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>
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:<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 &phi; = constant. We can re-write a different form of this equation by substituting <math>\frac{\partial \phi}{\partial t} = M</math> and <math>\frac{\partial \phi}{\partial y} = N</math>. This yields <math>(\frac{\partial \phi}{\partial t}) dt + (\frac{\partial \phi}{\partial y}) dy = 0</math>.
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To find the solution of this equation, we assume that the solution is &phi; = 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 &phi;, we integrate M with respect to t and N with respect to y. This will give us two different equations. To find &phi; , we
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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 &phi; write the sum of each term found in each equation. For terms that appear in both equations, only write them once.
  
Go through the example to find &phi; by integrating, then check that
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To solve the expression for <math>y/,</math>, use the quadratic formula.
:<math>\frac{\partial \phi}{\partial t} = M</math>
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and
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:<math>\frac{\partial \phi}{\partial y} = N</math>
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and that any function &phi; = some constant, when turned into the corresponding dy/dt, satisfies the original equation.  Be sure to emphasize that one must check first that
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:<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>
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To find <math>y</math>, first set <math>M = \frac{\partial \phi}{\partial t}</math> and <math>N = \frac{\partial \phi}{\partial y}</math>. Then manipulate to get <math>M \partial t = \partial \phi</math> and <math>N \partial y = \partial \phi</math>. Integrate both sides, compare the results for <math>\phi</math>, and combine the terms into one equation (for terms that show up in both expressions, only write once in the combined expression.) To solve the expression for <math>y</math>, plug into 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.