# 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:

- $y = C + \map f C$

where $C$ is a constant.

## Proof

We have:

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

Differentiating the equation 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

- 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**Clairaut's equation**