Clairaut' equation

In mathematics, a differential equation of the form

$$y = xy'+f(y')$$

is called the Clairaut's equation. To solve such a problem, we differentiate with respect to $x$, yielding

$$y' = y' + xy'' + f'(y')y''$$

Then $(x+f'(y'))y'' = 0$. Thus, either $y'' = 0$ or $x+f'(y') = 0$. In the former case, $y' = c$ for some constant. Substituting this into the Clairaut's equation, we have the family of straight line functions given by $y(x) = cx+f(c)$. The latter case $x+f'(y') = 0$. We let $\left\{\begin{array}{l} x = -f'(t)\\ y=-tf'(t)+f(t) \end{array}\right.$, where $t$ is a parameter. Then

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-tf''(t)}{-f''(t)} = -t$$

Thus, $y = -tf'(t) + f(t) = \frac{dy}{dx}x + f(\frac{dy}{dx}) = xp + f(p)$. Then this is a solution to the Clairaut's equation. If $x = -f'(t)$ has a solution $t = h(x)$, then $y = xh(x) + f(h(x))$ and a singular solution to $y = xp + g(p)$.

A differntial equation of the form

$$y = xf(y') + g(y'),$$

where $f$ and $g$ are function of $y'$, is called D'Alembert equation. To solve the D'Alembert's equation, we differentiate with respect to $x$.

$$y' = f'(y') + xf'(y')y'' + g'(y')y''$$

Rewriting

$$y' - f'(y') = (xf'(y') + g'(y'))y''$$

Now write this equation as

$$\frac{dx}{dy'} = \frac{f'(y')}{y'-f'(y')} + \frac{g'(y')}{y' -f'(y')}$$

Then this is a linear differential equation with dependent variable $x$ and independent variable in $y$.

We use the following symbol for simplicity.

$$p = y' = \frac{dy}{dx}$$

Then the Clairaut's equation is expressed in the form

$$y = xp + f(p)$$

and the D'Alembert's equation is expressed in the form

$$y = xf(p) + g(p)$$

Example 1..21   Solve the following differential equation.

$$p^2 + (2x - y^2)p - 2xy^2 = 0$$

SOLUTION We solve for $p$. By factorization, we obtain $(p + 2x)(p - y^2) = 0$. This equation is satisfied by either $p + 2x = 0$ or $p - y^2 = 0$. Note that the general solution of $p + 2x = 0$ is $y + x^2 + C = 0$. Also, $p - y^2 = 0$ is separable differential equation. Thus we obtaine the general solution $xy+Cy + 1 = 0$. Form this the general solution is

$$(y + x^2 + C)(xy + Cy + 1) = 0$$

Example 1..22   Solve the defferential equation $xp^2 - 2yp = x$.

SOLUTION Solve for $p$. Then by the quadratic formula, we have

$$p = \frac{y \pm \sqrt{y^2 +x^2}}{x}$$

Since $p = \frac{dy}{dx}$, this is homogeneous. Let $y = vx$. Then $v + x\frac{dv}{dx} = v \pm \sqrt{1 + v^2}$. Thus

$$\int\frac{dv}{\sqrt{1 + v^2}} = \pm \int \frac{dx}{x}$$

$$\log\vert v + \sqrt{1 + v^2}\vert = \pm \log\vert x\vert + C$$

$$\log\vert\frac{y}{x} + \sqrt{1 + (\frac{y}{x})^2}\vert \mp \log\vert x\vert = C$$

Therefore,

$$y \mp \sqrt{x^2 + y^2} = C$$

From this, we have the general solution

$$(y - \sqrt{x^2 + y^2} - C)(y + \sqrt{x^2 + y^2} - C) = 0$$

or

$$y = \frac{C^2 - x^2}{2C} = \frac{C}{2} - \frac{x^2}{2C}$$

Example 1..23   Solve the differntial equation $y = xp + p^{-1}$, where $y(1) = 2$

SOLUTION This is a Clairaut's equation. We first differentiate with respect to $x$.

$$p = p + x\frac{dp}{dx} - p^{-2}\frac{dp}{dx} = p+(x-p^{-2})\frac{dp}{dx}$$

Simplifying to get

$$(x-p^{-2})\frac{dp}{dx} = 0$$

From this, we have $\frac{dp}{dx} = 0$ or $x = p^{-2}$. For the first case, we have $p = c$ a some constant. Thus

$$y = xc + \frac{1}{c}$$

Since $y(1) = 1$, we have $2 = 1 + \frac{1}{c}$. Thus

$$y = x + 1$$

For the second case, put $x = p^{-2}$ into the original equation. Then

$$y = p^{-1} + p{-1} = 2p^{-1}$$

Rewrite this,

$$p = \frac{2}{y}$$

Thus

$$ydy = 2dx$$

$$\frac{y^2}{2} = 2x + c$$

Thus,

$$y^2 = 4x + c$$

Since $y(1) = 2$, we have $4 = 4+c$. Therefore, the singular solution is

$$y^2 = 4x$$

Example 1..24   Solve the differential equation $yp^2 + 2xp = y$.

SOLUTION We first solve for $x$. Then $x = y(\frac{1}{2p} - \frac{p}{2})$. Now differentiate with respect to $y$, Then

$$\frac{1}{p} = p(\frac{1}{2p} - \frac{p}{2}) - y(\frac{1}{2p^2} + \frac{1}{2})\frac{dp}{dy}$$

Thus,

$$\frac{dp}{dy} = -\frac{\frac{1+p^2}{2p}}{y(\frac{1+p^2}{2p^2})} = -\frac{p}{y}$$

Simplifying to get $\frac{dp}{p} = -\frac{dy}{y}$. Then $\log\vert p\vert = -\log\vert y\vert + C$, $py = C$. From this, we have $yp^2 + 2xp = y$. Delete $p$ and we have the general solution

$$C^2 + 2xC = y^2$$

Example 1..25   Solve the differential equation $x = 2p + \log\vert p\vert$.

SOLUTION We differentiate with respect to $y$.

$$\frac{1}{p} = 2\frac{dp}{dy} + \frac{1}{p}\frac{dp}{dy} = (2 + \frac{1}{p})\frac{dp}{dy}$$

Then

$$(2p+1)dp = dy$$

Integrate with respect $y$

$$p^2 + p + C = y$$

Thus the general solution is given $p$ as a parameter,

$$x = 2p + \log\vert p\vert,\ y = p^2 + p + C$$