Math 2.0

Ordinary differential equations Elementary methods of solution Example

Clairaut's equation

Builds on Singular solutionGeneral solution and particular solution

Example

Let ff be a twice continuously differentiable function on an open interval JJ. Clairaut's equation is the implicit first-order equation y=x y′+f(y′),y = x\, y' + f(y'), for an unknown yy with y′∈Jy' \in J. For each constant C∈JC \in J, the line y=Cx+f(C)y = C x + f(C) is a solution. These lines form the general solution. If f′′≠0f'' \ne 0 on JJ, the curve with the parametric form x=−f′(p)x = -f'(p), y=f(p)−pf′(p)y = f(p) - p f'(p) for p∈Jp \in J is also a solution. It is the envelope of the lines, and it is a singular solution.

Remarks

Differentiation of the equation gives y′′(x+f′(y′))=0y''\left(x + f'(y')\right) = 0. The case y′′=0y'' = 0 gives the lines, and the case x+f′(y′)=0x + f'(y') = 0 gives the envelope. Example: for f(p)=p2f(p) = p^2, the lines are y=Cx+C2y = C x + C^2, and the singular solution is the parabola y=−x2/4y = -x^2/4. Each line of the general solution is tangent to the singular solution at one point.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Clairaut%27s_equation, Solution
About this page
Tag0T6D4PE51C
URIm2://0T6D4PE51C
Kindexample
WritingW2 usable
MSC34A09
Instance ofExplicit and implicit differential equations
UsesSingular solution, General solution and particular solution
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