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Exact differential equation (view source)
Revision as of 18:04, August 2, 2010
, 18:04, August 2, 2010Created page with '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: <math>M y' + N = 0 \<…'
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:
<math>M y' + N = 0 \</math> or <math>M dy + N dt = 0 \</math>
The solution is φ = constant, where φ is determined by integrating M and N.
:<math>\phi(t, y) = \int_0^t M(s, 0) ds + \int_0^y N(t, s) ds</math>
Go through the example to find φ by integrating, then check that
:<math>\frac{\partial \phi}{\partial t} = M</math>
and
:<math>\frac{\partial \phi}{\partial y} = N</math>
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
:<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.)
<math>M(t,y) + N(t,y)y' = 0</math> or <math>M(t,y) dt + N(t,y) dy = 0\,</math>
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.
[[Category:Calculus]]
[[Category:Differential Equations]]
Suppose you are given an equation of the form:
<math>M y' + N = 0 \</math> or <math>M dy + N dt = 0 \</math>
The solution is φ = constant, where φ is determined by integrating M and N.
:<math>\phi(t, y) = \int_0^t M(s, 0) ds + \int_0^y N(t, s) ds</math>
Go through the example to find φ by integrating, then check that
:<math>\frac{\partial \phi}{\partial t} = M</math>
and
:<math>\frac{\partial \phi}{\partial y} = N</math>
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
:<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.)
<math>M(t,y) + N(t,y)y' = 0</math> or <math>M(t,y) dt + N(t,y) dy = 0\,</math>
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.
[[Category:Calculus]]
[[Category:Differential Equations]]