# MATHEMATICA TUTORIAL, Part 1.2: Clairaut equations (2023)

Clairaut's equation (or the Clairaut equation) is a differential equation of the form

$y(x) = x \,\frac{{\text d} y}{{\text d}x} + f \left( \frac{{\text d} y}{{\text d}x} \right) ,$

where f is continuously differentiable function. It is named after the French mathematician Alexis Clairaut (1713--1765), who introduced it in 1734.

To solve Clairaut's equation, one differentiates with respect to x, yielding

$\left[ x + f' \left( \frac{{\text d} y}{{\text d}x} \right) \right] \frac{{\text d}^2 y}{{\text d}x^2} = 0 .$

Hence, either $$\displaystyle \frac{{\text d}^2 y}{{\text d}x^2} = 0$$ or $$\displaystyle x + f' \left( \frac{{\text d} y}{{\text d}x} \right) =0.$$ In the former case, the first derivative must be a constant, $$c = {\text d}y/{\text d}x .$$ Substituting this into the Clairaut's equation, one obtains the family of straight line functions given by

$y(x) = C\,x + f\left( C \right) ,$

the so-called general solution of Clairaut's equation.

The latter case,

$x + f' \left( \frac{{\text d} y}{{\text d}x} \right) = 0$

defines only one solution y(x), the so-called singular solution, whose graph is the envelope of the graphs of the general solutions. The singular solution is usually represented using parametric notation, as (x(p), y(p)), where p = dy/dx.

Example 1: With p = dy/dx, Clairaut's equation

$y = x\,y' + \frac{y'}{4 + (y')^2} \qquad\mbox{or} \qquad y = x\,p + \frac{p}{4 + p^2} \tag{1.1}$

has the general solution

$y = x\,c + \frac{c}{4 + c^2} , \tag{1.2}$

(Video) Mathematica for Difference Equations and Differential Equations (Mathematica Basics at the Start)

where c is an arbitrary constant. We plot a family of solutions from the general solution: fun[f_, x_, a_List, x0_, x1_, c_, yp_, al_, off_, opts : OptionsPattern[]] := With[{p = c (x0 + x1)}, Plot[Evaluate[f[x, #] & /@ a], {x, x0, x1}, Epilog -> ({Text[Framed[g[x, #]], {p, yp f[p, #]}, {0, off}], Arrow[{{p, yp f[p, #]}, {al p, f[al p, #]}}]} & /@ a), Evaluate@FilterRules[{opts}, Options[Plot]]]] g[x_, a_] := a x + a/(a^2 + 4) fun[g, x, {-1, 0, 1, 2, 3, 4, 5}, -2, 2, s, y, arrow, offset, ImageSize -> 500] A family of solutions. Mathematica code

Clairaut's equations usually have singular solutions. To determine them, we rewrite the general solution in another form:

$y \left( 4 + c^2 \right) \left( y - cx \right) = c . -((4 + 3 c^2) x) + 2 c y = 1$

Upon differentiating with respect to c gives

$-((4 + 3 c^2) x) + 2 c y = 1 .$

D[(y - c*x)*(4 + c^2), c]

-((4 + 3 c^2) x) + 2 c y

This quadratic equation can be solved:

$c = \frac{1}{3x} \left[ y \pm \sqrt{y^2 -3x - 12 x^2} \right[ . \tag{1.3}$

Solve[-((4 + 3 c^2) x) + 2 c y == 1, c]

{{c -> (y - Sqrt[-3 x - 12 x^2 + y^2])/(3 x)}, {c -> ( y + Sqrt[-3 x - 12 x^2 + y^2])/(3 x)}}

Each braanch of quadratic solution is substituting into the general solution (1.2). Mathematica helps to visialize singular solutions *the vertical line x = 0 must excluded)

c = (y - Sqrt[-3 x - 12 x^2 + y^2])/(3 x);
Simplify[c*x + c/(4 + c^2) - y]

1/3 (-2 y - Sqrt[-3 x - 12 x^2 + y^2] + ( y - Sqrt[-3 x - 12 x^2 + y^2])/( x (4 + (y - Sqrt[-3 x - 12 x^2 + y^2])^2/(9 x^2))))

ContourPlot[ -2 y - Sqrt[-3 x - 12 x^2 + y^2] + ( y - Sqrt[-3 x - 12 x^2 + y^2])/( x (4 + (y - Sqrt[-3 x - 12 x^2 + y^2])^2/(9 x^2))), {x, -4, 4}, {y, 0, 20}]

and for another branch:

c = (y + Sqrt[-3 x - 12 x^2 + y^2])/(3 x);
Simplify[c*x + c/(4 + c^2)-y]

1/3 (-2 y + Sqrt[-3 x - 12 x^2 + y^2] + ( y + Sqrt[-3 x - 12 x^2 + y^2])/( x (4 + (y + Sqrt[-3 x - 12 x^2 + y^2])^2/(9 x^2))))

ContourPlot[-2 y + Sqrt[-3 x - 12 x^2 + y^2] + ( y + Sqrt[-3 x - 12 x^2 + y^2])/( x (4 + (y + Sqrt[-3 x - 12 x^2 + y^2])^2/(9 x^2))), {x, -4, 4}, {y, 0, 20}]

(Video) 1. Basics for Mathematica | GE3 Differential Equations Practical | Mathematica | DU Generic 3 Lab  Implicit plot for negative branch. Implicit plot for positive branch.

Example 2: We consider another Clairaut's equation

$y = x\,y' + \left( y' \right)^2 .$

Differentiating with respect to x, we find

$y' = x\,y'' + y' + 2\,y'\,y'' ,$

which rearranges to

$0 = x\,y'' + 2\,y'\,y'' .$

If we divide by y'' ≠ 0, then

$2\,y' = -x \qquad \Longrightarrow \qquad y = - \frac{x^2}{4} + c,$

for some constant c of integration. As differentiating initially may have added solutions, then we now check whether these solutions are valid:

$\mbox{RHS} = x \left( - \frac{x}{2} \right) + \left( - \frac{x}{2} \right)^2 = - \frac{x^2}{4} + c = \mbox{LHS if} \quad c=0,$

which we see to be true only when c = 0.

If we return to the case when y'' = 0, then we have a linear solution y = Ax + B, for some constants A and B. If we substitute this into Clairaut’s equation, it gives

$A\,x + B = x\,A + A^2 \qquad \Longrightarrow \qquad B = A^2 ,$

and we obtain a family of solutions (called the general solution):

$y = x\,A + A^2 .$

It has the envelope (singular solution) $$\displaystyle y = - \frac{x^2}{4} .$$

If we substitute $$\displaystyle y = z - \frac{x^2}{4}$$ into Clairaut's equation, we obtain a differential equation

(Video) The Riccati Equation Lesson

$z- \frac{x^2}{4} = x \left( z' - \frac{x^2}{4} \right) + \left( z' - \frac{x^2}{4} \right)^2 \qquad \Longrightarrow \qquad z = \left( z' \right)^2 .$

Solving the differential equation for z, we get its general solution depending on two arbitrary constants:

$z(x) = \begin{cases} \frac{1}{4} \left( x- a \right)^2 , & \ x \leqslant a, \\ 0 , & \ a \leqslant x \leqslant b , \\ \frac{1}{4} \left( x- b \right)^2 , & \ x \geqslant b, \end{cases}$

where 𝑎 and b are constants satisfying -∞ ≤ 𝑎 ≤ b ≤ ∞. ■

Generalized Clairaut's equation

The differential equation

$f \left( x\,\frac{{\text d} y}{{\text d}x}-y \right) = g \left( \frac{{\text d} y}{{\text d}x} \right) ,$

where f and g are some given smooth functions, can be solved in exactly the same manner as Clairaut's equation to obtain the general solution

$f \left( x\,C-y \right) = g \left( C \right) ,$

where C is an arbitrary constant. The generalized Clairaut's equation may also have a singular solution. If it does, it can be obtained bydifferentiating the above equation with respect to x to obtain\begin{equation} y''\left[ f'(xy'-y)x-g'(y') \right] =0.\label{clair:c}\end{equation}If the first term in the above equation is zero, then the generalized Clairaut's equation is recovered. If the second term in the equation is zero, then two equations can be solved together to eliminate y'. The resulting equation for y=y(x) will have no arbitrary constants and so will be a singular solution.

Example 3: Consider the generalized Clairaut's equation

$(xy'-y)^2-(y')^2-1=0.$

Since this is a generalized Clairaut's equation with f(p) = p² and g(p) = p² -1, its general solution becomes

$(x\,C-y)^2 = C^2 +1 \qquad \mbox{or}\qquad y=C\,x\pm\sqrt{C^2-1} ,$

where C > 1 is an arbitrary constant.

To find the singular solution, we differentiate the given equation with respect to x to obtain \begin{equation} y''[2(xy'-2)x-2y']=0.\end{equation}If the second term is set equal to zero, then we find\begin{equation} y'=\frac{x y}{x^2-1}.\label{clair:g}\end{equation} We determine thesingular solution to be%\begin{equation} x^2+y^2=1.\label{clair:h}\end{equation}Note that this equation is not derivable from the general solution for any choice of C. ■

The Lagrange differential equation

(Video) Differential Equations Exam 1 Review Problems and Solutions

A differential equation of type

$y = x\, f(y' ) + g(y' ) ,$

where f and g are given functions differentiable on a certain interval, is called the Lagrange equation. It is convenient to denote by p = dy/dx, the derivative of the unknown function. Then differentiating the equation $$\displaystyle y = x\,f(p) + g(p), we obtain ${\text d}y = x\,{\text d}f(p) + f(p)\,{\text d}x + {\text d}g(p) .$ Using the relation \( \displaystyle p\,{\text d}x = {\text d}y,$$ we get

$p\,{\text d}x = x\,{\text d}f(p) + f(p)\,{\text d}x + {\text d}g(p) .$

Now move fdx to the left-hand side

$\left[ p - f(p) \right] {\text d}x = x\,{\text d}f(p) + {\text d}g(p)$

and divide both sides by dp:

$\left[ p - f(p) \right] \frac{{\text d}x}{{\text d}p} = x\,\frac{{\text d}f}{{\text d}p} + \frac{{\text d}g}{{\text d}p} .$

Assuming pf(p), we obtain

$\frac{{\text d}x}{{\text d}p} = \frac{1}{p - f(p)} \left[ x\,\frac{{\text d}f}{{\text d}p} + \frac{{\text d}g}{{\text d}p} \right] .$

This is a first-order differential equation, and its solution x(p) should be achievabledepending on the details of the functions f(p) and g(p). Even its singular solution may be found be setting the singular value p = f(p) in the original differential equation.

Example 4: Consider the differential equation

$y = 3p\,x + 7\,\ln p, \qquad p = {\text d}y/{\text d}x .$

With f(p) = 3p and g(p) = 5lnp, we obtain

$\frac{{\text d}x}{{\text d}p} = -\frac{1}{2p} \left[ 3x + \frac{7}{p} \right] .$

This is a first order differential equation. Its solution becomes

$x(p) = -\frac{7}{p} + \frac{C}{p^{1/2}} ,$

where C is an arbitrary constant. Inserting this expression into the given Lagrange equation, we obtain

$x(p) = 7 \left[ \ln p -3 \right] + 3\,\frac{C}{p^{1/2}} .$

The last two equations provide the parametric representation of the general solution y = y(x). A singular solution is y(x) = 0. ■

(Video) Calculus Students Be Like… #shorts #math #mathematics #enginerdmath

## Videos

1. Riccati Differential Equations | Concept with Examples
(MindOverBody)
2. Differential Equations: Lecture 2.5 Solutions by Substitutions
(The Math Sorcerer)
3. Introduction to Ordinary Differential Equations
(Houston Math Prep)
4. 1.1 Differential Equations and Mathematical Models
(Professor Cornell)
5. Applications of first order equations -- differential equations 8
(MathMajor )
6. how to solve this quadratic equation fast…
(TabletClass Math)
Top Articles
Latest Posts
Article information

Author: Sen. Ignacio Ratke

Last Updated: 09/26/2023

Views: 6407

Rating: 4.6 / 5 (76 voted)

Author information

Name: Sen. Ignacio Ratke

Birthday: 1999-05-27

Address: Apt. 171 8116 Bailey Via, Roberthaven, GA 58289

Phone: +2585395768220