Mixing
Find a first integral of an ODE system
$begingroup$
I have the system
$$ left{begin{array}{ccl}
dot x & = & 2xy \
dot y & = & x+y^2 \
end{array} right. $$
and I need to find a first integral $H$ of the system. This is easy if the equation
$(x+y^2)dx-2xy dy = 0 $ is exact (i.e the divergence of the field is 0). In this case, the divergence is $4y$ so I need to find an integrating factor $mu(x,y) $ . But I don't know how to do it, because it doesn't work if I use that $mu$ is just a function of $x$ or $y$.
Any help is welcome!
differential-equations vector-fields integrating-factor
asked Jan 1 at 19:16
RelativoRelativo
37519
$endgroup$
add a comment |
$begingroup$
I have the system
$$ left{begin{array}{ccl}
dot x & = & 2xy \
dot y & = & x+y^2 \
end{array} right. $$
and I need to find a first integral $H$ of the system. This is easy if the equation
$(x+y^2)dx-2xy dy = 0 $ is exact (i.e the divergence of the field is 0). In this case, the divergence is $4y$ so I need to find an integrating factor $mu(x,y) $ . But I don't know how to do it, because it doesn't work if I use that $mu$ is just a function of $x$ or $y$.
Any help is welcome!
differential-equations vector-fields integrating-factor
asked Jan 1 at 19:16
RelativoRelativo
37519
$endgroup$
add a comment |
$begingroup$
I have the system
$$ left{begin{array}{ccl}
dot x & = & 2xy \
dot y & = & x+y^2 \
end{array} right. $$
and I need to find a first integral $H$ of the system. This is easy if the equation
$(x+y^2)dx-2xy dy = 0 $ is exact (i.e the divergence of the field is 0). In this case, the divergence is $4y$ so I need to find an integrating factor $mu(x,y) $ . But I don't know how to do it, because it doesn't work if I use that $mu$ is just a function of $x$ or $y$.
Any help is welcome!
differential-equations vector-fields integrating-factor
asked Jan 1 at 19:16
RelativoRelativo
37519
$endgroup$
I have the system
$$ left{begin{array}{ccl}
dot x & = & 2xy \
dot y & = & x+y^2 \
end{array} right. $$
and I need to find a first integral $H$ of the system. This is easy if the equation
$(x+y^2)dx-2xy dy = 0 $ is exact (i.e the divergence of the field is 0). In this case, the divergence is $4y$ so I need to find an integrating factor $mu(x,y) $ . But I don't know how to do it, because it doesn't work if I use that $mu$ is just a function of $x$ or $y$.
Any help is welcome!
differential-equations vector-fields integrating-factor
differential-equations vector-fields integrating-factor
asked Jan 1 at 19:16
RelativoRelativo
37519
asked Jan 1 at 19:16
RelativoRelativo
37519
asked Jan 1 at 19:16
RelativoRelativo
37519
asked Jan 1 at 19:16
RelativoRelativo
37519
asked Jan 1 at 19:16
RelativoRelativo
37519
37519
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Collect terms in the differential form and combine them into differentials of increasingly complex functions to reduce the number of terms,
$$
0=frac12d(x^2)+y^2dx - xd(y^2)=frac12d(x^2)-x^2d(frac{y^2}x).
$$
With $u=x^2$ and $v=frac{y^2}{x}$ this reads as
$$
0=frac12du-udv=-u,d!left(-frac12ln u+vright),
$$
which gives $F=-frac12ln u+v=-ln|x|+frac{y^2}x$ as first integral. Test by differentiating along a solution
$$
frac{dF}{dt}=-frac{dot x}{x}-frac{y^2dot x}{x^2}+frac{2ydot y}{x}
=-2y-frac{2y^3}{x}+frac{2y(x+y^2)}{x}=0
$$
answered Jan 1 at 20:30
LutzLLutzL
56.8k42054
$endgroup$
add a comment |
$begingroup$
$$frac{dy}{dx}=frac{x+y^2}{2xy}$$
$$2yfrac{dy}{dx}=1+frac{y^2}{x}$$
Let $Y=y^2$
$$frac{dY}{dx}=1+frac{Y}{x}$$
This is a first order linear ODE easy to solve.
$$Y=xln|x|+cx$$
$$y=pmsqrt{xln|x|+cx}$$
answered Jan 2 at 5:35
JJacquelinJJacquelin
42.8k21750
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Collect terms in the differential form and combine them into differentials of increasingly complex functions to reduce the number of terms,
$$
0=frac12d(x^2)+y^2dx - xd(y^2)=frac12d(x^2)-x^2d(frac{y^2}x).
$$
With $u=x^2$ and $v=frac{y^2}{x}$ this reads as
$$
0=frac12du-udv=-u,d!left(-frac12ln u+vright),
$$
which gives $F=-frac12ln u+v=-ln|x|+frac{y^2}x$ as first integral. Test by differentiating along a solution
$$
frac{dF}{dt}=-frac{dot x}{x}-frac{y^2dot x}{x^2}+frac{2ydot y}{x}
=-2y-frac{2y^3}{x}+frac{2y(x+y^2)}{x}=0
$$
answered Jan 1 at 20:30
LutzLLutzL
56.8k42054
$endgroup$
add a comment |
$begingroup$
Collect terms in the differential form and combine them into differentials of increasingly complex functions to reduce the number of terms,
$$
0=frac12d(x^2)+y^2dx - xd(y^2)=frac12d(x^2)-x^2d(frac{y^2}x).
$$
With $u=x^2$ and $v=frac{y^2}{x}$ this reads as
$$
0=frac12du-udv=-u,d!left(-frac12ln u+vright),
$$
which gives $F=-frac12ln u+v=-ln|x|+frac{y^2}x$ as first integral. Test by differentiating along a solution
$$
frac{dF}{dt}=-frac{dot x}{x}-frac{y^2dot x}{x^2}+frac{2ydot y}{x}
=-2y-frac{2y^3}{x}+frac{2y(x+y^2)}{x}=0
$$
answered Jan 1 at 20:30
LutzLLutzL
56.8k42054
$endgroup$
add a comment |
$begingroup$
Collect terms in the differential form and combine them into differentials of increasingly complex functions to reduce the number of terms,
$$
0=frac12d(x^2)+y^2dx - xd(y^2)=frac12d(x^2)-x^2d(frac{y^2}x).
$$
With $u=x^2$ and $v=frac{y^2}{x}$ this reads as
$$
0=frac12du-udv=-u,d!left(-frac12ln u+vright),
$$
which gives $F=-frac12ln u+v=-ln|x|+frac{y^2}x$ as first integral. Test by differentiating along a solution
$$
frac{dF}{dt}=-frac{dot x}{x}-frac{y^2dot x}{x^2}+frac{2ydot y}{x}
=-2y-frac{2y^3}{x}+frac{2y(x+y^2)}{x}=0
$$
answered Jan 1 at 20:30
LutzLLutzL
56.8k42054
$endgroup$
Collect terms in the differential form and combine them into differentials of increasingly complex functions to reduce the number of terms,
$$
0=frac12d(x^2)+y^2dx - xd(y^2)=frac12d(x^2)-x^2d(frac{y^2}x).
$$
With $u=x^2$ and $v=frac{y^2}{x}$ this reads as
$$
0=frac12du-udv=-u,d!left(-frac12ln u+vright),
$$
which gives $F=-frac12ln u+v=-ln|x|+frac{y^2}x$ as first integral. Test by differentiating along a solution
$$
frac{dF}{dt}=-frac{dot x}{x}-frac{y^2dot x}{x^2}+frac{2ydot y}{x}
=-2y-frac{2y^3}{x}+frac{2y(x+y^2)}{x}=0
$$
answered Jan 1 at 20:30
LutzLLutzL
56.8k42054
answered Jan 1 at 20:30
LutzLLutzL
56.8k42054
answered Jan 1 at 20:30
LutzLLutzL
56.8k42054
answered Jan 1 at 20:30
LutzLLutzL
56.8k42054
56.8k42054
add a comment |
add a comment |
$begingroup$
$$frac{dy}{dx}=frac{x+y^2}{2xy}$$
$$2yfrac{dy}{dx}=1+frac{y^2}{x}$$
Let $Y=y^2$
$$frac{dY}{dx}=1+frac{Y}{x}$$
This is a first order linear ODE easy to solve.
$$Y=xln|x|+cx$$
$$y=pmsqrt{xln|x|+cx}$$
answered Jan 2 at 5:35
JJacquelinJJacquelin
42.8k21750
$endgroup$
add a comment |
$begingroup$
$$frac{dy}{dx}=frac{x+y^2}{2xy}$$
$$2yfrac{dy}{dx}=1+frac{y^2}{x}$$
Let $Y=y^2$
$$frac{dY}{dx}=1+frac{Y}{x}$$
This is a first order linear ODE easy to solve.
$$Y=xln|x|+cx$$
$$y=pmsqrt{xln|x|+cx}$$
answered Jan 2 at 5:35
JJacquelinJJacquelin
42.8k21750
$endgroup$
add a comment |
$begingroup$
$$frac{dy}{dx}=frac{x+y^2}{2xy}$$
$$2yfrac{dy}{dx}=1+frac{y^2}{x}$$
Let $Y=y^2$
$$frac{dY}{dx}=1+frac{Y}{x}$$
This is a first order linear ODE easy to solve.
$$Y=xln|x|+cx$$
$$y=pmsqrt{xln|x|+cx}$$
answered Jan 2 at 5:35
JJacquelinJJacquelin
42.8k21750
$endgroup$
$$frac{dy}{dx}=frac{x+y^2}{2xy}$$
$$2yfrac{dy}{dx}=1+frac{y^2}{x}$$
Let $Y=y^2$
$$frac{dY}{dx}=1+frac{Y}{x}$$
This is a first order linear ODE easy to solve.
$$Y=xln|x|+cx$$
$$y=pmsqrt{xln|x|+cx}$$
answered Jan 2 at 5:35
JJacquelinJJacquelin
42.8k21750
answered Jan 2 at 5:35
JJacquelinJJacquelin
42.8k21750
answered Jan 2 at 5:35
JJacquelinJJacquelin
42.8k21750
answered Jan 2 at 5:35
JJacquelinJJacquelin
42.8k21750
42.8k21750
add a comment |
add a comment |
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