Why do exact differential equations have to equal zero?
$begingroup$
Why is it that in order to solve an exact differential equation, the total differential must be zero? In other words, the following function must be equal to a constant
$ phi (x,y) = c $
Then $ d phi = frac { partial phi}{ partial x} dx + frac { partial phi}{ partial y} dy = 0$
Then a differential equation of the form $ M dx + N dy = 0 $ can be solved if $ frac { partial M}{ partial y} = frac { partial N}{ partial x}$
Then its called an 'exact differential equation'
But why in order to do this, there must be a condition where
$ phi (x,y) = c $ ?
Why can't it be something else? Thanks
ordinary-differential-equations
asked Jan 24 at 20:11
khaled014zkhaled014z
1769
$endgroup$
add a comment |
$begingroup$
Why is it that in order to solve an exact differential equation, the total differential must be zero? In other words, the following function must be equal to a constant
$ phi (x,y) = c $
Then $ d phi = frac { partial phi}{ partial x} dx + frac { partial phi}{ partial y} dy = 0$
Then a differential equation of the form $ M dx + N dy = 0 $ can be solved if $ frac { partial M}{ partial y} = frac { partial N}{ partial x}$
Then its called an 'exact differential equation'
But why in order to do this, there must be a condition where
$ phi (x,y) = c $ ?
Why can't it be something else? Thanks
ordinary-differential-equations
asked Jan 24 at 20:11
khaled014zkhaled014z
1769
$endgroup$
add a comment |
$begingroup$
Why is it that in order to solve an exact differential equation, the total differential must be zero? In other words, the following function must be equal to a constant
$ phi (x,y) = c $
Then $ d phi = frac { partial phi}{ partial x} dx + frac { partial phi}{ partial y} dy = 0$
Then a differential equation of the form $ M dx + N dy = 0 $ can be solved if $ frac { partial M}{ partial y} = frac { partial N}{ partial x}$
Then its called an 'exact differential equation'
But why in order to do this, there must be a condition where
$ phi (x,y) = c $ ?
Why can't it be something else? Thanks
ordinary-differential-equations
asked Jan 24 at 20:11
khaled014zkhaled014z
1769
$endgroup$
Why is it that in order to solve an exact differential equation, the total differential must be zero? In other words, the following function must be equal to a constant
$ phi (x,y) = c $
Then $ d phi = frac { partial phi}{ partial x} dx + frac { partial phi}{ partial y} dy = 0$
Then a differential equation of the form $ M dx + N dy = 0 $ can be solved if $ frac { partial M}{ partial y} = frac { partial N}{ partial x}$
Then its called an 'exact differential equation'
But why in order to do this, there must be a condition where
$ phi (x,y) = c $ ?
Why can't it be something else? Thanks
ordinary-differential-equations
ordinary-differential-equations
asked Jan 24 at 20:11
khaled014zkhaled014z
1769
asked Jan 24 at 20:11
khaled014zkhaled014z
1769
asked Jan 24 at 20:11
khaled014zkhaled014z
1769
asked Jan 24 at 20:11
khaled014zkhaled014z
1769
asked Jan 24 at 20:11
khaled014zkhaled014z
1769
1769
add a comment |
add a comment |
1 Answer
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The idea is to represent your solutions as level curves of a certain function $Phi$. If that is the case, you should have $dPhi=0$ along integral curves and the curves themselves will have the form of level curves $Phi(x,y)=const.$
answered Jan 24 at 20:15
GReyesGReyes
1,82015
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add a comment |
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1 Answer
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$begingroup$
The idea is to represent your solutions as level curves of a certain function $Phi$. If that is the case, you should have $dPhi=0$ along integral curves and the curves themselves will have the form of level curves $Phi(x,y)=const.$
answered Jan 24 at 20:15
GReyesGReyes
1,82015
$endgroup$
add a comment |
$begingroup$
The idea is to represent your solutions as level curves of a certain function $Phi$. If that is the case, you should have $dPhi=0$ along integral curves and the curves themselves will have the form of level curves $Phi(x,y)=const.$
answered Jan 24 at 20:15
GReyesGReyes
1,82015
$endgroup$
add a comment |
$begingroup$
The idea is to represent your solutions as level curves of a certain function $Phi$. If that is the case, you should have $dPhi=0$ along integral curves and the curves themselves will have the form of level curves $Phi(x,y)=const.$
answered Jan 24 at 20:15
GReyesGReyes
1,82015
$endgroup$
The idea is to represent your solutions as level curves of a certain function $Phi$. If that is the case, you should have $dPhi=0$ along integral curves and the curves themselves will have the form of level curves $Phi(x,y)=const.$
answered Jan 24 at 20:15
GReyesGReyes
1,82015
answered Jan 24 at 20:15
GReyesGReyes
1,82015
answered Jan 24 at 20:15
GReyesGReyes
1,82015
answered Jan 24 at 20:15
GReyesGReyes
1,82015
1,82015
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