# Difference between revisions of "Exact differential equation"

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Suppose you are given an equation of the form: | Suppose you are given an equation of the form: | ||

− | :<math>M(t,y) + N(t,y)y' = 0\,</math> or <math>M(t,y) dt + N(t,y) dy = 0\,</math> | + | :<math>M(t,y) + N(t,y)y' = 0\,</math> or <math>M(t,y) dt + N(t,y) dy = 0\,</math> (1) |

Before we begin solving it, we must first check that the equation is exact. This means that: | Before we begin solving it, we must first check that the equation is exact. This means that: | ||

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<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math> | <math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math> | ||

− | To find the solution of this equation, we assume that the solution is φ = constant. We | + | To find the solution of this equation, we assume that the solution is φ = constant. We assume that <math>\frac{\partial \phi}{\partial t} = M</math> and <math>\frac{\partial \phi}{\partial y} = N</math>. (If we substitute M and N back into (1), it yields <math>(\frac{\partial \phi}{\partial t}) dt + (\frac{\partial \phi}{\partial y}) dy = 0</math>, which makes sense.) |

− | To find | + | To find <math>y/,</math>, manipulate the substitutions of M and N to get <math>M \partial t = \partial \phi</math> and <math>N \partial y = \partial \phi</math>. Integrate both sides. To get the main function φ write the sum of each term found in each equation. For terms that appear in both equations, only write them once. |

− | + | To solve the expression for <math>y/,</math>, use the quadratic formula. | |

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[[Category:Calculus]] | [[Category:Calculus]] | ||

[[Category:Differential Equations]] | [[Category:Differential Equations]] |

## Revision as of 13:53, 2 August 2010

An **exact differential equation** is a differential equation that can be solved in the following manner.

Suppose you are given an equation of the form:

- or (1)

Before we begin solving it, we must first check that the equation is exact. This means that:

To find the solution of this equation, we assume that the solution is φ = constant. We assume that and . (If we substitute M and N back into (1), it yields , which makes sense.)

To find , manipulate the substitutions of M and N to get and . Integrate both sides. To get the main function φ write the sum of each term found in each equation. For terms that appear in both equations, only write them once.

To solve the expression for , use the quadratic formula.