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Why is it legitimate to solve the differential equation $frac{dy}{dx}=frac{y}{x}$ by taking $int frac{1}{y}...
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Answers to this question Homogeneous differential equation $frac{dy}{dx} = frac{y}{x}$ solution? assert that to find a solution to the differential equation $$dfrac{dy}{dx} = dfrac{y}{x}$$ we may rearrange and integrate $$int frac{1}{y} dy=int frac{1}{x} dx.$$ If we perform the integration we get $log y=log x+c$ or $$y=kx$$ for constants $c,k in mathbb{R}$. I've seen others use methods like this before too, but I'm unsure why it works.
Question: Why is it legitimate to solve the differential equation in this way?
ordinary-differential-equations
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 4 '13 at 1:03
Rebecca J. StonesRebecca J. Stones
21.1k22781
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add a comment |
$begingroup$
Answers to this question Homogeneous differential equation $frac{dy}{dx} = frac{y}{x}$ solution? assert that to find a solution to the differential equation $$dfrac{dy}{dx} = dfrac{y}{x}$$ we may rearrange and integrate $$int frac{1}{y} dy=int frac{1}{x} dx.$$ If we perform the integration we get $log y=log x+c$ or $$y=kx$$ for constants $c,k in mathbb{R}$. I've seen others use methods like this before too, but I'm unsure why it works.
Question: Why is it legitimate to solve the differential equation in this way?
ordinary-differential-equations
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 4 '13 at 1:03
Rebecca J. StonesRebecca J. Stones
21.1k22781
$endgroup$
$begingroup$
@Amzoti I don't think your link explains why the method is valid. In particular, I think the OP is asking what $dy/y = dx/x$ means. (And if it has no formal meaning, how can the argument be valid?)
$endgroup$
– Trevor Wilson
Sep 4 '13 at 1:15
$begingroup$
Yes the ODE. In other words then, why is it legitimate to solve separable ODEs in this manner? Or perhaps, why does it work? We can manipulate the equation in all sorts of ways, most of which won't be useful; why do this?
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:15
$begingroup$
@Rebecca In short, this calculational technique is the u-substitution theorem which more or less is the integration-version of the chain rule. Ultimately, the legitimacy is evidenced by the success of the method.
$endgroup$
– James S. Cook
Sep 4 '13 at 1:15
1
$begingroup$
@TrevorWilson: Fair enough, she can look at math.bd.psu.edu/faculty/jprevite/251f11/250bookSec2.1.pdf for the actual proof of the method.
$endgroup$
– Amzoti
Sep 4 '13 at 1:17
$begingroup$
It is justified by the chain rule
$endgroup$
– oldrinb
Sep 4 '13 at 12:43
add a comment |
$begingroup$
Answers to this question Homogeneous differential equation $frac{dy}{dx} = frac{y}{x}$ solution? assert that to find a solution to the differential equation $$dfrac{dy}{dx} = dfrac{y}{x}$$ we may rearrange and integrate $$int frac{1}{y} dy=int frac{1}{x} dx.$$ If we perform the integration we get $log y=log x+c$ or $$y=kx$$ for constants $c,k in mathbb{R}$. I've seen others use methods like this before too, but I'm unsure why it works.
Question: Why is it legitimate to solve the differential equation in this way?
ordinary-differential-equations
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 4 '13 at 1:03
Rebecca J. StonesRebecca J. Stones
21.1k22781
$endgroup$
Answers to this question Homogeneous differential equation $frac{dy}{dx} = frac{y}{x}$ solution? assert that to find a solution to the differential equation $$dfrac{dy}{dx} = dfrac{y}{x}$$ we may rearrange and integrate $$int frac{1}{y} dy=int frac{1}{x} dx.$$ If we perform the integration we get $log y=log x+c$ or $$y=kx$$ for constants $c,k in mathbb{R}$. I've seen others use methods like this before too, but I'm unsure why it works.
Question: Why is it legitimate to solve the differential equation in this way?
ordinary-differential-equations
ordinary-differential-equations
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 4 '13 at 1:03
Rebecca J. StonesRebecca J. Stones
21.1k22781
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 4 '13 at 1:03
Rebecca J. StonesRebecca J. Stones
21.1k22781
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Sep 4 '13 at 1:03
Rebecca J. StonesRebecca J. Stones
21.1k22781
asked Sep 4 '13 at 1:03
Rebecca J. StonesRebecca J. Stones
21.1k22781
asked Sep 4 '13 at 1:03
Rebecca J. StonesRebecca J. Stones
21.1k22781
21.1k22781
$begingroup$
@Amzoti I don't think your link explains why the method is valid. In particular, I think the OP is asking what $dy/y = dx/x$ means. (And if it has no formal meaning, how can the argument be valid?)
$endgroup$
– Trevor Wilson
Sep 4 '13 at 1:15
$begingroup$
Yes the ODE. In other words then, why is it legitimate to solve separable ODEs in this manner? Or perhaps, why does it work? We can manipulate the equation in all sorts of ways, most of which won't be useful; why do this?
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:15
$begingroup$
@Rebecca In short, this calculational technique is the u-substitution theorem which more or less is the integration-version of the chain rule. Ultimately, the legitimacy is evidenced by the success of the method.
$endgroup$
– James S. Cook
Sep 4 '13 at 1:15
1
$begingroup$
@TrevorWilson: Fair enough, she can look at math.bd.psu.edu/faculty/jprevite/251f11/250bookSec2.1.pdf for the actual proof of the method.
$endgroup$
– Amzoti
Sep 4 '13 at 1:17
$begingroup$
It is justified by the chain rule
$endgroup$
– oldrinb
Sep 4 '13 at 12:43
add a comment |
$begingroup$
@Amzoti I don't think your link explains why the method is valid. In particular, I think the OP is asking what $dy/y = dx/x$ means. (And if it has no formal meaning, how can the argument be valid?)
$endgroup$
– Trevor Wilson
Sep 4 '13 at 1:15
$begingroup$
Yes the ODE. In other words then, why is it legitimate to solve separable ODEs in this manner? Or perhaps, why does it work? We can manipulate the equation in all sorts of ways, most of which won't be useful; why do this?
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:15
$begingroup$
@Rebecca In short, this calculational technique is the u-substitution theorem which more or less is the integration-version of the chain rule. Ultimately, the legitimacy is evidenced by the success of the method.
$endgroup$
– James S. Cook
Sep 4 '13 at 1:15
1
$begingroup$
@TrevorWilson: Fair enough, she can look at math.bd.psu.edu/faculty/jprevite/251f11/250bookSec2.1.pdf for the actual proof of the method.
$endgroup$
– Amzoti
Sep 4 '13 at 1:17
$begingroup$
It is justified by the chain rule
$endgroup$
– oldrinb
Sep 4 '13 at 12:43
$begingroup$
@Amzoti I don't think your link explains why the method is valid. In particular, I think the OP is asking what $dy/y = dx/x$ means. (And if it has no formal meaning, how can the argument be valid?)
$endgroup$
– Trevor Wilson
Sep 4 '13 at 1:15
$begingroup$
@Amzoti I don't think your link explains why the method is valid. In particular, I think the OP is asking what $dy/y = dx/x$ means. (And if it has no formal meaning, how can the argument be valid?)
$endgroup$
– Trevor Wilson
Sep 4 '13 at 1:15
$begingroup$
Yes the ODE. In other words then, why is it legitimate to solve separable ODEs in this manner? Or perhaps, why does it work? We can manipulate the equation in all sorts of ways, most of which won't be useful; why do this?
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:15
$begingroup$
Yes the ODE. In other words then, why is it legitimate to solve separable ODEs in this manner? Or perhaps, why does it work? We can manipulate the equation in all sorts of ways, most of which won't be useful; why do this?
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:15
$begingroup$
@Rebecca In short, this calculational technique is the u-substitution theorem which more or less is the integration-version of the chain rule. Ultimately, the legitimacy is evidenced by the success of the method.
$endgroup$
– James S. Cook
Sep 4 '13 at 1:15
$begingroup$
@Rebecca In short, this calculational technique is the u-substitution theorem which more or less is the integration-version of the chain rule. Ultimately, the legitimacy is evidenced by the success of the method.
$endgroup$
– James S. Cook
Sep 4 '13 at 1:15
1
1
$begingroup$
@TrevorWilson: Fair enough, she can look at math.bd.psu.edu/faculty/jprevite/251f11/250bookSec2.1.pdf for the actual proof of the method.
$endgroup$
– Amzoti
Sep 4 '13 at 1:17
$begingroup$
@TrevorWilson: Fair enough, she can look at math.bd.psu.edu/faculty/jprevite/251f11/250bookSec2.1.pdf for the actual proof of the method.
$endgroup$
– Amzoti
Sep 4 '13 at 1:17
$begingroup$
It is justified by the chain rule
$endgroup$
– oldrinb
Sep 4 '13 at 12:43
$begingroup$
It is justified by the chain rule
$endgroup$
– oldrinb
Sep 4 '13 at 12:43
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You start with
$$
y'=frac{y}{x}implies frac{y'}{y}=frac{1}{x}impliesintfrac{y'dx}{y}=int frac{dx}{x},
$$
and you make the change of variables in the first integral, which results in what you've written
$$
intfrac{dy}{y}=int frac{dx}{x}
$$
answered Sep 4 '13 at 1:15
ArtemArtem
11.5k32245
$endgroup$
$begingroup$
Thanks! I quite like this answer in particular, as it highlights the change of variables (which seems to be the crux of the technique). Thanks for the other answers too!
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:34
add a comment |
$begingroup$
To be honest I think it's BS to teach separable variables like this without the Riemann-Stieljes integral.
The way I solve them is by doing what actually is done: integrate with respect to $x$ on both sides.
Remember that $y$ is a function (on the variable $x$). So your differential equation is, for all $x$ in a certain interval, $y'(x)=dfrac{y(x)}{x}$ or equivalently $dfrac{y'(x)}{y(x)}=dfrac {1}{x}$and integrating with respect to $x$ you get the desired result.
In my opinion integrating with respect to $y$ is nothing more than a cheap trick, the same way $dfrac{dy}{dx}=1iff dy=dx$ is a cheap trick. It works only because of some higher math.
More generally, if you can rewrite your DE as $g(y(x))y'(x)=f(x)$ for some functions $f$ and $g$ that have antiderivatives, $F$ and $G$, in the given interval, then $g(y(x))y'(x)=f(x)iff G(y(x))=F(x)+C$, for some $Cin Bbb R$. (To establish $Longleftarrow$ just differentiate). And if we're lucky enough for $G$ to be invertible, we get $y(x)=G^{-1}left(F(x)+Cright)$. If $G$ isn't invertible, then hopefully the implicit function theorem will yield the solutions to the DE implicitly by the equation $G(y(x))=F(x)+C$.
In your example $g$ is the function $tmapsto dfrac{1}{t}$ which has $tto log (|t|)$ as an antiderivative. (Don't forget the absolute value).
answered Sep 4 '13 at 1:19
Git GudGit Gud
28.9k1050101
$endgroup$
1
$begingroup$
I agree, and this caused me a lot of confusion when I was first learning ODEs.
$endgroup$
– littleO
Sep 4 '13 at 1:20
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@GitGud Know of a short paper that one can read that will explain this with the needed measure theory you refered to?
$endgroup$
– yiyi
Sep 4 '13 at 1:58
2
$begingroup$
Why is measure theory needed here? Generally the problems encountered in a first course in ODEs have functions nice enough that the Riemann integral suffices to handle them.
$endgroup$
– Potato
Sep 4 '13 at 2:28
1
$begingroup$
But in any case, as @GitGud showed in his answer, we can solve this ODE just fine without using Riemann-Stieltjes integration or anything fancy. (In fact, we only need to know that if $f' = g'$ on some open integral, then $f$ and $g$ differ by a constant on that interval.)
$endgroup$
– littleO
Sep 4 '13 at 20:41
1
$begingroup$
No problem. Nice answer, by the way.
$endgroup$
– Potato
Sep 4 '13 at 21:48
|
show 6 more comments
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I don't like the notation that's often used when solving ODEs. I'd prefer to write the solution like this:
begin{align}
& y'(x) = frac{y(x)}{x} quad text{for all }x > 0 \
implies & frac{y'(x)}{y(x)} = frac{1}{x} quad text{for all }x > 0 \
implies & log y(x) = log x + C quad text{for all } x > 0 ,(text{for some } C in mathbb R).
end{align}
(I'm assuming $y(x) > 0$ for all $x>0$.)
answered Sep 4 '13 at 1:16
littleOlittleO
30.4k648111
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add a comment |
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dy/dx = y/x => (dy/dx)*1/y = 1/x ... integrating both sides wrt to dx gives
∫1/y(dy/dx)*dx = ∫(1/x)dx => ∫(1/y)dy = ∫(1/x)dx
So we are not integrating each side with respect to different variables, but with respect to x, and (dy/dx)*dx = dy, which then allows integration by the separate variables.
answered Jan 31 at 21:30
JoeJoe
1
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That's a good answer! Make it even better by using mathjax formatting to make it more readable :)
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– YiFan
Jan 31 at 21:34
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You start with
$$
y'=frac{y}{x}implies frac{y'}{y}=frac{1}{x}impliesintfrac{y'dx}{y}=int frac{dx}{x},
$$
and you make the change of variables in the first integral, which results in what you've written
$$
intfrac{dy}{y}=int frac{dx}{x}
$$
answered Sep 4 '13 at 1:15
ArtemArtem
11.5k32245
$endgroup$
$begingroup$
Thanks! I quite like this answer in particular, as it highlights the change of variables (which seems to be the crux of the technique). Thanks for the other answers too!
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:34
add a comment |
$begingroup$
You start with
$$
y'=frac{y}{x}implies frac{y'}{y}=frac{1}{x}impliesintfrac{y'dx}{y}=int frac{dx}{x},
$$
and you make the change of variables in the first integral, which results in what you've written
$$
intfrac{dy}{y}=int frac{dx}{x}
$$
answered Sep 4 '13 at 1:15
ArtemArtem
11.5k32245
$endgroup$
$begingroup$
Thanks! I quite like this answer in particular, as it highlights the change of variables (which seems to be the crux of the technique). Thanks for the other answers too!
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:34
add a comment |
$begingroup$
You start with
$$
y'=frac{y}{x}implies frac{y'}{y}=frac{1}{x}impliesintfrac{y'dx}{y}=int frac{dx}{x},
$$
and you make the change of variables in the first integral, which results in what you've written
$$
intfrac{dy}{y}=int frac{dx}{x}
$$
answered Sep 4 '13 at 1:15
ArtemArtem
11.5k32245
$endgroup$
You start with
$$
y'=frac{y}{x}implies frac{y'}{y}=frac{1}{x}impliesintfrac{y'dx}{y}=int frac{dx}{x},
$$
and you make the change of variables in the first integral, which results in what you've written
$$
intfrac{dy}{y}=int frac{dx}{x}
$$
answered Sep 4 '13 at 1:15
ArtemArtem
11.5k32245
answered Sep 4 '13 at 1:15
ArtemArtem
11.5k32245
answered Sep 4 '13 at 1:15
ArtemArtem
11.5k32245
answered Sep 4 '13 at 1:15
ArtemArtem
11.5k32245
11.5k32245
$begingroup$
Thanks! I quite like this answer in particular, as it highlights the change of variables (which seems to be the crux of the technique). Thanks for the other answers too!
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:34
add a comment |
$begingroup$
Thanks! I quite like this answer in particular, as it highlights the change of variables (which seems to be the crux of the technique). Thanks for the other answers too!
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:34
$begingroup$
Thanks! I quite like this answer in particular, as it highlights the change of variables (which seems to be the crux of the technique). Thanks for the other answers too!
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:34
$begingroup$
Thanks! I quite like this answer in particular, as it highlights the change of variables (which seems to be the crux of the technique). Thanks for the other answers too!
$endgroup$
– Rebecca J. Stones
Sep 4 '13 at 1:34
add a comment |
$begingroup$
To be honest I think it's BS to teach separable variables like this without the Riemann-Stieljes integral.
The way I solve them is by doing what actually is done: integrate with respect to $x$ on both sides.
Remember that $y$ is a function (on the variable $x$). So your differential equation is, for all $x$ in a certain interval, $y'(x)=dfrac{y(x)}{x}$ or equivalently $dfrac{y'(x)}{y(x)}=dfrac {1}{x}$and integrating with respect to $x$ you get the desired result.
In my opinion integrating with respect to $y$ is nothing more than a cheap trick, the same way $dfrac{dy}{dx}=1iff dy=dx$ is a cheap trick. It works only because of some higher math.
More generally, if you can rewrite your DE as $g(y(x))y'(x)=f(x)$ for some functions $f$ and $g$ that have antiderivatives, $F$ and $G$, in the given interval, then $g(y(x))y'(x)=f(x)iff G(y(x))=F(x)+C$, for some $Cin Bbb R$. (To establish $Longleftarrow$ just differentiate). And if we're lucky enough for $G$ to be invertible, we get $y(x)=G^{-1}left(F(x)+Cright)$. If $G$ isn't invertible, then hopefully the implicit function theorem will yield the solutions to the DE implicitly by the equation $G(y(x))=F(x)+C$.
In your example $g$ is the function $tmapsto dfrac{1}{t}$ which has $tto log (|t|)$ as an antiderivative. (Don't forget the absolute value).
answered Sep 4 '13 at 1:19
Git GudGit Gud
28.9k1050101
$endgroup$
1
$begingroup$
I agree, and this caused me a lot of confusion when I was first learning ODEs.
$endgroup$
– littleO
Sep 4 '13 at 1:20
$begingroup$
@GitGud Know of a short paper that one can read that will explain this with the needed measure theory you refered to?
$endgroup$
– yiyi
Sep 4 '13 at 1:58
2
$begingroup$
Why is measure theory needed here? Generally the problems encountered in a first course in ODEs have functions nice enough that the Riemann integral suffices to handle them.
$endgroup$
– Potato
Sep 4 '13 at 2:28
1
$begingroup$
But in any case, as @GitGud showed in his answer, we can solve this ODE just fine without using Riemann-Stieltjes integration or anything fancy. (In fact, we only need to know that if $f' = g'$ on some open integral, then $f$ and $g$ differ by a constant on that interval.)
$endgroup$
– littleO
Sep 4 '13 at 20:41
1
$begingroup$
No problem. Nice answer, by the way.
$endgroup$
– Potato
Sep 4 '13 at 21:48
|
show 6 more comments
$begingroup$
To be honest I think it's BS to teach separable variables like this without the Riemann-Stieljes integral.
The way I solve them is by doing what actually is done: integrate with respect to $x$ on both sides.
Remember that $y$ is a function (on the variable $x$). So your differential equation is, for all $x$ in a certain interval, $y'(x)=dfrac{y(x)}{x}$ or equivalently $dfrac{y'(x)}{y(x)}=dfrac {1}{x}$and integrating with respect to $x$ you get the desired result.
In my opinion integrating with respect to $y$ is nothing more than a cheap trick, the same way $dfrac{dy}{dx}=1iff dy=dx$ is a cheap trick. It works only because of some higher math.
More generally, if you can rewrite your DE as $g(y(x))y'(x)=f(x)$ for some functions $f$ and $g$ that have antiderivatives, $F$ and $G$, in the given interval, then $g(y(x))y'(x)=f(x)iff G(y(x))=F(x)+C$, for some $Cin Bbb R$. (To establish $Longleftarrow$ just differentiate). And if we're lucky enough for $G$ to be invertible, we get $y(x)=G^{-1}left(F(x)+Cright)$. If $G$ isn't invertible, then hopefully the implicit function theorem will yield the solutions to the DE implicitly by the equation $G(y(x))=F(x)+C$.
In your example $g$ is the function $tmapsto dfrac{1}{t}$ which has $tto log (|t|)$ as an antiderivative. (Don't forget the absolute value).
answered Sep 4 '13 at 1:19
Git GudGit Gud
28.9k1050101
$endgroup$
1
$begingroup$
I agree, and this caused me a lot of confusion when I was first learning ODEs.
$endgroup$
– littleO
Sep 4 '13 at 1:20
$begingroup$
@GitGud Know of a short paper that one can read that will explain this with the needed measure theory you refered to?
$endgroup$
– yiyi
Sep 4 '13 at 1:58
2
$begingroup$
Why is measure theory needed here? Generally the problems encountered in a first course in ODEs have functions nice enough that the Riemann integral suffices to handle them.
$endgroup$
– Potato
Sep 4 '13 at 2:28
1
$begingroup$
But in any case, as @GitGud showed in his answer, we can solve this ODE just fine without using Riemann-Stieltjes integration or anything fancy. (In fact, we only need to know that if $f' = g'$ on some open integral, then $f$ and $g$ differ by a constant on that interval.)
$endgroup$
– littleO
Sep 4 '13 at 20:41
1
$begingroup$
No problem. Nice answer, by the way.
$endgroup$
– Potato
Sep 4 '13 at 21:48
|
show 6 more comments
$begingroup$
To be honest I think it's BS to teach separable variables like this without the Riemann-Stieljes integral.
The way I solve them is by doing what actually is done: integrate with respect to $x$ on both sides.
Remember that $y$ is a function (on the variable $x$). So your differential equation is, for all $x$ in a certain interval, $y'(x)=dfrac{y(x)}{x}$ or equivalently $dfrac{y'(x)}{y(x)}=dfrac {1}{x}$and integrating with respect to $x$ you get the desired result.
In my opinion integrating with respect to $y$ is nothing more than a cheap trick, the same way $dfrac{dy}{dx}=1iff dy=dx$ is a cheap trick. It works only because of some higher math.
More generally, if you can rewrite your DE as $g(y(x))y'(x)=f(x)$ for some functions $f$ and $g$ that have antiderivatives, $F$ and $G$, in the given interval, then $g(y(x))y'(x)=f(x)iff G(y(x))=F(x)+C$, for some $Cin Bbb R$. (To establish $Longleftarrow$ just differentiate). And if we're lucky enough for $G$ to be invertible, we get $y(x)=G^{-1}left(F(x)+Cright)$. If $G$ isn't invertible, then hopefully the implicit function theorem will yield the solutions to the DE implicitly by the equation $G(y(x))=F(x)+C$.
In your example $g$ is the function $tmapsto dfrac{1}{t}$ which has $tto log (|t|)$ as an antiderivative. (Don't forget the absolute value).
answered Sep 4 '13 at 1:19
Git GudGit Gud
28.9k1050101
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To be honest I think it's BS to teach separable variables like this without the Riemann-Stieljes integral.
The way I solve them is by doing what actually is done: integrate with respect to $x$ on both sides.
Remember that $y$ is a function (on the variable $x$). So your differential equation is, for all $x$ in a certain interval, $y'(x)=dfrac{y(x)}{x}$ or equivalently $dfrac{y'(x)}{y(x)}=dfrac {1}{x}$and integrating with respect to $x$ you get the desired result.
In my opinion integrating with respect to $y$ is nothing more than a cheap trick, the same way $dfrac{dy}{dx}=1iff dy=dx$ is a cheap trick. It works only because of some higher math.
More generally, if you can rewrite your DE as $g(y(x))y'(x)=f(x)$ for some functions $f$ and $g$ that have antiderivatives, $F$ and $G$, in the given interval, then $g(y(x))y'(x)=f(x)iff G(y(x))=F(x)+C$, for some $Cin Bbb R$. (To establish $Longleftarrow$ just differentiate). And if we're lucky enough for $G$ to be invertible, we get $y(x)=G^{-1}left(F(x)+Cright)$. If $G$ isn't invertible, then hopefully the implicit function theorem will yield the solutions to the DE implicitly by the equation $G(y(x))=F(x)+C$.
In your example $g$ is the function $tmapsto dfrac{1}{t}$ which has $tto log (|t|)$ as an antiderivative. (Don't forget the absolute value).
answered Sep 4 '13 at 1:19
Git GudGit Gud
28.9k1050101
edited Sep 4 '13 at 21:24
answered Sep 4 '13 at 1:19
Git GudGit Gud
28.9k1050101
answered Sep 4 '13 at 1:19
Git GudGit Gud
28.9k1050101
answered Sep 4 '13 at 1:19
Git GudGit Gud
28.9k1050101
28.9k1050101
1
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I agree, and this caused me a lot of confusion when I was first learning ODEs.
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– littleO
Sep 4 '13 at 1:20
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@GitGud Know of a short paper that one can read that will explain this with the needed measure theory you refered to?
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– yiyi
Sep 4 '13 at 1:58
2
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Why is measure theory needed here? Generally the problems encountered in a first course in ODEs have functions nice enough that the Riemann integral suffices to handle them.
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– Potato
Sep 4 '13 at 2:28
1
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But in any case, as @GitGud showed in his answer, we can solve this ODE just fine without using Riemann-Stieltjes integration or anything fancy. (In fact, we only need to know that if $f' = g'$ on some open integral, then $f$ and $g$ differ by a constant on that interval.)
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– littleO
Sep 4 '13 at 20:41
1
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No problem. Nice answer, by the way.
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– Potato
Sep 4 '13 at 21:48
|
show 6 more comments
1
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I agree, and this caused me a lot of confusion when I was first learning ODEs.
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– littleO
Sep 4 '13 at 1:20
$begingroup$
@GitGud Know of a short paper that one can read that will explain this with the needed measure theory you refered to?
$endgroup$
– yiyi
Sep 4 '13 at 1:58
2
$begingroup$
Why is measure theory needed here? Generally the problems encountered in a first course in ODEs have functions nice enough that the Riemann integral suffices to handle them.
$endgroup$
– Potato
Sep 4 '13 at 2:28
1
$begingroup$
But in any case, as @GitGud showed in his answer, we can solve this ODE just fine without using Riemann-Stieltjes integration or anything fancy. (In fact, we only need to know that if $f' = g'$ on some open integral, then $f$ and $g$ differ by a constant on that interval.)
$endgroup$
– littleO
Sep 4 '13 at 20:41
1
$begingroup$
No problem. Nice answer, by the way.
$endgroup$
– Potato
Sep 4 '13 at 21:48
1
1
$begingroup$
I agree, and this caused me a lot of confusion when I was first learning ODEs.
$endgroup$
– littleO
Sep 4 '13 at 1:20
$begingroup$
I agree, and this caused me a lot of confusion when I was first learning ODEs.
$endgroup$
– littleO
Sep 4 '13 at 1:20
$begingroup$
@GitGud Know of a short paper that one can read that will explain this with the needed measure theory you refered to?
$endgroup$
– yiyi
Sep 4 '13 at 1:58
$begingroup$
@GitGud Know of a short paper that one can read that will explain this with the needed measure theory you refered to?
$endgroup$
– yiyi
Sep 4 '13 at 1:58
2
2
$begingroup$
Why is measure theory needed here? Generally the problems encountered in a first course in ODEs have functions nice enough that the Riemann integral suffices to handle them.
$endgroup$
– Potato
Sep 4 '13 at 2:28
$begingroup$
Why is measure theory needed here? Generally the problems encountered in a first course in ODEs have functions nice enough that the Riemann integral suffices to handle them.
$endgroup$
– Potato
Sep 4 '13 at 2:28
1
1
$begingroup$
But in any case, as @GitGud showed in his answer, we can solve this ODE just fine without using Riemann-Stieltjes integration or anything fancy. (In fact, we only need to know that if $f' = g'$ on some open integral, then $f$ and $g$ differ by a constant on that interval.)
$endgroup$
– littleO
Sep 4 '13 at 20:41
$begingroup$
But in any case, as @GitGud showed in his answer, we can solve this ODE just fine without using Riemann-Stieltjes integration or anything fancy. (In fact, we only need to know that if $f' = g'$ on some open integral, then $f$ and $g$ differ by a constant on that interval.)
$endgroup$
– littleO
Sep 4 '13 at 20:41
1
1
$begingroup$
No problem. Nice answer, by the way.
$endgroup$
– Potato
Sep 4 '13 at 21:48
$begingroup$
No problem. Nice answer, by the way.
$endgroup$
– Potato
Sep 4 '13 at 21:48
|
show 6 more comments
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I don't like the notation that's often used when solving ODEs. I'd prefer to write the solution like this:
begin{align}
& y'(x) = frac{y(x)}{x} quad text{for all }x > 0 \
implies & frac{y'(x)}{y(x)} = frac{1}{x} quad text{for all }x > 0 \
implies & log y(x) = log x + C quad text{for all } x > 0 ,(text{for some } C in mathbb R).
end{align}
(I'm assuming $y(x) > 0$ for all $x>0$.)
answered Sep 4 '13 at 1:16
littleOlittleO
30.4k648111
$endgroup$
add a comment |
$begingroup$
I don't like the notation that's often used when solving ODEs. I'd prefer to write the solution like this:
begin{align}
& y'(x) = frac{y(x)}{x} quad text{for all }x > 0 \
implies & frac{y'(x)}{y(x)} = frac{1}{x} quad text{for all }x > 0 \
implies & log y(x) = log x + C quad text{for all } x > 0 ,(text{for some } C in mathbb R).
end{align}
(I'm assuming $y(x) > 0$ for all $x>0$.)
answered Sep 4 '13 at 1:16
littleOlittleO
30.4k648111
$endgroup$
add a comment |
$begingroup$
I don't like the notation that's often used when solving ODEs. I'd prefer to write the solution like this:
begin{align}
& y'(x) = frac{y(x)}{x} quad text{for all }x > 0 \
implies & frac{y'(x)}{y(x)} = frac{1}{x} quad text{for all }x > 0 \
implies & log y(x) = log x + C quad text{for all } x > 0 ,(text{for some } C in mathbb R).
end{align}
(I'm assuming $y(x) > 0$ for all $x>0$.)
answered Sep 4 '13 at 1:16
littleOlittleO
30.4k648111
$endgroup$
I don't like the notation that's often used when solving ODEs. I'd prefer to write the solution like this:
begin{align}
& y'(x) = frac{y(x)}{x} quad text{for all }x > 0 \
implies & frac{y'(x)}{y(x)} = frac{1}{x} quad text{for all }x > 0 \
implies & log y(x) = log x + C quad text{for all } x > 0 ,(text{for some } C in mathbb R).
end{align}
(I'm assuming $y(x) > 0$ for all $x>0$.)
answered Sep 4 '13 at 1:16
littleOlittleO
30.4k648111
answered Sep 4 '13 at 1:16
littleOlittleO
30.4k648111
answered Sep 4 '13 at 1:16
littleOlittleO
30.4k648111
answered Sep 4 '13 at 1:16
littleOlittleO
30.4k648111
30.4k648111
add a comment |
add a comment |
$begingroup$
dy/dx = y/x => (dy/dx)*1/y = 1/x ... integrating both sides wrt to dx gives
∫1/y(dy/dx)*dx = ∫(1/x)dx => ∫(1/y)dy = ∫(1/x)dx
So we are not integrating each side with respect to different variables, but with respect to x, and (dy/dx)*dx = dy, which then allows integration by the separate variables.
answered Jan 31 at 21:30
JoeJoe
1
$endgroup$
$begingroup$
That's a good answer! Make it even better by using mathjax formatting to make it more readable :)
$endgroup$
– YiFan
Jan 31 at 21:34
add a comment |
$begingroup$
dy/dx = y/x => (dy/dx)*1/y = 1/x ... integrating both sides wrt to dx gives
∫1/y(dy/dx)*dx = ∫(1/x)dx => ∫(1/y)dy = ∫(1/x)dx
So we are not integrating each side with respect to different variables, but with respect to x, and (dy/dx)*dx = dy, which then allows integration by the separate variables.
answered Jan 31 at 21:30
JoeJoe
1
$endgroup$
$begingroup$
That's a good answer! Make it even better by using mathjax formatting to make it more readable :)
$endgroup$
– YiFan
Jan 31 at 21:34
add a comment |
$begingroup$
dy/dx = y/x => (dy/dx)*1/y = 1/x ... integrating both sides wrt to dx gives
∫1/y(dy/dx)*dx = ∫(1/x)dx => ∫(1/y)dy = ∫(1/x)dx
So we are not integrating each side with respect to different variables, but with respect to x, and (dy/dx)*dx = dy, which then allows integration by the separate variables.
answered Jan 31 at 21:30
JoeJoe
1
$endgroup$
dy/dx = y/x => (dy/dx)*1/y = 1/x ... integrating both sides wrt to dx gives
∫1/y(dy/dx)*dx = ∫(1/x)dx => ∫(1/y)dy = ∫(1/x)dx
So we are not integrating each side with respect to different variables, but with respect to x, and (dy/dx)*dx = dy, which then allows integration by the separate variables.
answered Jan 31 at 21:30
JoeJoe
1
answered Jan 31 at 21:30
JoeJoe
1
answered Jan 31 at 21:30
JoeJoe
1
answered Jan 31 at 21:30
JoeJoe
1
1
$begingroup$
That's a good answer! Make it even better by using mathjax formatting to make it more readable :)
$endgroup$
– YiFan
Jan 31 at 21:34
add a comment |
$begingroup$
That's a good answer! Make it even better by using mathjax formatting to make it more readable :)
$endgroup$
– YiFan
Jan 31 at 21:34
$begingroup$
That's a good answer! Make it even better by using mathjax formatting to make it more readable :)
$endgroup$
– YiFan
Jan 31 at 21:34
$begingroup$
That's a good answer! Make it even better by using mathjax formatting to make it more readable :)
$endgroup$
– YiFan
Jan 31 at 21:34
add a comment |
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$begingroup$
@Amzoti I don't think your link explains why the method is valid. In particular, I think the OP is asking what $dy/y = dx/x$ means. (And if it has no formal meaning, how can the argument be valid?)
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– Trevor Wilson
Sep 4 '13 at 1:15
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Yes the ODE. In other words then, why is it legitimate to solve separable ODEs in this manner? Or perhaps, why does it work? We can manipulate the equation in all sorts of ways, most of which won't be useful; why do this?
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– Rebecca J. Stones
Sep 4 '13 at 1:15
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@Rebecca In short, this calculational technique is the u-substitution theorem which more or less is the integration-version of the chain rule. Ultimately, the legitimacy is evidenced by the success of the method.
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– James S. Cook
Sep 4 '13 at 1:15
1
$begingroup$
@TrevorWilson: Fair enough, she can look at math.bd.psu.edu/faculty/jprevite/251f11/250bookSec2.1.pdf for the actual proof of the method.
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– Amzoti
Sep 4 '13 at 1:17
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It is justified by the chain rule
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– oldrinb
Sep 4 '13 at 12:43