Confusion about exact DEs
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If I understand the definition of an exact de correctly, if $M(x,y)dx+N(x,y)dy=0 $ and $M_y=N_x$, then
$$f(x,y)=int M(x,y)dx + g(y)$$
and
$$g'(y)=int M(x,y)dx-N(x,y)$$
1) Is my interpretation correct?
2) Is M always the expression in the given equation preceding dx? Is N always the expression preceding Dy?
calculus differential-equations multivariable-calculus
asked Nov 19 at 18:12
Joe Stavitsky
2461317
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If I understand the definition of an exact de correctly, if $M(x,y)dx+N(x,y)dy=0 $ and $M_y=N_x$, then
$$f(x,y)=int M(x,y)dx + g(y)$$
and
$$g'(y)=int M(x,y)dx-N(x,y)$$
1) Is my interpretation correct?
2) Is M always the expression in the given equation preceding dx? Is N always the expression preceding Dy?
calculus differential-equations multivariable-calculus
asked Nov 19 at 18:12
Joe Stavitsky
2461317
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If I understand the definition of an exact de correctly, if $M(x,y)dx+N(x,y)dy=0 $ and $M_y=N_x$, then
$$f(x,y)=int M(x,y)dx + g(y)$$
and
$$g'(y)=int M(x,y)dx-N(x,y)$$
1) Is my interpretation correct?
2) Is M always the expression in the given equation preceding dx? Is N always the expression preceding Dy?
calculus differential-equations multivariable-calculus
asked Nov 19 at 18:12
Joe Stavitsky
2461317
If I understand the definition of an exact de correctly, if $M(x,y)dx+N(x,y)dy=0 $ and $M_y=N_x$, then
$$f(x,y)=int M(x,y)dx + g(y)$$
and
$$g'(y)=int M(x,y)dx-N(x,y)$$
1) Is my interpretation correct?
2) Is M always the expression in the given equation preceding dx? Is N always the expression preceding Dy?
calculus differential-equations multivariable-calculus
calculus differential-equations multivariable-calculus
asked Nov 19 at 18:12
Joe Stavitsky
2461317
asked Nov 19 at 18:12
Joe Stavitsky
2461317
edited Nov 20 at 12:31
asked Nov 19 at 18:12
Joe Stavitsky
2461317
asked Nov 19 at 18:12
Joe Stavitsky
2461317
asked Nov 19 at 18:12
Joe Stavitsky
2461317
2461317
add a comment |
add a comment |
1 Answer
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1.
Your interpretation is partially correct.
$$
F(x, y) := int M(x,y)dx + g(y)
$$
However:
$$
g'(y) := N(x,y) - frac{partial}{partial y} int M(x,y)dx
$$
This is because $frac{partial F}{partial y} = N$.
$$
F(x, y) := int M(x,y)dx + g(y)
implies
N(x, y) = frac{partial}{partial y}int M(x,y)dx + frac{partial}{partial y}g(y) \ = frac{partial}{partial y}int M(x,y)dx + g'(y)
$$
- Yes, $M(x, y)$ precedes $dx$ and $N(x, y)$ precedes $dy$
answered Nov 19 at 23:51
chaseklvk
515
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
1.
Your interpretation is partially correct.
$$
F(x, y) := int M(x,y)dx + g(y)
$$
However:
$$
g'(y) := N(x,y) - frac{partial}{partial y} int M(x,y)dx
$$
This is because $frac{partial F}{partial y} = N$.
$$
F(x, y) := int M(x,y)dx + g(y)
implies
N(x, y) = frac{partial}{partial y}int M(x,y)dx + frac{partial}{partial y}g(y) \ = frac{partial}{partial y}int M(x,y)dx + g'(y)
$$
- Yes, $M(x, y)$ precedes $dx$ and $N(x, y)$ precedes $dy$
answered Nov 19 at 23:51
chaseklvk
515
add a comment |
up vote
1
down vote
1.
Your interpretation is partially correct.
$$
F(x, y) := int M(x,y)dx + g(y)
$$
However:
$$
g'(y) := N(x,y) - frac{partial}{partial y} int M(x,y)dx
$$
This is because $frac{partial F}{partial y} = N$.
$$
F(x, y) := int M(x,y)dx + g(y)
implies
N(x, y) = frac{partial}{partial y}int M(x,y)dx + frac{partial}{partial y}g(y) \ = frac{partial}{partial y}int M(x,y)dx + g'(y)
$$
- Yes, $M(x, y)$ precedes $dx$ and $N(x, y)$ precedes $dy$
answered Nov 19 at 23:51
chaseklvk
515
add a comment |
up vote
1
down vote
up vote
1
down vote
1.
Your interpretation is partially correct.
$$
F(x, y) := int M(x,y)dx + g(y)
$$
However:
$$
g'(y) := N(x,y) - frac{partial}{partial y} int M(x,y)dx
$$
This is because $frac{partial F}{partial y} = N$.
$$
F(x, y) := int M(x,y)dx + g(y)
implies
N(x, y) = frac{partial}{partial y}int M(x,y)dx + frac{partial}{partial y}g(y) \ = frac{partial}{partial y}int M(x,y)dx + g'(y)
$$
- Yes, $M(x, y)$ precedes $dx$ and $N(x, y)$ precedes $dy$
answered Nov 19 at 23:51
chaseklvk
515
1.
Your interpretation is partially correct.
$$
F(x, y) := int M(x,y)dx + g(y)
$$
However:
$$
g'(y) := N(x,y) - frac{partial}{partial y} int M(x,y)dx
$$
This is because $frac{partial F}{partial y} = N$.
$$
F(x, y) := int M(x,y)dx + g(y)
implies
N(x, y) = frac{partial}{partial y}int M(x,y)dx + frac{partial}{partial y}g(y) \ = frac{partial}{partial y}int M(x,y)dx + g'(y)
$$
- Yes, $M(x, y)$ precedes $dx$ and $N(x, y)$ precedes $dy$
answered Nov 19 at 23:51
chaseklvk
515
answered Nov 19 at 23:51
chaseklvk
515
answered Nov 19 at 23:51
chaseklvk
515
answered Nov 19 at 23:51
chaseklvk
515
515
add a comment |
add a comment |
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