Difference between revisions of "Exact differential equation"

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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>
 
:<math>M(t,y) + N(t,y)y' = 0\,</math>  or  <math>M(t,y) dt + N(t,y) dy = 0\,</math>
  
To find the solution of this equation, we assume that the solution is &phi; = constant. This means that:
+
Before we begin solving it, we must first check that the equation is exact. This means that:
<math>\frac{\partial \phi}{\partial t} = M</math> and <math>\frac{\partial \phi}{\partial y} = N</math>, since <math>(\frac{\partial \phi}{\partial t}) dt + (\frac{\partial \phi}{\partial y}) dy = 0</math>
+
  
&phi; is found by integrating M and N:
+
<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>
:<math>\phi(t, y) = \int_0^t M(s, 0) ds + \int_0^y N(t, s) ds</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>, 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
  
 
Go through the example to find &phi; by integrating, then check that
 
Go through the example to find &phi; by integrating, then check that
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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
 
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
 
:<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>
 
:<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>
(That's the condition for "exactness" of the differential form M dt + N dy.)
 
 
 
where <math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>.
 
  
 
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.
 
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.

Revision as of 13:46, 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

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 can re-write a different form of this equation by substituting and . This yields , which makes sense.

to find φ, we integrate M with respect to t and N with respect to y. This will give us two different equations. To find φ , we

Go through the example to find φ by integrating, then check that

and

and that any function φ = some constant, when turned into the corresponding dy/dt, satisfies the original equation. Be sure to emphasize that one must check first that

To find , first set and . Then manipulate to get and . Integrate both sides, compare the results for , 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 , plug into the quadratic formula.