Clairaut y = px − arcsin (p)

Problem (29)

y = p x - \arcsin (p)

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

\begin{aligned} (1) & F & = y - p x + \arcsin (p) \\ = 0 \end{aligned}

Then

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

The above gives

\begin{aligned} 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{aligned}

Substituting ﬁrst root in (1) gives

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

Hence

y = \sqrt{x^{2} - 1} - \arcsin (\frac{1}{x} \sqrt{x^{2} - 1}) x > 0

For the other root

\begin{aligned} 0 & = y - (- \frac{1}{x} \sqrt{x^{2} - 1}) x + \arcsin (- \frac{1}{x} \sqrt{x^{2} - 1}) \\ y & = - \sqrt{x^{2} - 1} + \arcsin (\frac{1}{x} \sqrt{x^{2} - 1}) x > 0 \end{aligned}

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

y = c_{1} x - \arcsin (c_{1})

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