Clairaut’s equation

The ordinary differential equation

$y=x{\displaystyle \frac{dy}{dx}}+\psi \left({\displaystyle \frac{dy}{dx}}\right),$(1)

where $\psi$ is a given differentiable real function, is called Clairaut’s equation.

For solving the equation we use an auxiliary variable  $p=:\frac{dy}{dx}$  and write (1) as

$$y=px+\psi (p).$$

Differentiating this equation gives

$$p=x\frac{dp}{dx}+p+{\psi }^{\prime }(p)\frac{dp}{dx},$$

or

$$[x+{\psi }^{\prime }(p)]\frac{dp}{dx}=\mathrm{\hspace{0.33em}0}.$$

The zero rule of product now yields the alternatives

${\displaystyle \frac{dp}{dx}}=\mathrm{\hspace{0.33em}0}$(2)

and

$x+{\psi }^{\prime }(p)=\mathrm{\hspace{0.33em}0}.$(3)

Integrating (2) we get  $p=C$ (), and substituting this in (1) gives the general solution

$y=Cx+\psi (C)$(4)

which presents a family of straight lines.

If (3) allows to solve $p$ in of $x$,  $p=p(x)$,  we can write (1) as

$y=xp(x)+\psi (p(x)),$(5)

which is easy to see satisfying (1).  The solution (5) may not be gotten from (4) using any value of $C$.  It is a singular solution which may be obtained by eliminating the parameter $p$ from the equations

$$y=px+\psi (p),x+{\psi }^{\prime }(p)=\mathrm{\hspace{0.33em}0}.$$

Thus the singular solution presents the envelope of the family (4).

Example.  The Clairaut’s equation

$$y=x\frac{dy}{dx}+\frac{a\frac{dy}{dx}}{\sqrt{1+{(\frac{dy}{dx})}^2}}$$

has the general solution

$$y=Cx+\frac{Ca}{\sqrt{1+C^2}}$$

and the singular solution

$$\{\begin{array}{cc}x=-\frac{a}{{(1+p^2)}^{3/2}},\hfill & \\ y=-\frac{a{p}^3}{{(1+p^2)}^{3/2}}\hfill & \end{array}$$

in a parametric form.  Eliminating the parametre $p$ yields the form

$$\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2},$$

which can be recognized to be the equation of an astroid.  The envelope (see “determining envelope (http://planetmath.org/DeterminingEnvelope)”) of the lines is only the left half of this curve ($x\leqq 0$).  The usual parametric of the astroid is  $x=a{\cos }^3\phi$,  $y=a{\sin }^3\phi$ ().