Clairaut's Differential Equation

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Theorem

Clairaut's differential equation is a first order ordinary differential equation which can be put into the form:

$y = x y' + \map f {y'}$


Its general solution is:

$y = C x + \map f C$

where $C$ is a constant.


Proof

We have:

$y = x y' + \map f {y'}$

Differentiating the equation with respect to $x$ we have:

\(\displaystyle y'\) \(=\) \(\displaystyle y' + x y'' + y'' \map {f'} {y'}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(=\) \(\displaystyle \map {y''} {x + \map {f'} {y'} }\)


Proof for General Solution

The first solution is:

\(\displaystyle y''\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y'\) \(=\) \(\displaystyle C_1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle C_1 x + C_2\)


By substituting into the original equation, we obtain:

\(\displaystyle C_1 x + C_2\) \(=\) \(\displaystyle x C_1 + \map f {C_1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle C_2\) \(=\) \(\displaystyle \map f {C_1}\)


Hence the result:

$y = C_1 x + \map f {C_1}$

$\blacksquare$


Source of Name

This entry was named for Alexis Claude Clairaut.


Sources