Mixing
Explanation of differing solutions for an exact ODE
$begingroup$
I was solving the following ODE
$$ y cos(xy) + x cos(xy) y' = 0$$
My initial intuition to solving it is to observe that it can be factored as:
$$ cos(xy) (y + xy') = 0$$
So over $x,y$ that don't satisfy $cos(xy) = 0$ we have
$$ y + xy' = 0$$
And this is an exact eqation which has solutions of the form
$$ xy = C$$
Now the video series that the ODE originated from takes a different approach. It notices that $$frac{partial}{partial y} [y cos(xy)] = cos(xy)- xy sin(xy) , frac{partial}{partial x} [xcos(xy)] = cos(xy) - xy sin(xy) $$
So therefore the solutions are of the form $ sin(xy) = C$
clearly my approach doesn't agree with the classical approach, but my only difference was dividing out a term which is 0 in a set non-dense in $mathbb{R}^2$ so I don't understand why my answer differs so much. And my intuition that $xy$ should be constant for all points where $cos(xy) ne 0$ can't possibly be true so I feel i'm deeply missing some intuition here.
ordinary-differential-equations pde
asked Jan 27 at 22:11
frogeyedpeasfrogeyedpeas
7,63172054
$endgroup$
add a comment |
$begingroup$
I was solving the following ODE
$$ y cos(xy) + x cos(xy) y' = 0$$
My initial intuition to solving it is to observe that it can be factored as:
$$ cos(xy) (y + xy') = 0$$
So over $x,y$ that don't satisfy $cos(xy) = 0$ we have
$$ y + xy' = 0$$
And this is an exact eqation which has solutions of the form
$$ xy = C$$
Now the video series that the ODE originated from takes a different approach. It notices that $$frac{partial}{partial y} [y cos(xy)] = cos(xy)- xy sin(xy) , frac{partial}{partial x} [xcos(xy)] = cos(xy) - xy sin(xy) $$
So therefore the solutions are of the form $ sin(xy) = C$
clearly my approach doesn't agree with the classical approach, but my only difference was dividing out a term which is 0 in a set non-dense in $mathbb{R}^2$ so I don't understand why my answer differs so much. And my intuition that $xy$ should be constant for all points where $cos(xy) ne 0$ can't possibly be true so I feel i'm deeply missing some intuition here.
ordinary-differential-equations pde
asked Jan 27 at 22:11
frogeyedpeasfrogeyedpeas
7,63172054
$endgroup$
add a comment |
$begingroup$
I was solving the following ODE
$$ y cos(xy) + x cos(xy) y' = 0$$
My initial intuition to solving it is to observe that it can be factored as:
$$ cos(xy) (y + xy') = 0$$
So over $x,y$ that don't satisfy $cos(xy) = 0$ we have
$$ y + xy' = 0$$
And this is an exact eqation which has solutions of the form
$$ xy = C$$
Now the video series that the ODE originated from takes a different approach. It notices that $$frac{partial}{partial y} [y cos(xy)] = cos(xy)- xy sin(xy) , frac{partial}{partial x} [xcos(xy)] = cos(xy) - xy sin(xy) $$
So therefore the solutions are of the form $ sin(xy) = C$
clearly my approach doesn't agree with the classical approach, but my only difference was dividing out a term which is 0 in a set non-dense in $mathbb{R}^2$ so I don't understand why my answer differs so much. And my intuition that $xy$ should be constant for all points where $cos(xy) ne 0$ can't possibly be true so I feel i'm deeply missing some intuition here.
ordinary-differential-equations pde
asked Jan 27 at 22:11
frogeyedpeasfrogeyedpeas
7,63172054
$endgroup$
I was solving the following ODE
$$ y cos(xy) + x cos(xy) y' = 0$$
My initial intuition to solving it is to observe that it can be factored as:
$$ cos(xy) (y + xy') = 0$$
So over $x,y$ that don't satisfy $cos(xy) = 0$ we have
$$ y + xy' = 0$$
And this is an exact eqation which has solutions of the form
$$ xy = C$$
Now the video series that the ODE originated from takes a different approach. It notices that $$frac{partial}{partial y} [y cos(xy)] = cos(xy)- xy sin(xy) , frac{partial}{partial x} [xcos(xy)] = cos(xy) - xy sin(xy) $$
So therefore the solutions are of the form $ sin(xy) = C$
clearly my approach doesn't agree with the classical approach, but my only difference was dividing out a term which is 0 in a set non-dense in $mathbb{R}^2$ so I don't understand why my answer differs so much. And my intuition that $xy$ should be constant for all points where $cos(xy) ne 0$ can't possibly be true so I feel i'm deeply missing some intuition here.
ordinary-differential-equations pde
ordinary-differential-equations pde
asked Jan 27 at 22:11
frogeyedpeasfrogeyedpeas
7,63172054
asked Jan 27 at 22:11
frogeyedpeasfrogeyedpeas
7,63172054
asked Jan 27 at 22:11
frogeyedpeasfrogeyedpeas
7,63172054
asked Jan 27 at 22:11
frogeyedpeasfrogeyedpeas
7,63172054
asked Jan 27 at 22:11
frogeyedpeasfrogeyedpeas
7,63172054
7,63172054
add a comment |
add a comment |
1 Answer
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$begingroup$
Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.
answered Jan 27 at 22:19
Eclipse SunEclipse Sun
7,8401438
$endgroup$
$begingroup$
oh damn, since we can just take $arcsin$ of both sides.
$endgroup$
– frogeyedpeas
Jan 27 at 22:38
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.
answered Jan 27 at 22:19
Eclipse SunEclipse Sun
7,8401438
$endgroup$
$begingroup$
oh damn, since we can just take $arcsin$ of both sides.
$endgroup$
– frogeyedpeas
Jan 27 at 22:38
add a comment |
$begingroup$
Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.
answered Jan 27 at 22:19
Eclipse SunEclipse Sun
7,8401438
$endgroup$
$begingroup$
oh damn, since we can just take $arcsin$ of both sides.
$endgroup$
– frogeyedpeas
Jan 27 at 22:38
add a comment |
$begingroup$
Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.
answered Jan 27 at 22:19
Eclipse SunEclipse Sun
7,8401438
$endgroup$
Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.
answered Jan 27 at 22:19
Eclipse SunEclipse Sun
7,8401438
answered Jan 27 at 22:19
Eclipse SunEclipse Sun
7,8401438
answered Jan 27 at 22:19
Eclipse SunEclipse Sun
7,8401438
answered Jan 27 at 22:19
Eclipse SunEclipse Sun
7,8401438
7,8401438
$begingroup$
oh damn, since we can just take $arcsin$ of both sides.
$endgroup$
– frogeyedpeas
Jan 27 at 22:38
add a comment |
$begingroup$
oh damn, since we can just take $arcsin$ of both sides.
$endgroup$
– frogeyedpeas
Jan 27 at 22:38
$begingroup$
oh damn, since we can just take $arcsin$ of both sides.
$endgroup$
– frogeyedpeas
Jan 27 at 22:38
$begingroup$
oh damn, since we can just take $arcsin$ of both sides.
$endgroup$
– frogeyedpeas
Jan 27 at 22:38
add a comment |
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