Difference between revisions of "Exact differential equation"

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An '''exact differential equation''' is a differential equation that can be solved in the following manner.
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An '''exact differential equation''' is a first order [[differential equation]] that has the form:
  
Suppose you are given an equation of the form:
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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>
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and is also exact.<ref name=MathematicalMethods>K.F. Riley, M.P. Hobson, S.J. Bence, ''Mathematical Methods for Physics and Engineering'', Cambridge University Press, 3<sup>rd</sup> ed., 2006</ref> This means that:
  
or
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:<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>
  
:<math>M(t,y) dt + N(t,y) dy = 0\,</math>
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If a first order differential equation is exact, then a [[conservative field]] exists and a [[potential energy|scalar potential]] can be defined.<ref name=mathworld>[http://mathworld.wolfram.com/ExactFirst-OrderOrdinaryDifferentialEquation.html Exact Equations] from mathworld.wolfram.com</ref>
  
To find the solution of this equation, we assume that the solution is &phi; = constant. This means that:
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==Solution==
<math>\frac{\partial \phi}{\partial t} = M</math> and <math>\frac{\partial \phi}{\partial y} = N</math>
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To find the solution of this equation, we consider the fact that the equation has the form of an exact differential of a [[function]] &phi;(y,t). The total differential of &phi;(y,t) is then:
  
&phi; is found by integrating M and N:
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:<math>
:<math>\phi(t, y) = \int_0^t M(s, 0) ds + \int_0^y N(t, s) ds</math>
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d\phi (y,t) = \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial t} dt
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</math>
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Comparing to our original differential equation, <math>\frac{\partial \phi}{\partial t} = M</math> and <math>\frac{\partial \phi}{\partial y} = N</math> and also that this implies that d&phi;(y,t)=0 and &phi;(y,t)=constant. To solve the equation, we [[Integration|integrate]] both sides of the total differential with respect to one of the variables, say t:
  
Go through the example to find &phi; by integrating, then check that
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:<math>
:<math>\frac{\partial \phi}{\partial t} = M</math>
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\phi(y,t) = \int M(t,y) \, dt + F(y)
and
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</math>
:<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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where F(y) is a "constant of integration". (In reality, F(y) is a [[function]], but when we differentiate it with respect t, we treat y as constant and so its [[derivative]] becomes zero.) In order to solve for F(y), we differentiate &phi;(y,t) with respect to the other variable, in this case y, and set the result equal to N(t,y).
:<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>
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(That's the condition for "exactness" of the differential form M dt + N dy.)
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==Non-Exact Case==
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If the two functions, M(t,y) and N(t,y), instead have the property:
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 +
:<math>
 +
\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}
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</math>
 +
 
 +
then the equation is said to be "inexact" and the above method cannot be used directly to solve it. However, it can be used if a function, &mu;(t,y) can be found such that, when the differential equation is multiplied by &mu;(t,y), it becomes exact. Such a function is called an "integrating function".<ref name=mathworld/> This would transform the condition for exactness to:
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 +
:<math>
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\frac{\partial}{\partial y} \left( \mu(y,t) M \right) = \frac{\partial}{\partial t} \left( \mu(y,t) N \right)
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</math>
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 +
meaning that the integrating function can be found as:
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 +
:<math>
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\mu = \frac{N\frac{\partial \mu}{\partial t}-M\frac{\partial \mu}{\partial y}}{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial t}}
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</math>
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 +
This isn't particularly useful by itself as we still have to solve for &mu; as a function of two variables. Often, the integrating function can be expressed as only one of the two variables, meaning we would only need to solve an ordinary differential equation rather than a partial differential equation. Otherwise, the integrating factor can sometimes be guessed or another method of finding a solution can be employed.
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==Example==
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Consider the equation:<ref name=MathematicalMethods/>
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 +
:<math>
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t \frac{dy}{dt}+3t+y=0
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</math>
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 +
Here we have M(t,y)=3t+y and N(t,y)=t. As:
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 +
:<math>
 +
\frac{\partial M}{\partial y} = 1 = \frac{\partial N}{\partial t}
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</math>
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the equation is exact. Integrating M(t,y) with respect to t produces:
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 +
:<math>
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\phi(t,y) = \int (3t+y) \, dt + F(y) = \frac{3t^2}{2} + ty + F(y)
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</math>
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and so:
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 +
:<math>
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\frac{\partial \phi}{\partial y} = N(t,y)
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</math>
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 +
:<math>
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t + F'(y) = t
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</math>
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Therefore F'(y)=0 which means f(y) is a constant, a. Therefore the solution is:
  
 +
:<math>
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\phi(t,y) = \frac{3t^2}{2} + ty = \text{constant}
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</math>
  
where <math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>.
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where the constant a has been combined with the constant of the condition &phi;(t,y)=constant.
  
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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==References==
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{{reflist}}
  
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==See also==
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*[[Differential equations]]
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*[[Separation of variables]]
  
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[[Category:Mathematics]]
 
[[Category:Calculus]]
 
[[Category:Calculus]]
 
[[Category:Differential Equations]]
 
[[Category:Differential Equations]]

Latest revision as of 08:35, 14 September 2017

An exact differential equation is a first order differential equation that has the form:

or

and is also exact.[1] This means that:

If a first order differential equation is exact, then a conservative field exists and a scalar potential can be defined.[2]

Solution

To find the solution of this equation, we consider the fact that the equation has the form of an exact differential of a function φ(y,t). The total differential of φ(y,t) is then:

Comparing to our original differential equation, and and also that this implies that dφ(y,t)=0 and φ(y,t)=constant. To solve the equation, we integrate both sides of the total differential with respect to one of the variables, say t:

where F(y) is a "constant of integration". (In reality, F(y) is a function, but when we differentiate it with respect t, we treat y as constant and so its derivative becomes zero.) In order to solve for F(y), we differentiate φ(y,t) with respect to the other variable, in this case y, and set the result equal to N(t,y).

Non-Exact Case

If the two functions, M(t,y) and N(t,y), instead have the property:

then the equation is said to be "inexact" and the above method cannot be used directly to solve it. However, it can be used if a function, μ(t,y) can be found such that, when the differential equation is multiplied by μ(t,y), it becomes exact. Such a function is called an "integrating function".[2] This would transform the condition for exactness to:

meaning that the integrating function can be found as:

This isn't particularly useful by itself as we still have to solve for μ as a function of two variables. Often, the integrating function can be expressed as only one of the two variables, meaning we would only need to solve an ordinary differential equation rather than a partial differential equation. Otherwise, the integrating factor can sometimes be guessed or another method of finding a solution can be employed.

Example

Consider the equation:[1]

Here we have M(t,y)=3t+y and N(t,y)=t. As:

the equation is exact. Integrating M(t,y) with respect to t produces:

and so:

Therefore F'(y)=0 which means f(y) is a constant, a. Therefore the solution is:

where the constant a has been combined with the constant of the condition φ(t,y)=constant.

References

  1. 1.0 1.1 K.F. Riley, M.P. Hobson, S.J. Bence, Mathematical Methods for Physics and Engineering, Cambridge University Press, 3rd ed., 2006
  2. 2.0 2.1 Exact Equations from mathworld.wolfram.com

See also