SkipCaptcha
1,279
edits
Changes
Jump to navigation
Jump to search
Line 1:
Line 1: − + + + + + + + + + + + + + + + + − Suppose you are given an [[equation]] of the form:+ + + − + − + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + − 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 the substitution <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. This will give us <math>\phi(t)\,</math> and <math>\phi(y)\,</math>. To get <math>\phi(t, y)\,</math>, 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]].+ + + + + +
Exact differential equation (view source)
Revision as of 13:35, September 14, 2017
, 13:35, September 14, 2017Rewrote to improve clarity and explain solution method, added example
An '''exact differential equation''' is a [[differential equation]] that can be solved in the following manner.
An '''exact differential equation''' is a first order [[differential equation]] that has the form:
:<math>M(t,y) + N(t,y)y' = 0\,</math> or <math>M(t,y) dt + N(t,y) dy = 0\,</math>
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:
:<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>
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>
==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:
:<math>
d\phi (y,t) = \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial t} dt
</math>
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 dφ(y,t)=0 and φ(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:
:<math>
\phi(y,t) = \int M(t,y) \, dt + F(y)
</math>
:<math>M(t,y) + N(t,y)y' = 0\,</math> or <math>M(t,y) dt + N(t,y) dy = 0\,</math>
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).
(we will call this equation 1)
==Non-Exact Case==
If the two functions, M(t,y) and N(t,y), instead have the property:
:<math>
\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}
</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, 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".<ref name=mathworld/> This would transform the condition for exactness to:
:<math>
\frac{\partial}{\partial y} \left( \mu(y,t) M \right) = \frac{\partial}{\partial t} \left( \mu(y,t) N \right)
</math>
meaning that the integrating function can be found as:
:<math>
\mu = \frac{N\frac{\partial \mu}{\partial t}-M\frac{\partial \mu}{\partial y}}{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial t}}
</math>
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:<ref name=MathematicalMethods/>
:<math>
t \frac{dy}{dt}+3t+y=0
</math>
Here we have M(t,y)=3t+y and N(t,y)=t. As:
:<math>
\frac{\partial M}{\partial y} = 1 = \frac{\partial N}{\partial t}
</math>
the equation is exact. Integrating M(t,y) with respect to t produces:
:<math>
\phi(t,y) = \int (3t+y) \, dt + F(y) = \frac{3t^2}{2} + ty + F(y)
</math>
and so:
<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>
:<math>
\frac{\partial \phi}{\partial y} = N(t,y)
</math>
:<math>
t + F'(y) = t
</math>
Therefore F'(y)=0 which means f(y) is a constant, a. Therefore the solution is:
:<math>
\phi(t,y) = \frac{3t^2}{2} + ty = \text{constant}
</math>
where the constant a has been combined with the constant of the condition φ(t,y)=constant.
==References==
{{reflist}}
==See also==
*[[Differential equations]]
*[[Separation of variables]]
[[Category:Mathematics]]
[[Category:Calculus]]
[[Category:Calculus]]
[[Category:Differential Equations]]
[[Category:Differential Equations]]