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Problems with an exact differential eqution
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Consider the following differential equation
$$
left(frac1x-fracy^2(x-y)^2right)dx=left(frac1y-fracx^2(x-y)^2right)dy
$$
I want to find its general solution. I get that this equation is exact but, I trying to solve it for this method and it seems not work. Can someone give me a hit?
differential-equations
edited 52 mins ago
JoeTaxpayer
2,17121226
asked 4 hours ago
Gödel
1,201319
add a comment |Â
up vote
2
down vote
favorite
Consider the following differential equation
$$
left(frac1x-fracy^2(x-y)^2right)dx=left(frac1y-fracx^2(x-y)^2right)dy
$$
I want to find its general solution. I get that this equation is exact but, I trying to solve it for this method and it seems not work. Can someone give me a hit?
differential-equations
edited 52 mins ago
JoeTaxpayer
2,17121226
asked 4 hours ago
Gödel
1,201319
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the following differential equation
$$
left(frac1x-fracy^2(x-y)^2right)dx=left(frac1y-fracx^2(x-y)^2right)dy
$$
I want to find its general solution. I get that this equation is exact but, I trying to solve it for this method and it seems not work. Can someone give me a hit?
differential-equations
edited 52 mins ago
JoeTaxpayer
2,17121226
asked 4 hours ago
Gödel
1,201319
Consider the following differential equation
$$
left(frac1x-fracy^2(x-y)^2right)dx=left(frac1y-fracx^2(x-y)^2right)dy
$$
I want to find its general solution. I get that this equation is exact but, I trying to solve it for this method and it seems not work. Can someone give me a hit?
differential-equations
differential-equations
edited 52 mins ago
JoeTaxpayer
2,17121226
asked 4 hours ago
Gödel
1,201319
edited 52 mins ago
JoeTaxpayer
2,17121226
asked 4 hours ago
Gödel
1,201319
edited 52 mins ago
JoeTaxpayer
2,17121226
edited 52 mins ago
JoeTaxpayer
2,17121226
edited 52 mins ago
JoeTaxpayer
2,17121226
2,17121226
asked 4 hours ago
Gödel
1,201319
asked 4 hours ago
Gödel
1,201319
asked 4 hours ago
Gödel
1,201319
1,201319
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
2
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accepted
We have an exact differential equation in the form $Mdx +Ndy = 0$, with
$$M equiv frac1x - left(fracyx-yright)^2, N equiv left(fracxx-yright)^2 - frac1y.$$
If we let the implicit solution be $F(x, y) = c$, since the equation is exact, it is known that $F_x = M, F_y = N.$
$$F_x = M Rightarrow, F = int left ( frac1x - left(fracyx-yright)^2 right )dx,$$
$$Rightarrow F = ln|x| + fracy^2x-y + phi(y),$$
where $phi(y)$ is a function of y.
$F_y = N,$
$$Rightarrow fracy(2x-y)(x-y)^2 +phi'(y) = fracx^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = fracx^2-2xy + y^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = 1 - frac1y,$$
$$phi(y) = y - ln|y|+C.$$
Thus, the implicit solution is $$F(x, y) = lnleft|fracxyright| + fracxyx-y = c.$$
edited 2 hours ago
Moo
4,8233920
answered 4 hours ago
math783625
637
add a comment |Â
up vote
1
down vote
The differential is indeed exact
$$left(frac1x-fracy^2(x-y)^2right)dx-left(frac1y-fracx^2(x-y)^2right)dy=0$$
Note that
$$frac dxx=dln (x)$$
$$frac dyy=dln (y)$$
And also that
$$
beginalign
E=&-fracy^2(x-y)^2dx+fracx^2(x-y)^2dy \
E=&frac-y^2dx+x^2dy(x-y)^2\
E=&frac-y^2dx+x^2dyx^2y^2frac (xy)^2(x-y)^2\
E=&(d(frac 1x- frac 1y))frac (xy)^2(x-y)^2\
E=&(frac xyy-x)^2d(frac y-xxy)  \
endalign
$$
$$ text Since we have frac dvv^2=-dleft(frac 1v right ) implies E=-d(frac xyy-x)$$
Therefore we have
$$d ln x -d ln y +d(frac xyx-y)=0$$
$$boxedln (frac xy)+frac xyx-y=K$$
answered 3 hours ago
Isham
11.7k3929
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We have an exact differential equation in the form $Mdx +Ndy = 0$, with
$$M equiv frac1x - left(fracyx-yright)^2, N equiv left(fracxx-yright)^2 - frac1y.$$
If we let the implicit solution be $F(x, y) = c$, since the equation is exact, it is known that $F_x = M, F_y = N.$
$$F_x = M Rightarrow, F = int left ( frac1x - left(fracyx-yright)^2 right )dx,$$
$$Rightarrow F = ln|x| + fracy^2x-y + phi(y),$$
where $phi(y)$ is a function of y.
$F_y = N,$
$$Rightarrow fracy(2x-y)(x-y)^2 +phi'(y) = fracx^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = fracx^2-2xy + y^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = 1 - frac1y,$$
$$phi(y) = y - ln|y|+C.$$
Thus, the implicit solution is $$F(x, y) = lnleft|fracxyright| + fracxyx-y = c.$$
edited 2 hours ago
Moo
4,8233920
answered 4 hours ago
math783625
637
add a comment |Â
up vote
2
down vote
accepted
We have an exact differential equation in the form $Mdx +Ndy = 0$, with
$$M equiv frac1x - left(fracyx-yright)^2, N equiv left(fracxx-yright)^2 - frac1y.$$
If we let the implicit solution be $F(x, y) = c$, since the equation is exact, it is known that $F_x = M, F_y = N.$
$$F_x = M Rightarrow, F = int left ( frac1x - left(fracyx-yright)^2 right )dx,$$
$$Rightarrow F = ln|x| + fracy^2x-y + phi(y),$$
where $phi(y)$ is a function of y.
$F_y = N,$
$$Rightarrow fracy(2x-y)(x-y)^2 +phi'(y) = fracx^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = fracx^2-2xy + y^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = 1 - frac1y,$$
$$phi(y) = y - ln|y|+C.$$
Thus, the implicit solution is $$F(x, y) = lnleft|fracxyright| + fracxyx-y = c.$$
edited 2 hours ago
Moo
4,8233920
answered 4 hours ago
math783625
637
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We have an exact differential equation in the form $Mdx +Ndy = 0$, with
$$M equiv frac1x - left(fracyx-yright)^2, N equiv left(fracxx-yright)^2 - frac1y.$$
If we let the implicit solution be $F(x, y) = c$, since the equation is exact, it is known that $F_x = M, F_y = N.$
$$F_x = M Rightarrow, F = int left ( frac1x - left(fracyx-yright)^2 right )dx,$$
$$Rightarrow F = ln|x| + fracy^2x-y + phi(y),$$
where $phi(y)$ is a function of y.
$F_y = N,$
$$Rightarrow fracy(2x-y)(x-y)^2 +phi'(y) = fracx^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = fracx^2-2xy + y^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = 1 - frac1y,$$
$$phi(y) = y - ln|y|+C.$$
Thus, the implicit solution is $$F(x, y) = lnleft|fracxyright| + fracxyx-y = c.$$
edited 2 hours ago
Moo
4,8233920
answered 4 hours ago
math783625
637
We have an exact differential equation in the form $Mdx +Ndy = 0$, with
$$M equiv frac1x - left(fracyx-yright)^2, N equiv left(fracxx-yright)^2 - frac1y.$$
If we let the implicit solution be $F(x, y) = c$, since the equation is exact, it is known that $F_x = M, F_y = N.$
$$F_x = M Rightarrow, F = int left ( frac1x - left(fracyx-yright)^2 right )dx,$$
$$Rightarrow F = ln|x| + fracy^2x-y + phi(y),$$
where $phi(y)$ is a function of y.
$F_y = N,$
$$Rightarrow fracy(2x-y)(x-y)^2 +phi'(y) = fracx^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = fracx^2-2xy + y^2(x-y)^2 - frac1y,$$
$$Rightarrow phi'(y) = 1 - frac1y,$$
$$phi(y) = y - ln|y|+C.$$
Thus, the implicit solution is $$F(x, y) = lnleft|fracxyright| + fracxyx-y = c.$$
edited 2 hours ago
Moo
4,8233920
answered 4 hours ago
math783625
637
edited 2 hours ago
Moo
4,8233920
edited 2 hours ago
Moo
4,8233920
edited 2 hours ago
Moo
4,8233920
4,8233920
answered 4 hours ago
math783625
637
answered 4 hours ago
math783625
637
answered 4 hours ago
math783625
637
637
add a comment |Â
add a comment |Â
up vote
1
down vote
The differential is indeed exact
$$left(frac1x-fracy^2(x-y)^2right)dx-left(frac1y-fracx^2(x-y)^2right)dy=0$$
Note that
$$frac dxx=dln (x)$$
$$frac dyy=dln (y)$$
And also that
$$
beginalign
E=&-fracy^2(x-y)^2dx+fracx^2(x-y)^2dy \
E=&frac-y^2dx+x^2dy(x-y)^2\
E=&frac-y^2dx+x^2dyx^2y^2frac (xy)^2(x-y)^2\
E=&(d(frac 1x- frac 1y))frac (xy)^2(x-y)^2\
E=&(frac xyy-x)^2d(frac y-xxy)  \
endalign
$$
$$ text Since we have frac dvv^2=-dleft(frac 1v right ) implies E=-d(frac xyy-x)$$
Therefore we have
$$d ln x -d ln y +d(frac xyx-y)=0$$
$$boxedln (frac xy)+frac xyx-y=K$$
answered 3 hours ago
Isham
11.7k3929
add a comment |Â
up vote
1
down vote
The differential is indeed exact
$$left(frac1x-fracy^2(x-y)^2right)dx-left(frac1y-fracx^2(x-y)^2right)dy=0$$
Note that
$$frac dxx=dln (x)$$
$$frac dyy=dln (y)$$
And also that
$$
beginalign
E=&-fracy^2(x-y)^2dx+fracx^2(x-y)^2dy \
E=&frac-y^2dx+x^2dy(x-y)^2\
E=&frac-y^2dx+x^2dyx^2y^2frac (xy)^2(x-y)^2\
E=&(d(frac 1x- frac 1y))frac (xy)^2(x-y)^2\
E=&(frac xyy-x)^2d(frac y-xxy)  \
endalign
$$
$$ text Since we have frac dvv^2=-dleft(frac 1v right ) implies E=-d(frac xyy-x)$$
Therefore we have
$$d ln x -d ln y +d(frac xyx-y)=0$$
$$boxedln (frac xy)+frac xyx-y=K$$
answered 3 hours ago
Isham
11.7k3929
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The differential is indeed exact
$$left(frac1x-fracy^2(x-y)^2right)dx-left(frac1y-fracx^2(x-y)^2right)dy=0$$
Note that
$$frac dxx=dln (x)$$
$$frac dyy=dln (y)$$
And also that
$$
beginalign
E=&-fracy^2(x-y)^2dx+fracx^2(x-y)^2dy \
E=&frac-y^2dx+x^2dy(x-y)^2\
E=&frac-y^2dx+x^2dyx^2y^2frac (xy)^2(x-y)^2\
E=&(d(frac 1x- frac 1y))frac (xy)^2(x-y)^2\
E=&(frac xyy-x)^2d(frac y-xxy)  \
endalign
$$
$$ text Since we have frac dvv^2=-dleft(frac 1v right ) implies E=-d(frac xyy-x)$$
Therefore we have
$$d ln x -d ln y +d(frac xyx-y)=0$$
$$boxedln (frac xy)+frac xyx-y=K$$
answered 3 hours ago
Isham
11.7k3929
The differential is indeed exact
$$left(frac1x-fracy^2(x-y)^2right)dx-left(frac1y-fracx^2(x-y)^2right)dy=0$$
Note that
$$frac dxx=dln (x)$$
$$frac dyy=dln (y)$$
And also that
$$
beginalign
E=&-fracy^2(x-y)^2dx+fracx^2(x-y)^2dy \
E=&frac-y^2dx+x^2dy(x-y)^2\
E=&frac-y^2dx+x^2dyx^2y^2frac (xy)^2(x-y)^2\
E=&(d(frac 1x- frac 1y))frac (xy)^2(x-y)^2\
E=&(frac xyy-x)^2d(frac y-xxy)  \
endalign
$$
$$ text Since we have frac dvv^2=-dleft(frac 1v right ) implies E=-d(frac xyy-x)$$
Therefore we have
$$d ln x -d ln y +d(frac xyx-y)=0$$
$$boxedln (frac xy)+frac xyx-y=K$$
answered 3 hours ago
Isham
11.7k3929
edited 3 hours ago
answered 3 hours ago
Isham
11.7k3929
answered 3 hours ago
Isham
11.7k3929
answered 3 hours ago
Isham
11.7k3929
11.7k3929
add a comment |Â
add a comment |Â
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