Difference between revisions of "Exact differential equation"

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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 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 <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.
 
  
To solve the expression for <math>y/,</math>, use the quadratic formula.
+
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.
 +
 
 +
 
 +
To solve the expression for <math>y</math>, use the quadratic formula.
  
  
 
[[Category:Calculus]]
 
[[Category:Calculus]]
 
[[Category:Differential Equations]]
 
[[Category:Differential Equations]]

Revision as of 13:55, 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 (equation 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.