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Line 9: − + − + − 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> − − 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.
Exact differential equation (view source)
Revision as of 18:53, August 2, 2010
, 18:53, August 2, 2010no edit summary
Suppose you are given an equation of the form:
Suppose you are given an equation of the form:
:<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> (1)
Before we begin solving it, we must first check that the equation is exact. This means that:
Before we begin solving it, we must first check that the equation is exact. This means 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>
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 <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>.
To find the solution of this equation, we assume that the solution is φ = constant. We assume that <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 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
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. To get the main function φ 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.
[[Category:Calculus]]
[[Category:Calculus]]
[[Category:Differential Equations]]
[[Category:Differential Equations]]