2.25 Example 25 Clairaut \(y=xp+a\sqrt {1+p^{2}}\)

\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.