2}-4\left ( -1\right ) \left ( y-x^{2}\right ) & =0\\ x^{2}+4y-4x^{2} & =0\\ y & =\frac {3}{4}x^{2}\end{align*}

Which agrees with the p-discriminant curve. Hence this is a singular solution. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

Figure 12: Singular solution envelope with general solution curves

2.16 \(p^{2}\left ( 1-y\right ) ^{2}-2+y=0\)

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

Hence

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

We first check that \(\frac {\partial F}{\partial y}=-2\left ( y^{\prime }\right ) ^{2}\left ( 1-y\right ) +1\neq 0\). Since the ode is quadratic in \(p\), we can use the more direct method which is the discriminant of the quadratic equation instead of the elimination method. The discriminant of (1) is

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

Comparing this to \(ET^{2}C=0\) shows that \(y=2\) is \(E\), i.e. the envelope singular solution which also satisfies the ode, while \(y=1\) is \(T\) which is Tac locus. This does not satisfy the ode but it plotted below to show geometrically what it means.

The primitive can be found to be

\[ \Psi \left ( x,y,c\right ) =4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9\left ( x+c\right ) ^{2}=0 \]

Since this is also quadratic in \(c\), we can use the more direct method which is the discriminant of the quadratic equation instead of the elimination method. Hence

\begin{align*} 4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9\left ( x^{2}+c^{2}+2cx\right ) & =0\\ 4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9x^{2}-9c^{2}-18cx & =0 \end{align*}

Therefore

\begin{align*} b^{2}-4aC & =0\\ \left ( -18x\right ) ^{2}-4\left ( -9\right ) \left ( 4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9x^{2}\right ) & =0\\ \left ( -18x\right ) ^{2}+144\left ( 2-y\right ) \left ( y+1\right ) ^{2}-324x^{2} & =0\\ 144\left ( 2-y\right ) \left ( y+1\right ) ^{2} & =0\\ \left ( 2-y\right ) \left ( y+1\right ) ^{2} & =0 \end{align*}

Comparing to general form \(EN^{2}C^{3}=0\) shows that \(y=2\) is \(E\) which agrees with what was found using p-discriminant as expected, and \(y=-1\) is \(N\) which is nodal locus \(N\) since it shows to power of two. Notice that \(N\) can only show up using c-discriminant method.

The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\), and also show the nodal locus \(N\) shown as solid green line and \(T\) (Tac locus) drawn as solid blue line.  

The above result of \(E,N,T\) can also be found using elimination method. But since the equation are already quadratic in \(p,c\), it can be easier to just use the quadratic equation discriminant directly. As metioned before, elimination method is more general since it works for any equation and not just quadratic.

Figure 13: Singular solution envelope with general solution curves

2.17 \(\left ( y-xp\right ) ^{2}-p^{2}=1\)

\begin{align*} \left ( y-xy^{\prime }\right ) ^{2}-\left ( y^{\prime }\right ) ^{2} & =1\\ \left ( y-xp\right ) ^{2}-p^{2} & =1\\ F & =1+p^{2}-\left ( y-xp\right ) ^{2}\\ & =1+p^{2}-\left ( y^{2}+x^{2}p^{2}-2yxp\right ) \\ & =1+p^{2}-y^{2}-x^{2}p^{2}+2yxp\\ & =p^{2}\left ( 1-x^{2}\right ) +p\left ( 2yx\right ) +1-y^{2}\\ & =0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}=2\left ( y-xp\right ) \neq 0\). This is quadratic in \(p\). Hence

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

Both of these solutions verify the ode. Hence they are both singular solutions. The primitive can be found to be

\begin{align*} \Psi \left ( x,y,c\right ) & =y-xc\pm \sqrt {1+c^{2}}\\ & =0 \end{align*}

Eliminating \(c\). First solution gives

\begin{align*} \Psi _{1}\left ( x,y,c\right ) & =y-xc+\sqrt {1+c^{2}}=0\\ \frac {\partial \Psi _{1}\left ( x,y,c\right ) }{\partial c} & =-x+\frac {1}{2}\frac {2c}{\sqrt {1+c^{2}}}=0 \end{align*}

Second equation gives \(-x+\frac {1}{2}\frac {2c}{\sqrt {1+c^{2}}}=0\) or \(c=x\sqrt {\frac {1}{1-x^{2}}}\). Substituting into the first equation above gives

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

Which is given by p-discriminant above. Hence it is singular solution. If we try \(\Psi _{2}\left ( x,y,c\right ) =y-xc-\sqrt {1+c^{2}}=0\) we also can verify the second singular solution. Hence

\[ y=\pm \sqrt {1-x^{2}}\]

The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

Figure 14: Singular solution envelope with general solution curves

2.18 \(y=xp+ap\left ( 1-p\right ) \)

\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.

Figure 15: Singular solution envelope with general solution curves

2.19 \(p^{3}-4xyp+8y^{2}=0\)

\begin{align*} \left ( y^{\prime }\right ) ^{3}-4xyy^{\prime }+8y^{2} & =0\\ p^{3}-4xyp+8y^{2} & =0 \end{align*}

Applying p-discriminant method gives

\begin{align*} F & =p^{3}-4xyp+8y^{2}=0\\ \frac {\partial F}{\partial p} & =3p^{2}-4xy=0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}=-4xy^{\prime }+16y\neq 0\).  Now we apply p-discriminant. Eliminating \(p\). Second equation gives \(p=\pm \left ( \frac {4xy}{3}\right ) ^{\frac {1}{2}}\). Substituting first solution in the first equation gives

\begin{align*} \left ( \frac {4xy}{3}\right ) ^{\frac {3}{2}}-4xy\left ( \frac {4xy}{3}\right ) ^{\frac {1}{2}}+8y^{2} & =0\\ y_{s} & =\frac {4}{27}x^{3}\end{align*}

Which satisfies the ode. The general solution can be found to be

\[ \Psi \left ( x,y,c\right ) =y-\frac {1}{4}\frac {x^{2}}{c}+\frac {1}{8}\frac {x}{c^{2}}-\frac {1}{64c^{3}}=0 \]

Hence

\begin{align*} \Psi \left ( x,y,c\right ) & =y-\frac {1}{4}\frac {x^{2}}{c}+\frac {1}{8}\frac {x}{c^{2}}-\frac {1}{64c^{3}}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =\frac {1}{4}\frac {x^{2}}{c^{2}}-\frac {1}{4}\frac {x}{c^{3}}+\frac {3}{64c^{4}}=0 \end{align*}

Eliminating \(c\). Second equation gives \(c=\frac {1}{4x}\) or \(c=\frac {3}{4x}\). Substituting \(c=\frac {1}{4x}\) in the first equation above gives

\begin{align*} y-\frac {1}{4}\frac {x^{2}}{\left ( \frac {1}{4x}\right ) }+\frac {1}{8}\frac {x}{\left ( \frac {1}{4x}\right ) ^{2}}-\frac {1}{64\left ( \frac {1}{4x}\right ) ^{3}} & =0\\ y_{s} & =0 \end{align*}

Which satisfies the ode. But \(y=0\) can be obtained from the general solution above when \(c=\infty \) so it is not singular solution. Substituting \(c=\frac {3}{4x}\) in the first equation above gives

\begin{align*} y-\frac {1}{4}\frac {x^{2}}{\left ( \frac {3}{4x}\right ) }+\frac {1}{8}\frac {x}{\left ( \frac {3}{4x}\right ) ^{2}}-\frac {1}{64\left ( \frac {3}{4x}\right ) ^{3}} & =0\\ y & =\frac {4}{27}x^{3}\end{align*}

Which is the same obtained by 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 \(c\).

Figure 16: Singular solution envelope with general solution curves

2.20 \(4xp^{2}=\left ( 2x-1\right ) ^{2}\)

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

We first check that \(\frac {\partial F}{\partial y}\) and see it is zero. Hence no singular solution exists.

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

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

Checking

\begin{align*} \frac {\partial F}{\partial y} & =\frac {x}{2}\frac {-2y}{\sqrt {1-y^{2}}}\\ & =-\frac {xy}{\sqrt {1-y^{2}}}\end{align*}

Not zero. But this is linear in \(p\). Hence no singular solution will exist using p-discriminant. Lets see. Applying p-discriminant method gives

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

The p-discriminant does not yield result since it gives \(1=0\). Lets try C-discriminant.

\begin{align*} \Psi \left ( x,y,c\right ) & =y-\sin \left ( x^{2}+2c\right ) =0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =-2\cos \left ( x^{2}+2c\right ) =0 \end{align*}

Hence \(x^{2}+2c=\frac {\pi }{2}\) (there are infinite solutions). Hence \(c=\frac {\pi }{4}-\frac {x^{2}}{2}\). Substituting in the first equation gives

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

And if we took \(x^{2}+2c=-\frac {\pi }{2}\) then we now obtain \(y_{s}=-1\). Now we check that \(y=\pm 1\) satisfy the ode itself. We see that they do.  This is an example where c-discriminant found singular solution but not p-discriminant. This is strange as books say that same \(E\) should result using both methods. Need to look more into this example.

The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\). p-discriminant does not yield result but C-discriminant does.

Figure 17: Singular solution envelope with general solution curves

2.22 \(p^{2}+2xp-y=0\)

\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.

2.23 \(p^{2}\left ( 2-3y\right ) ^{2}-4\left ( 1-y\right ) =0\)

\begin{align*} \left ( y^{\prime }\right ) ^{2}\left ( 2-3y\right ) ^{2}-4\left ( 1-y\right ) & =0\\ p^{2}\left ( 2-3y\right ) ^{2}-4\left ( 1-y\right ) & =0 \end{align*}

Since quadratic in \(p\) then

\begin{align*} b^{2}-4ac & =0\\ 0-4\left ( 2-3y\right ) ^{2}\left ( -4\left ( 1-y\right ) \right ) & =0\\ \left ( 2-3y\right ) ^{2}\left ( 1-y\right ) & =0 \end{align*}

Comparing to \(ET^{2}C=0\) shows that \(y=\frac {2}{3}\) is Tac locus and \(y=1\) is \(E\) since it verifies the ode (\(C\) will not verify the ode).

The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c.\)

Figure 18: Singular solution envelope with general solution curves

2.24 \(8p^{3}x=y\left ( 12p^{2}-9\right ) \)

Problem (25)

\begin{align*} 8\left ( y^{\prime }\right ) ^{3}x & =y\left ( 12\left ( y^{\prime }\right ) ^{2}-9\right ) \\ 8p^{3}x & =y\left ( 12p^{2}-9\right ) \end{align*}

Hence

\begin{align} F & =8p^{3}x-y\left ( 12p^{2}-9\right ) =0\tag {1}\\ \frac {\partial F}{\partial p} & =p^{2}x-yp=0 \tag {2}\end{align}

We first check that \(\frac {\partial F}{\partial y}=12\left ( y^{\prime }\right ) ^{2}-9\neq 0\).  Now we apply p-discriminant. Eliminating \(p\). EQ (2) gives \(p\left ( px-y\right ) =0\) or \(p=0,px-y=0\). Hence

\begin{align*} p_{1} & =0\\ p_{2} & =\frac {y}{x}\end{align*}

Substituting into the EQ (1) gives these candidate singular solutions

\begin{align*} y & =0\\ 8\left ( \frac {y}{x}\right ) ^{3}x-y\left ( 12\left ( \frac {y}{x}\right ) ^{2}-9\right ) & =0 \end{align*}

Or

\begin{align*} y & =0\\ 8\frac {y^{3}}{x^{2}}-12\frac {y^{3}}{x^{2}}+9y & =0 \end{align*}

Or

\begin{align*} y & =0\\ 8\frac {y^{2}}{x^{2}}-12\frac {y^{2}}{x^{2}}+9 & =0 \end{align*}

Or

\begin{align} y & =0\tag {13}\\ y & =-\frac {3}{2}x\tag {4}\\ y & =\frac {3}{2}x \tag {5}\end{align}

We now have to check if this solution satisfies the ode. We see it does. Now we have to find the general solution (also called the primitive). This comes out to be

\begin{align} \Psi \left ( x,y,c\right ) & =y-\frac {\left ( x+3c_{1}\right ) ^{\frac {3}{2}}}{3\sqrt {c_{1}}}\tag {6}\\ & =y+\frac {\left ( x+3c_{1}\right ) ^{\frac {3}{2}}}{3\sqrt {c_{1}}} \tag {7}\end{align}

Looking at (7)

\begin{align*} \Psi \left ( x,y,c\right ) & =0=y+\frac {\left ( x+3c_{1}\right ) ^{\frac {3}{2}}}{3\sqrt {c_{1}}}\\ \frac {\partial \Psi }{\partial c} & =0=\frac {3\sqrt {3c_{1}+x}}{2\sqrt {c}}-\frac {\left ( 3c_{1}+x\right ) ^{\frac {3}{2}}}{6c_{1}^{\frac {3}{2}}}\end{align*}

Eliminating \(c\). Second equation gives \(c_{1}=-\frac {1}{3}x\) or \(x=\frac {x}{6}\,\). Using \(x=\frac {x}{6}\) and Substituting into the first equation above gives

\begin{align*} 0 & =y+\frac {\left ( x+3\frac {x}{6}\right ) ^{\frac {3}{2}}}{3\sqrt {\frac {x}{6}}}\\ y & =-\frac {\left ( x+\frac {x}{2}\right ) ^{\frac {3}{2}}}{3\sqrt {\frac {x}{6}}}\end{align*}

Hence the c-discriminant method gives

\begin{align} y_{s} & =0\tag {2}\\ y_{s} & =3\nonumber \end{align}

Now we take the common \(y_{s}\) from the p-discriminant and the c-discriminant from (1,2). We see that \(y_{s}=3\) is common. Hence

\[ y_{s}=3 \]

And \(y_{s}=0\) is removed. We also see that \(y_{s}=0\) does not even satisfy the ode. But even if it did, it is removed since it is not common with the p-discriminant .

If there is no common \(y_{s}\) found from applying the two method (p-discriminant and the c-discriminant) then it means there is no singular solution. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

2.25 Clairaut \(y=xp+a\sqrt {1+p^{2}}\)

Problem (26)

\begin{equation} y=xp+a\sqrt {1+p^{2}} \tag {1}\end{equation}

In this problem, will use new method to find singular solution. Which is to write the equation in rational form and find the discriminant from \(b^{2}-4ac\) from the quadratic equation. This is simpler method than using elimination as was done in all the problems above. I just learned this method from old book by Daniel A. Murray.  But this only works if equation for \(p\) and \(c\) can be written as quadratic equation. In General, the method of elimination works for all cases.

In this quadratic method, we write the ode as quadratic in \(p\) and find the discriminant and set it to zero.

\begin{align*} -y+xp+a\sqrt {1+p^{2}} & =0\\ \frac {y}{a}-\frac {xp}{a} & =\sqrt {1+p^{2}}\\ 1+p^{2} & =\frac {y^{2}}{a^{2}}+\frac {x^{2}p^{2}}{a^{2}}-2\frac {xyp}{a^{2}}\\ p^{2}\left ( 1-\frac {x^{2}}{a^{2}}\right ) +p\left ( 2\frac {xy}{a^{2}}\right ) +1-\frac {y^{2}}{a^{2}} & =0 \end{align*}

Hence p-discriminant is given by \(b^{2}-4ac=0\) or

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

The singular solutions are given by the above using p-discriminant method. Now we do the same using c-discriminant. The general solution is

\[ \Psi \left ( x,y,c\right ) =y-cx-a\sqrt {1+c^{2}}=0 \]

Writing the above as

\begin{align*} \sqrt {1+c^{2}} & =\frac {y-cx}{a}\\ 1+c^{2} & =\frac {\left ( y-cx\right ) ^{2}}{a^{2}}\\ a^{2}c^{2}+a^{2} & =y^{2}+c^{2}x^{2}-2cxy\\ c^{2}\left ( a^{2}-x^{2}\right ) +2cxy+a^{2}-y^{2} & =0 \end{align*}

The discriminant zero condition is \(b^{2}-4ac=0\). Or

\begin{align} \left ( 2xy\right ) ^{2}-4\left ( a^{2}-x^{2}\right ) \left ( a^{2}-y^{2}\right ) & =0\nonumber \\ 4x^{2}y^{2}-4\left ( a^{4}-a^{2}y^{2}-a^{2}x^{2}+x^{2}y^{2}\right ) & =0\nonumber \\ 4x^{2}y^{2}-4a^{4}+4a^{2}y^{2}+4a^{2}x^{2}-4x^{2}y^{2} & =0\nonumber \\ -4a^{4}+4a^{2}y^{2}+4a^{2}x^{2} & =0\nonumber \\ -a^{2}+x^{2}+y^{2} & =0\nonumber \\ x^{2}+y^{2} & =a^{2} \tag {3}\end{align}

Which satisfies the ode and is the same singular solution as found using the p-discriminant. Now will do the same but using elimination method to see if we get same result as above. We will use c-discriminant and p-discriminant. Both should give same singular solution as above, which is \(x^{2}+y^{2}=a^{2}\). To use c-discriminant we start with general solution which is

\begin{align} \Psi \left ( x,y,c\right ) & =y-cx-a\sqrt {1+c^{2}}=0\tag {4}\\ \frac {\partial \Psi }{\partial c} & =-x-\frac {ac}{\sqrt {1+c^{2}}}=0 \tag {5}\end{align}

Eliminating \(c\). From (5) \(-x\sqrt {1+c^{2}}-ac=0\) or \(\left ( ac\right ) ^{2}=x^{2}\left ( 1+c^{2}\right ) \) or \(a^{2}c^{2}-x^{2}c^{2}-x^{2}=0\) or \(c^{2}\left ( a^{2}-x^{2}\right ) =x^{2}\) or \(c=\frac {\pm x}{\sqrt {a^{2}-x^{2}}}\). Substituting \(\frac {x}{\sqrt {a^{2}-x^{2}}}\) into (4) gives

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

Which does not satisfy the ode. Trying now \(c=\frac {-x}{\sqrt {a^{2}-x^{2}}}\) then (4) gives

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

Which satisfies the ode. This is the same as found earlier. Now we will use p-discriminant elimination method.

\begin{align} F & =y-xp-a\sqrt {1+p^{2}}=0\tag {6}\\ \frac {\partial F}{\partial p} & =x-\frac {ap}{\sqrt {1+p^{2}}}=0 \tag {7}\end{align}

Eliminating \(p\) from (7).

\begin{align*} x\sqrt {1+p^{2}}-ap & =0\\ \sqrt {1+p^{2}} & =\frac {ap}{x}\\ 1+p^{2} & =\frac {a^{2}p^{2}}{x^{2}}\\ \frac {a^{2}p^{2}}{x^{2}}-p^{2}-1 & =0\\ a^{2}p^{2}-x^{2}p^{2}-x^{2} & =0\\ p^{2}\left ( a^{2}-x^{2}\right ) & =x^{2}\\ p & =\pm \frac {x}{\sqrt {a^{2}-x^{2}}}\end{align*}

Using \(p=\frac {x}{\sqrt {a^{2}-x^{2}}}\) in (6) gives

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

Which does not satisfy the ode. Using \(p=\frac {-x}{\sqrt {a^{2}-x^{2}}}\) in (6) gives

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

Hence \(y^{2}=a^{2}-x^{2}\). This is the same as found earlier.

The above shows that using Elimination or quadratic equation discriminant gives same singular solution and both p  or c discriminant give same result. The elimination method is more general as it works for equations in \(p,c\) which are not quadratic. If the equation in \(p,c\,\) come out to be quadratic, then it is better not to use elimination and use direct discriminant as it is simpler.

The following is a plot of general solution and the above singular solution for \(a=1\).

Figure 19: Singular solution envelope with general solution curves

Since this is Clairaut, let us use the standard Clairaut ode method to find the singular solution also and compare. We should obtain the same singular solution as above.

As we know, Clairaut ode has form

\begin{equation} y=xf\left ( p\right ) +g\left ( p\right ) \tag {1}\end{equation}

Comparing the ode we are given \(y=xp+a\sqrt {1+p^{2}}\), we see that

\begin{align*} f\left ( p\right ) & =p\\ g\left ( p\right ) & =a\sqrt {1+p^{2}}\end{align*}

The singular solution comes from solving

\begin{equation} x+g^{\prime }\left ( p\right ) =0 \tag {2}\end{equation}

For \(p\) and then substituting this back into (1).  But (2) is

\begin{align*} x+\frac {d}{dp}\left ( a\sqrt {1+p^{2}}\right ) & =0\\ x+\frac {a}{2\sqrt {1+p^{2}}}2p & =0\\ x\sqrt {1+p^{2}}+ap & =0\\ \sqrt {1+p^{2}} & =-\frac {ap}{x}\\ 1+p^{2} & =\frac {a^{2}p^{2}}{x^{2}}\\ p^{2}\left ( 1-\frac {a^{2}}{x^{2}}\right ) & =-1\\ p^{2} & =\frac {-1}{1-\frac {a^{2}}{x^{2}}}\\ & =\frac {-x^{2}}{x^{2}-a^{2}}\\ & =\frac {x^{2}}{a^{2}-x^{2}}\end{align*}

Hence

\[ p=\pm \frac {x}{\sqrt {a^{2}-x^{2}}}\]

Substituting this into (1) gives singular solutions (using first solution)

\begin{align*} y & =xp+a\sqrt {1+p^{2}}\\ & =x\frac {x}{\sqrt {a^{2}-x^{2}}}+a\sqrt {1+\left ( \frac {x}{\sqrt {a^{2}-x^{2}}}\right ) ^{2}}\\ & =\frac {x^{2}}{\sqrt {a^{2}-x^{2}}}+a\sqrt {1+\frac {x^{2}}{a^{2}-x^{2}}}\\ & =\frac {x^{2}}{\sqrt {a^{2}-x^{2}}}+\frac {a}{\sqrt {a^{2}-x^{2}}}\sqrt {a^{2}-x^{2}+x^{2}}\\ & =\frac {x^{2}}{\sqrt {a^{2}-x^{2}}}+\frac {a^{2}}{\sqrt {a^{2}-x^{2}}}\\ & =\frac {x^{2}+a^{2}}{\sqrt {a^{2}-x^{2}}}\end{align*}

Which does not satisfy the ode. Trying the second root \(p=\frac {-x}{\sqrt {a^{2}-x^{2}}}\) then (1) becomes

\begin{align*} y & =xp+a\sqrt {1+p^{2}}\\ & =x\frac {-x}{\sqrt {a^{2}-x^{2}}}+a\sqrt {1+\left ( \frac {-x}{\sqrt {a^{2}-x^{2}}}\right ) ^{2}}\\ & =\frac {-x^{2}}{\sqrt {a^{2}-x^{2}}}+a\sqrt {1+\frac {x^{2}}{a^{2}-x^{2}}}\\ & =\frac {-x^{2}}{\sqrt {a^{2}-x^{2}}}+\frac {a}{\sqrt {a^{2}-x^{2}}}\sqrt {a^{2}-x^{2}+x^{2}}\\ & =\frac {-x^{2}}{\sqrt {a^{2}-x^{2}}}+\frac {a^{2}}{\sqrt {a^{2}-x^{2}}}\\ & =\frac {a^{2}-x^{2}}{\sqrt {a^{2}-x^{2}}}\\ & =\frac {a^{2}-x^{2}}{\sqrt {a^{2}-x^{2}}}\frac {\sqrt {a^{2}-x^{2}}}{\sqrt {a^{2}-x^{2}}}\\ & =\frac {a^{2}-x^{2}}{a^{2}-x^{2}}\sqrt {a^{2}-x^{2}}\\ & =\sqrt {a^{2}-x^{2}}\end{align*}

Hence \(y^{2}=a^{2}-x^{2}\) which is the same solution found using \(p,c\) elimination.

2.26 Clairaut \(y=xp-e^{p}\)

Problem (27)

\begin{equation} y=xp-e^{p} \tag {1}\end{equation}

Using elimination method.

\begin{align} F & =y-xp+e^{p}=0\tag {2}\\ \frac {\partial F}{\partial p} & =-x+e^{p}=0 \tag {3}\end{align}

We first check that \(\frac {\partial F}{\partial y}=1\neq 0\).  Now we apply p-discriminant. Eliminating \(p\). EQ (3) gives \(e^{p}=x\) or

\[ p=\ln \left ( x\right ) \]

Substituting into the EQ (2) gives

\begin{align*} y-x\ln \left ( x\right ) +e^{\ln \left ( x\right ) } & =0\\ y-x\ln \left ( x\right ) +x & =0\\ y & =x\ln \left ( x\right ) -x \end{align*}

We now have to check if this solution satisfies the ode. We see it does. Now we have to find the general solution (also called the primitive). This comes out to be

\begin{align*} \Psi \left ( x,y,c\right ) & =0=y-cx+e^{c}\\ \frac {\partial \Psi }{\partial c} & =0=-x+e^{c}\end{align*}

Eliminating \(c\). Second equation gives \(c=\ln x\) and Substituting into the first equation above gives

\begin{align*} 0 & =y-x\ln \left ( x\right ) +e^{\ln x}\\ 0 & =y-x\ln \left ( x\right ) +x\\ y & =x\ln \left ( x\right ) -x \end{align*}

Which is the same as p-discriminant.

Since this is Clairaut, let us use the standard Clairaut ode method to find the singular solution also and compare. As we know, Clairaut ode has form

\begin{equation} y=xf\left ( p\right ) +g\left ( p\right ) \tag {4}\end{equation}

Comparing the ode we are given \(y=xp-e^{p}\), we see that

\begin{align*} f\left ( p\right ) & =p\\ g\left ( p\right ) & =-e^{p}\end{align*}

The singular solution comes from solving

\begin{equation} x+g^{\prime }\left ( p\right ) =0 \tag {5}\end{equation}

For \(p\) and then substituting this back into (4).  But (5) is

\begin{align*} x+\frac {d}{dp}\left ( -e^{p}\right ) & =0\\ x-e^{p} & =0 \end{align*}

Hence

\[ p=\ln \left ( x\right ) \]

Substituting this into (4) gives singular solution

\begin{align*} y & =xp-e^{p}\\ & =x\ln x-e^{\ln x}\\ & =x\ln x-x \end{align*}

Which is same as found using elimination methods.

Figure 20: Singular solution envelope with general solution curves

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

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

2.29 Clairaut \(y=px-\arcsin \left ( p\right ) \)

Problem (29)

\[ y=px-\arcsin \left ( p\right ) \]

Since this is not quadratic in \(p\), we have to use elimination. Writing the ode as

\begin{align} F & =y-px+\arcsin \left ( p\right ) \tag {1}\\ & =0\nonumber \end{align}

Then

\begin{align*} \frac {\partial F}{\partial p} & =-x+\frac {1}{\sqrt {1-p^{2}}}\\ & =0 \end{align*}

The above gives

\begin{align*} 1 & =x\sqrt {1-p^{2}}\\ \frac {1}{x^{2}} & =1-p^{2}\\ p^{2} & =1-\frac {1}{x^{2}}\\ & =\frac {x^{2}-1}{x^{2}}\\ p & =\pm \frac {1}{x}\sqrt {x^{2}-1}\end{align*}

Substituting first root in (1) gives

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

Hence

\[ y=\sqrt {x^{2}-1}-\arcsin \left ( \frac {1}{x}\sqrt {x^{2}-1}\right ) \hspace {0.5in}x>0 \]

For the other root

\begin{align*} 0 & =y-\left ( -\frac {1}{x}\sqrt {x^{2}-1}\right ) x+\arcsin \left ( -\frac {1}{x}\sqrt {x^{2}-1}\right ) \\ y & =-\sqrt {x^{2}-1}+\arcsin \left ( \frac {1}{x}\sqrt {x^{2}-1}\right ) \hspace {0.5in}x>0 \end{align*}

These are the envelops. The general solution can be found to

\[ y=c_{1}x-\arcsin \left ( c_{1}\right ) \]

The following plot shows solution curves (in blue) for different values of \(c_{1}\) with the singular solutions in dashed red style.

Figure 23: Singular solution envelope with general solution curves

2.30 \(4xp^{2}=\left ( 3x-b\right ) ^{2}\)

Problem (30)

\[ 4xp^{2}-\left ( 3x-b\right ) ^{2}=0 \]

The ode is quadratic in \(p\). Using the quadratic discriminant \(b^{2}-4ac=0\) gives

\begin{align*} 0-4\left ( 4x\right ) \left ( -\left ( 3x-b\right ) ^{2}\right ) & =0\\ x\left ( 3x-b\right ) ^{2} & =0 \end{align*}

Hence \(x=0\) (y-axis) is solution. But this does not verify the ode (book is wrong, it said it verifies ode). So can not be envelope. The other solution is \(3x=b\) or \(x=\frac {b}{3}\). Recalling that the \(p\) discriminant has form \(ET^{2}C\) then this y line is Tac locus since it shows twice and \(x=0\) is \(C\) or cusp locus.

The solution to this ode can be found to be

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

Since this is quadratic, then using C-discriminant \(b^{2}-4ac=0\) gives

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

Hence \(x=0\) which is same as in \(p\)-discriminant and \(x=b\) which, recalling that C-discriminant is \(EN^{2}C^{3}\) then \(x=b\) is \(N\) or nodal locus.

The following plot shows solution curves (in blue) for different values of \(c_{1}\) with the above solutions. In this plot \(b=6\) we used.

Figure 24: Singular solution envelope with general solution curves

2.31 \(4p^{2}x\left ( x-A\right ) \left ( x-B\right ) =\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2}\)

Problem (31)

\[ 4p^{2}x\left ( x-A\right ) \left ( x-B\right ) -\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2}=0 \]

The ode is quadratic in \(p\). Using the quadratic discriminant \(b^{2}-4ac=0\) gives

\begin{align*} 0-4\left ( 4x\left ( x-A\right ) \left ( x-B\right ) \right ) \left ( -\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2}\right ) & =0\\ -16x\left ( x-A\right ) \left ( x-B\right ) \left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2} & =0\\ x\left ( x-A\right ) \left ( x-B\right ) \left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2} & =0 \end{align*}

Hence lines are \(x=0,x=A,x=B\). The ode is not satisfied by these (again, book says they are. Book must be wrong). So looking at \(ET^{2}C\), then these are Cusp locus. Looking at the second factor \(\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2}=0\), this gives \(\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) =0\) (factor is 2), or

\[ 3x^{2}-2x\left ( A+B\right ) +AB=0 \]

The solution is

\begin{equation} x=\frac {2\left ( A+B\right ) \pm \sqrt {4\left ( A+B\right ) ^{2}-12AB}}{6}\tag {1}\end{equation}

Since it shows as factor of two, and comparing to \(ET^{2}C\,\,\) shows the above is Tac locus.  The solution to the ode can be found to be

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

The following plot is made using \(A=3,B=6\). It shows \(x=0,x=A,x=B\) as lines parallel to the y axis. These are Cusp locus (book says these are Envelope, but these do not satisfy the ode. So something is wrong in book, or I am overlooking something). There are two additional vertical lines that comes from Eq (1). these are Tac locus. One of these two lines is real contact line. The other the book calls imaginary point of contact. But both are drawn below.

Figure 25: Singular solution envelope with general solution curves