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Exact differential equation
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Revision as of 18:35, August 2, 2010
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18:35, August 2, 2010
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Suppose you are given an equation of the form:
Suppose you are given an equation of the form:
−
<math>M y
'
+ N = 0 \</math> or <math>M
dy
+ N
dt
= 0 \</math>
+
<math>M
(t,
y
)
+ N
(t,y)y'
= 0\
,
</math> or <math>M
(t,y) dt
+ N
(t,y) dy
= 0\
,
</math>
−
The
solution is φ = constant, where φ is determined by integrating M and N.
+
To find the solution of this equation, we assume that 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>
:<math>\phi(t, y) = \int_0^t M(s, 0) ds + \int_0^y N(t, s) ds</math>
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:<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.)
(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>
PhyllisS
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