Why is $y'=fracyx$ exact? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Trigonometric Differential Equation 3Is okay to have different solution to differential equation?Solving Ordinary Differential Equation $y(x^4-y^2)dx+x(x^4+y^2)dy=0$How to prove this exact Differential Equation is exact?Why is this differential equation an exact differential equation?Exact differential equation ProblemSolve the following differential equation: $frac (ydx+xdy)(1-x^2y^2)+xdx=0$Solution to the differential equation $left(x cscleft(fracyxright)-yright) dx + xdy$?What is the integrating factor for $xdy + y(x+1)dx =0 $?Why do exact differential equations have to equal zero?
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Why is $y'=fracyx$ exact?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Trigonometric Differential Equation 3Is okay to have different solution to differential equation?Solving Ordinary Differential Equation $y(x^4-y^2)dx+x(x^4+y^2)dy=0$How to prove this exact Differential Equation is exact?Why is this differential equation an exact differential equation?Exact differential equation ProblemSolve the following differential equation: $frac (ydx+xdy)(1-x^2y^2)+xdx=0$Solution to the differential equation $left(x cscleft(fracyxright)-yright) dx + xdy$?What is the integrating factor for $xdy + y(x+1)dx =0 $?Why do exact differential equations have to equal zero?
$begingroup$
A well known theorem states that differential form $m dx+n dy=0$ is exact iff $partial_y m=partial_x n$. But why is $y'=fracyx$ ($equiv xdy-ydx=0$) exact?
ordinary-differential-equations
asked Apr 1 at 12:54
C.F.GC.F.G
1,4711821
$endgroup$
add a comment |
$begingroup$
A well known theorem states that differential form $m dx+n dy=0$ is exact iff $partial_y m=partial_x n$. But why is $y'=fracyx$ ($equiv xdy-ydx=0$) exact?
ordinary-differential-equations
asked Apr 1 at 12:54
C.F.GC.F.G
1,4711821
$endgroup$
$begingroup$
If you say it is exact, how can you find the solution according to steps of exact equation
$endgroup$
– E.H.E
Apr 1 at 13:02
$begingroup$
see $partial_y m=-partial_x n$
$endgroup$
– E.H.E
Apr 1 at 13:03
$begingroup$
I asked this question because l see in a book. (In book: First show that this equation is exact then solve it.)
$endgroup$
– C.F.G
Apr 1 at 13:19
$begingroup$
this $ xdy+ydx=0$ is exact but $ xdy-ydx=0$ not exact
$endgroup$
– E.H.E
Apr 1 at 13:20
$begingroup$
A similar situation, $y'=fracycos x+sin y+ysin x+xcos y+x$ is this exact?
$endgroup$
– C.F.G
Apr 1 at 17:32
add a comment |
$begingroup$
A well known theorem states that differential form $m dx+n dy=0$ is exact iff $partial_y m=partial_x n$. But why is $y'=fracyx$ ($equiv xdy-ydx=0$) exact?
ordinary-differential-equations
asked Apr 1 at 12:54
C.F.GC.F.G
1,4711821
$endgroup$
A well known theorem states that differential form $m dx+n dy=0$ is exact iff $partial_y m=partial_x n$. But why is $y'=fracyx$ ($equiv xdy-ydx=0$) exact?
ordinary-differential-equations
ordinary-differential-equations
asked Apr 1 at 12:54
C.F.GC.F.G
1,4711821
asked Apr 1 at 12:54
C.F.GC.F.G
1,4711821
asked Apr 1 at 12:54
C.F.GC.F.G
1,4711821
asked Apr 1 at 12:54
C.F.GC.F.G
1,4711821
asked Apr 1 at 12:54
C.F.GC.F.G
1,4711821
1,4711821
$begingroup$
If you say it is exact, how can you find the solution according to steps of exact equation
$endgroup$
– E.H.E
Apr 1 at 13:02
$begingroup$
see $partial_y m=-partial_x n$
$endgroup$
– E.H.E
Apr 1 at 13:03
$begingroup$
I asked this question because l see in a book. (In book: First show that this equation is exact then solve it.)
$endgroup$
– C.F.G
Apr 1 at 13:19
$begingroup$
this $ xdy+ydx=0$ is exact but $ xdy-ydx=0$ not exact
$endgroup$
– E.H.E
Apr 1 at 13:20
$begingroup$
A similar situation, $y'=fracycos x+sin y+ysin x+xcos y+x$ is this exact?
$endgroup$
– C.F.G
Apr 1 at 17:32
add a comment |
$begingroup$
If you say it is exact, how can you find the solution according to steps of exact equation
$endgroup$
– E.H.E
Apr 1 at 13:02
$begingroup$
see $partial_y m=-partial_x n$
$endgroup$
– E.H.E
Apr 1 at 13:03
$begingroup$
I asked this question because l see in a book. (In book: First show that this equation is exact then solve it.)
$endgroup$
– C.F.G
Apr 1 at 13:19
$begingroup$
this $ xdy+ydx=0$ is exact but $ xdy-ydx=0$ not exact
$endgroup$
– E.H.E
Apr 1 at 13:20
$begingroup$
A similar situation, $y'=fracycos x+sin y+ysin x+xcos y+x$ is this exact?
$endgroup$
– C.F.G
Apr 1 at 17:32
$begingroup$
If you say it is exact, how can you find the solution according to steps of exact equation
$endgroup$
– E.H.E
Apr 1 at 13:02
$begingroup$
If you say it is exact, how can you find the solution according to steps of exact equation
$endgroup$
– E.H.E
Apr 1 at 13:02
$begingroup$
see $partial_y m=-partial_x n$
$endgroup$
– E.H.E
Apr 1 at 13:03
$begingroup$
see $partial_y m=-partial_x n$
$endgroup$
– E.H.E
Apr 1 at 13:03
$begingroup$
I asked this question because l see in a book. (In book: First show that this equation is exact then solve it.)
$endgroup$
– C.F.G
Apr 1 at 13:19
$begingroup$
I asked this question because l see in a book. (In book: First show that this equation is exact then solve it.)
$endgroup$
– C.F.G
Apr 1 at 13:19
$begingroup$
this $ xdy+ydx=0$ is exact but $ xdy-ydx=0$ not exact
$endgroup$
– E.H.E
Apr 1 at 13:20
$begingroup$
this $ xdy+ydx=0$ is exact but $ xdy-ydx=0$ not exact
$endgroup$
– E.H.E
Apr 1 at 13:20
$begingroup$
A similar situation, $y'=fracycos x+sin y+ysin x+xcos y+x$ is this exact?
$endgroup$
– C.F.G
Apr 1 at 17:32
$begingroup$
A similar situation, $y'=fracycos x+sin y+ysin x+xcos y+x$ is this exact?
$endgroup$
– C.F.G
Apr 1 at 17:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Differential equations aren't exact or inexact; expressions of the form $F(x,,y)dx+G(x,,y)dy$ are. The difference is crucial, because the choice of $F,,G$ equivalent to a given equation won't be unique. What's exact in this case is $dleft(fracxyright)=frac1ydx-fracxy^2dy$, which is $0$ for solutions of $y^prime=fracyx$.
answered Apr 1 at 13:26
J.G.J.G.
33.6k23252
$endgroup$
add a comment |
$begingroup$
see what happens if you assume it exact
$$int xdy=xy+f(x)$$
$$fracpartial partial x(xy+f(x))=y+f'(x)=-y$$
$$f'(x)=-2y$$
that is impossible, so the equation is not exact
answered Apr 1 at 13:37
E.H.EE.H.E
16.7k11969
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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$begingroup$
Differential equations aren't exact or inexact; expressions of the form $F(x,,y)dx+G(x,,y)dy$ are. The difference is crucial, because the choice of $F,,G$ equivalent to a given equation won't be unique. What's exact in this case is $dleft(fracxyright)=frac1ydx-fracxy^2dy$, which is $0$ for solutions of $y^prime=fracyx$.
answered Apr 1 at 13:26
J.G.J.G.
33.6k23252
$endgroup$
add a comment |
$begingroup$
Differential equations aren't exact or inexact; expressions of the form $F(x,,y)dx+G(x,,y)dy$ are. The difference is crucial, because the choice of $F,,G$ equivalent to a given equation won't be unique. What's exact in this case is $dleft(fracxyright)=frac1ydx-fracxy^2dy$, which is $0$ for solutions of $y^prime=fracyx$.
answered Apr 1 at 13:26
J.G.J.G.
33.6k23252
$endgroup$
add a comment |
$begingroup$
Differential equations aren't exact or inexact; expressions of the form $F(x,,y)dx+G(x,,y)dy$ are. The difference is crucial, because the choice of $F,,G$ equivalent to a given equation won't be unique. What's exact in this case is $dleft(fracxyright)=frac1ydx-fracxy^2dy$, which is $0$ for solutions of $y^prime=fracyx$.
answered Apr 1 at 13:26
J.G.J.G.
33.6k23252
$endgroup$
Differential equations aren't exact or inexact; expressions of the form $F(x,,y)dx+G(x,,y)dy$ are. The difference is crucial, because the choice of $F,,G$ equivalent to a given equation won't be unique. What's exact in this case is $dleft(fracxyright)=frac1ydx-fracxy^2dy$, which is $0$ for solutions of $y^prime=fracyx$.
answered Apr 1 at 13:26
J.G.J.G.
33.6k23252
edited Apr 1 at 13:31
answered Apr 1 at 13:26
J.G.J.G.
33.6k23252
answered Apr 1 at 13:26
J.G.J.G.
33.6k23252
answered Apr 1 at 13:26
J.G.J.G.
33.6k23252
33.6k23252
add a comment |
add a comment |
$begingroup$
see what happens if you assume it exact
$$int xdy=xy+f(x)$$
$$fracpartial partial x(xy+f(x))=y+f'(x)=-y$$
$$f'(x)=-2y$$
that is impossible, so the equation is not exact
answered Apr 1 at 13:37
E.H.EE.H.E
16.7k11969
$endgroup$
add a comment |
$begingroup$
see what happens if you assume it exact
$$int xdy=xy+f(x)$$
$$fracpartial partial x(xy+f(x))=y+f'(x)=-y$$
$$f'(x)=-2y$$
that is impossible, so the equation is not exact
answered Apr 1 at 13:37
E.H.EE.H.E
16.7k11969
$endgroup$
add a comment |
$begingroup$
see what happens if you assume it exact
$$int xdy=xy+f(x)$$
$$fracpartial partial x(xy+f(x))=y+f'(x)=-y$$
$$f'(x)=-2y$$
that is impossible, so the equation is not exact
answered Apr 1 at 13:37
E.H.EE.H.E
16.7k11969
$endgroup$
see what happens if you assume it exact
$$int xdy=xy+f(x)$$
$$fracpartial partial x(xy+f(x))=y+f'(x)=-y$$
$$f'(x)=-2y$$
that is impossible, so the equation is not exact
answered Apr 1 at 13:37
E.H.EE.H.E
16.7k11969
answered Apr 1 at 13:37
E.H.EE.H.E
16.7k11969
answered Apr 1 at 13:37
E.H.EE.H.E
16.7k11969
answered Apr 1 at 13:37
E.H.EE.H.E
16.7k11969
16.7k11969
add a comment |
add a comment |
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$begingroup$
If you say it is exact, how can you find the solution according to steps of exact equation
$endgroup$
– E.H.E
Apr 1 at 13:02
$begingroup$
see $partial_y m=-partial_x n$
$endgroup$
– E.H.E
Apr 1 at 13:03
$begingroup$
I asked this question because l see in a book. (In book: First show that this equation is exact then solve it.)
$endgroup$
– C.F.G
Apr 1 at 13:19
$begingroup$
this $ xdy+ydx=0$ is exact but $ xdy-ydx=0$ not exact
$endgroup$
– E.H.E
Apr 1 at 13:20
$begingroup$
A similar situation, $y'=fracycos x+sin y+ysin x+xcos y+x$ is this exact?
$endgroup$
– C.F.G
Apr 1 at 17:32