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.