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:
\(\ds y'\) | \(=\) | \(\ds y' + x y'' + y'' \map {f'} {y'}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds \map {y''} {x + \map {f'} {y'} }\) |
Proof for General Solution
The first solution is:
\(\ds y''\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y'\) | \(=\) | \(\ds C_1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds C_1 x + C_2\) |
By substituting into the original equation, we obtain:
\(\ds C_1 x + C_2\) | \(=\) | \(\ds x C_1 + \map f {C_1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds C_2\) | \(=\) | \(\ds \map f {C_1}\) |
Hence the result:
- $y = C_1 x + \map f {C_1}$
$\blacksquare$
Also known as
Clairaut's Differential Equation is also known as Clairaut's Equation.
Source of Name
This entry was named for Alexis Claude Clairaut.
Sources
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Clairaut's equation
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Clairaut's equation
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Clairaut's equation