To solve $ frac dydx=frac 1sqrtx^2+y^2$
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How do we solve the differential equation $ dfrac dydx=dfrac 1sqrtx^2+y^2$ ?
calculus integration differential-equations
edited Aug 19 at 10:33
NoChance
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How do we solve the differential equation $ dfrac dydx=dfrac 1sqrtx^2+y^2$ ?
calculus integration differential-equations
edited Aug 19 at 10:33
NoChance
3,34021221
Where did you take this problem ? I am just curious : so simple differential equation leading (to me, at least) to a real nightmare !
– Claude Leibovici
May 8 '14 at 11:48
The solution has to be approximated numerically I guess. Wolfram Differential Equations does not provide a closed form.
– Nicky Hekster
May 8 '14 at 11:52
2
This is equivalent to $x'(y)=sqrtx^2+y^2$.
– Lucian
May 8 '14 at 13:03
1
$x(y)$ grows at least as fast as exponential, since distance to origin is no less than distance from axis. Eventually the y value will be so much less than the x value that asymptotically it should behave like $x = e^y$ or so, no?
– DanielV
May 8 '14 at 13:40
@ClaudeLeibovici Every truly nonlinear ODE is a real nightmare. Let me suggest you one more example: $y''=y^2+x$. If you find even one particular solution analytically, this would be an extremely significant mathematical result.
– Start wearing purple
May 9 '14 at 12:55
 |Â
show 2 more comments
up vote
15
down vote
favorite
up vote
15
down vote
favorite
How do we solve the differential equation $ dfrac dydx=dfrac 1sqrtx^2+y^2$ ?
calculus integration differential-equations
edited Aug 19 at 10:33
NoChance
3,34021221
How do we solve the differential equation $ dfrac dydx=dfrac 1sqrtx^2+y^2$ ?
calculus integration differential-equations
edited Aug 19 at 10:33
NoChance
3,34021221
edited Aug 19 at 10:33
NoChance
3,34021221
edited Aug 19 at 10:33
NoChance
3,34021221
edited Aug 19 at 10:33
NoChance
3,34021221
3,34021221
asked May 8 '14 at 10:59
user123733
Where did you take this problem ? I am just curious : so simple differential equation leading (to me, at least) to a real nightmare !
– Claude Leibovici
May 8 '14 at 11:48
The solution has to be approximated numerically I guess. Wolfram Differential Equations does not provide a closed form.
– Nicky Hekster
May 8 '14 at 11:52
2
This is equivalent to $x'(y)=sqrtx^2+y^2$.
– Lucian
May 8 '14 at 13:03
1
$x(y)$ grows at least as fast as exponential, since distance to origin is no less than distance from axis. Eventually the y value will be so much less than the x value that asymptotically it should behave like $x = e^y$ or so, no?
– DanielV
May 8 '14 at 13:40
@ClaudeLeibovici Every truly nonlinear ODE is a real nightmare. Let me suggest you one more example: $y''=y^2+x$. If you find even one particular solution analytically, this would be an extremely significant mathematical result.
– Start wearing purple
May 9 '14 at 12:55
 |Â
show 2 more comments
Where did you take this problem ? I am just curious : so simple differential equation leading (to me, at least) to a real nightmare !
– Claude Leibovici
May 8 '14 at 11:48
The solution has to be approximated numerically I guess. Wolfram Differential Equations does not provide a closed form.
– Nicky Hekster
May 8 '14 at 11:52
2
This is equivalent to $x'(y)=sqrtx^2+y^2$.
– Lucian
May 8 '14 at 13:03
1
$x(y)$ grows at least as fast as exponential, since distance to origin is no less than distance from axis. Eventually the y value will be so much less than the x value that asymptotically it should behave like $x = e^y$ or so, no?
– DanielV
May 8 '14 at 13:40
@ClaudeLeibovici Every truly nonlinear ODE is a real nightmare. Let me suggest you one more example: $y''=y^2+x$. If you find even one particular solution analytically, this would be an extremely significant mathematical result.
– Start wearing purple
May 9 '14 at 12:55
Where did you take this problem ? I am just curious : so simple differential equation leading (to me, at least) to a real nightmare !
– Claude Leibovici
May 8 '14 at 11:48
Where did you take this problem ? I am just curious : so simple differential equation leading (to me, at least) to a real nightmare !
– Claude Leibovici
May 8 '14 at 11:48
The solution has to be approximated numerically I guess. Wolfram Differential Equations does not provide a closed form.
– Nicky Hekster
May 8 '14 at 11:52
The solution has to be approximated numerically I guess. Wolfram Differential Equations does not provide a closed form.
– Nicky Hekster
May 8 '14 at 11:52
2
2
This is equivalent to $x'(y)=sqrtx^2+y^2$.
– Lucian
May 8 '14 at 13:03
This is equivalent to $x'(y)=sqrtx^2+y^2$.
– Lucian
May 8 '14 at 13:03
1
1
$x(y)$ grows at least as fast as exponential, since distance to origin is no less than distance from axis. Eventually the y value will be so much less than the x value that asymptotically it should behave like $x = e^y$ or so, no?
– DanielV
May 8 '14 at 13:40
$x(y)$ grows at least as fast as exponential, since distance to origin is no less than distance from axis. Eventually the y value will be so much less than the x value that asymptotically it should behave like $x = e^y$ or so, no?
– DanielV
May 8 '14 at 13:40
@ClaudeLeibovici Every truly nonlinear ODE is a real nightmare. Let me suggest you one more example: $y''=y^2+x$. If you find even one particular solution analytically, this would be an extremely significant mathematical result.
– Start wearing purple
May 9 '14 at 12:55
@ClaudeLeibovici Every truly nonlinear ODE is a real nightmare. Let me suggest you one more example: $y''=y^2+x$. If you find even one particular solution analytically, this would be an extremely significant mathematical result.
– Start wearing purple
May 9 '14 at 12:55
 |Â
show 2 more comments
2 Answers
2
active
oldest
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up vote
3
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Since this problem is not amenable to the standard repertoire of 1st order techniques, we might use some asymptotic analysis to understand the qualitative features of this ODE. Firstly, the derivative is always positive and bounded as $x$ or $y$ or both approach $pminfty$. The limiting differential equation in all of these cases is $y'=0$. And so in the asymptotic behavior is constant. On the other hand, the origin is a singularity and so we might ask what is the behavior near $(x,y)=(0,0)$. Suppose we investigate $y=csqrtx$ for $cinmathbbR$ near the origin. Then the differential equation near the origin becomes $y'=frac1sqrtx^2+c^2xapproxfrac1csqrtx$ and so $y(x)approx sqrtx/c$ and we see this matches our supposition up to a multiplicative constant. If we suppose that $y = cx^r$ near the origin and go through the same argument we find that $r=1/2$ necessarily. So the asymptotic behavior near the origin goes as the square root.
answered May 8 '14 at 17:13
JEM
917617
add a comment |Â
up vote
1
down vote
$dfracdydx=dfrac1sqrtx^2+y^2$
$dfracdxdy=sqrtx^2+y^2$
Apply the Euler substitution:
Let $u=x+sqrtx^2+y^2$ ,
Then $x=dfracu2-dfracy^22u$
$dfracdxdy=left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu$
$thereforeleft(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=u-left(dfracu2-dfracy^22uright)$
$left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=dfracu2+dfracy^22u$
$left(dfrac12+dfracy^22u^2right)dfracdudy=dfracu2+dfracy^2+2y2u$
$(u^2+y^2)dfracdudy=u^3+(y^2+2y)u$
Let $v=u^2$ ,
Then $dfracdvdy=2udfracdudy$
$thereforedfracu^2+y^22udfracdvdy=u^3+(y^2+2y)u$
$(u^2+y^2)dfracdvdy=2u^4+(2y^2+4y)u^2$
$(v+y^2)dfracdvdy=2v^2+(2y^2+4y)v$
Let $w=v+y^2$ ,
Then $v=w-y^2$
$dfracdvdy=dfracdwdy-2y$
$therefore wleft(dfracdwdy-2yright)=2(w-y^2)^2+(2y^2+4y)(w-y^2)$
$wdfracdwdy-2yw=2w^2+(4y-2y^2)w-4y^3$
$wdfracdwdy=2w^2+(6y-2y^2)w-4y^3$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=dfrac1z$ ,
Then $dfracdwdy=-dfrac1z^2dfracdzdy$
$therefore-dfrac1z^3dfracdzdy=dfrac2z^2+dfrac6y-2y^2z-4y^3$
$dfracdzdy=4y^3z^3+(2y^2-6y)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered May 10 '14 at 8:51
doraemonpaul
12.1k31660
4
Sorry, I don't understand the conclusion. Have you managed to find the general solution? If yes, could you please write it down explicitly?
– Start wearing purple
May 11 '14 at 22:09
I think OP should be a warned that before trying the method (which in this case will be very complicated and messy) that it is very unlikely that the Abel equation you end up with has a nice (or any) elementary solution. One indication of this (but does not prove anything) is that none of the best mathematical software packages (Mathematica, Maple) we have today is unable to solve it.
– Winther
Feb 6 '15 at 16:28
Consider a substitution such as $x=y tan(u)$ as suggested by Symbolab web site.
– NoChance
Aug 19 at 10:31
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Since this problem is not amenable to the standard repertoire of 1st order techniques, we might use some asymptotic analysis to understand the qualitative features of this ODE. Firstly, the derivative is always positive and bounded as $x$ or $y$ or both approach $pminfty$. The limiting differential equation in all of these cases is $y'=0$. And so in the asymptotic behavior is constant. On the other hand, the origin is a singularity and so we might ask what is the behavior near $(x,y)=(0,0)$. Suppose we investigate $y=csqrtx$ for $cinmathbbR$ near the origin. Then the differential equation near the origin becomes $y'=frac1sqrtx^2+c^2xapproxfrac1csqrtx$ and so $y(x)approx sqrtx/c$ and we see this matches our supposition up to a multiplicative constant. If we suppose that $y = cx^r$ near the origin and go through the same argument we find that $r=1/2$ necessarily. So the asymptotic behavior near the origin goes as the square root.
answered May 8 '14 at 17:13
JEM
917617
add a comment |Â
up vote
3
down vote
Since this problem is not amenable to the standard repertoire of 1st order techniques, we might use some asymptotic analysis to understand the qualitative features of this ODE. Firstly, the derivative is always positive and bounded as $x$ or $y$ or both approach $pminfty$. The limiting differential equation in all of these cases is $y'=0$. And so in the asymptotic behavior is constant. On the other hand, the origin is a singularity and so we might ask what is the behavior near $(x,y)=(0,0)$. Suppose we investigate $y=csqrtx$ for $cinmathbbR$ near the origin. Then the differential equation near the origin becomes $y'=frac1sqrtx^2+c^2xapproxfrac1csqrtx$ and so $y(x)approx sqrtx/c$ and we see this matches our supposition up to a multiplicative constant. If we suppose that $y = cx^r$ near the origin and go through the same argument we find that $r=1/2$ necessarily. So the asymptotic behavior near the origin goes as the square root.
answered May 8 '14 at 17:13
JEM
917617
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Since this problem is not amenable to the standard repertoire of 1st order techniques, we might use some asymptotic analysis to understand the qualitative features of this ODE. Firstly, the derivative is always positive and bounded as $x$ or $y$ or both approach $pminfty$. The limiting differential equation in all of these cases is $y'=0$. And so in the asymptotic behavior is constant. On the other hand, the origin is a singularity and so we might ask what is the behavior near $(x,y)=(0,0)$. Suppose we investigate $y=csqrtx$ for $cinmathbbR$ near the origin. Then the differential equation near the origin becomes $y'=frac1sqrtx^2+c^2xapproxfrac1csqrtx$ and so $y(x)approx sqrtx/c$ and we see this matches our supposition up to a multiplicative constant. If we suppose that $y = cx^r$ near the origin and go through the same argument we find that $r=1/2$ necessarily. So the asymptotic behavior near the origin goes as the square root.
answered May 8 '14 at 17:13
JEM
917617
Since this problem is not amenable to the standard repertoire of 1st order techniques, we might use some asymptotic analysis to understand the qualitative features of this ODE. Firstly, the derivative is always positive and bounded as $x$ or $y$ or both approach $pminfty$. The limiting differential equation in all of these cases is $y'=0$. And so in the asymptotic behavior is constant. On the other hand, the origin is a singularity and so we might ask what is the behavior near $(x,y)=(0,0)$. Suppose we investigate $y=csqrtx$ for $cinmathbbR$ near the origin. Then the differential equation near the origin becomes $y'=frac1sqrtx^2+c^2xapproxfrac1csqrtx$ and so $y(x)approx sqrtx/c$ and we see this matches our supposition up to a multiplicative constant. If we suppose that $y = cx^r$ near the origin and go through the same argument we find that $r=1/2$ necessarily. So the asymptotic behavior near the origin goes as the square root.
answered May 8 '14 at 17:13
JEM
917617
answered May 8 '14 at 17:13
JEM
917617
answered May 8 '14 at 17:13
JEM
917617
answered May 8 '14 at 17:13
JEM
917617
917617
add a comment |Â
add a comment |Â
up vote
1
down vote
$dfracdydx=dfrac1sqrtx^2+y^2$
$dfracdxdy=sqrtx^2+y^2$
Apply the Euler substitution:
Let $u=x+sqrtx^2+y^2$ ,
Then $x=dfracu2-dfracy^22u$
$dfracdxdy=left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu$
$thereforeleft(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=u-left(dfracu2-dfracy^22uright)$
$left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=dfracu2+dfracy^22u$
$left(dfrac12+dfracy^22u^2right)dfracdudy=dfracu2+dfracy^2+2y2u$
$(u^2+y^2)dfracdudy=u^3+(y^2+2y)u$
Let $v=u^2$ ,
Then $dfracdvdy=2udfracdudy$
$thereforedfracu^2+y^22udfracdvdy=u^3+(y^2+2y)u$
$(u^2+y^2)dfracdvdy=2u^4+(2y^2+4y)u^2$
$(v+y^2)dfracdvdy=2v^2+(2y^2+4y)v$
Let $w=v+y^2$ ,
Then $v=w-y^2$
$dfracdvdy=dfracdwdy-2y$
$therefore wleft(dfracdwdy-2yright)=2(w-y^2)^2+(2y^2+4y)(w-y^2)$
$wdfracdwdy-2yw=2w^2+(4y-2y^2)w-4y^3$
$wdfracdwdy=2w^2+(6y-2y^2)w-4y^3$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=dfrac1z$ ,
Then $dfracdwdy=-dfrac1z^2dfracdzdy$
$therefore-dfrac1z^3dfracdzdy=dfrac2z^2+dfrac6y-2y^2z-4y^3$
$dfracdzdy=4y^3z^3+(2y^2-6y)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered May 10 '14 at 8:51
doraemonpaul
12.1k31660
4
Sorry, I don't understand the conclusion. Have you managed to find the general solution? If yes, could you please write it down explicitly?
– Start wearing purple
May 11 '14 at 22:09
I think OP should be a warned that before trying the method (which in this case will be very complicated and messy) that it is very unlikely that the Abel equation you end up with has a nice (or any) elementary solution. One indication of this (but does not prove anything) is that none of the best mathematical software packages (Mathematica, Maple) we have today is unable to solve it.
– Winther
Feb 6 '15 at 16:28
Consider a substitution such as $x=y tan(u)$ as suggested by Symbolab web site.
– NoChance
Aug 19 at 10:31
add a comment |Â
up vote
1
down vote
$dfracdydx=dfrac1sqrtx^2+y^2$
$dfracdxdy=sqrtx^2+y^2$
Apply the Euler substitution:
Let $u=x+sqrtx^2+y^2$ ,
Then $x=dfracu2-dfracy^22u$
$dfracdxdy=left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu$
$thereforeleft(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=u-left(dfracu2-dfracy^22uright)$
$left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=dfracu2+dfracy^22u$
$left(dfrac12+dfracy^22u^2right)dfracdudy=dfracu2+dfracy^2+2y2u$
$(u^2+y^2)dfracdudy=u^3+(y^2+2y)u$
Let $v=u^2$ ,
Then $dfracdvdy=2udfracdudy$
$thereforedfracu^2+y^22udfracdvdy=u^3+(y^2+2y)u$
$(u^2+y^2)dfracdvdy=2u^4+(2y^2+4y)u^2$
$(v+y^2)dfracdvdy=2v^2+(2y^2+4y)v$
Let $w=v+y^2$ ,
Then $v=w-y^2$
$dfracdvdy=dfracdwdy-2y$
$therefore wleft(dfracdwdy-2yright)=2(w-y^2)^2+(2y^2+4y)(w-y^2)$
$wdfracdwdy-2yw=2w^2+(4y-2y^2)w-4y^3$
$wdfracdwdy=2w^2+(6y-2y^2)w-4y^3$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=dfrac1z$ ,
Then $dfracdwdy=-dfrac1z^2dfracdzdy$
$therefore-dfrac1z^3dfracdzdy=dfrac2z^2+dfrac6y-2y^2z-4y^3$
$dfracdzdy=4y^3z^3+(2y^2-6y)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered May 10 '14 at 8:51
doraemonpaul
12.1k31660
4
Sorry, I don't understand the conclusion. Have you managed to find the general solution? If yes, could you please write it down explicitly?
– Start wearing purple
May 11 '14 at 22:09
I think OP should be a warned that before trying the method (which in this case will be very complicated and messy) that it is very unlikely that the Abel equation you end up with has a nice (or any) elementary solution. One indication of this (but does not prove anything) is that none of the best mathematical software packages (Mathematica, Maple) we have today is unable to solve it.
– Winther
Feb 6 '15 at 16:28
Consider a substitution such as $x=y tan(u)$ as suggested by Symbolab web site.
– NoChance
Aug 19 at 10:31
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$dfracdydx=dfrac1sqrtx^2+y^2$
$dfracdxdy=sqrtx^2+y^2$
Apply the Euler substitution:
Let $u=x+sqrtx^2+y^2$ ,
Then $x=dfracu2-dfracy^22u$
$dfracdxdy=left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu$
$thereforeleft(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=u-left(dfracu2-dfracy^22uright)$
$left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=dfracu2+dfracy^22u$
$left(dfrac12+dfracy^22u^2right)dfracdudy=dfracu2+dfracy^2+2y2u$
$(u^2+y^2)dfracdudy=u^3+(y^2+2y)u$
Let $v=u^2$ ,
Then $dfracdvdy=2udfracdudy$
$thereforedfracu^2+y^22udfracdvdy=u^3+(y^2+2y)u$
$(u^2+y^2)dfracdvdy=2u^4+(2y^2+4y)u^2$
$(v+y^2)dfracdvdy=2v^2+(2y^2+4y)v$
Let $w=v+y^2$ ,
Then $v=w-y^2$
$dfracdvdy=dfracdwdy-2y$
$therefore wleft(dfracdwdy-2yright)=2(w-y^2)^2+(2y^2+4y)(w-y^2)$
$wdfracdwdy-2yw=2w^2+(4y-2y^2)w-4y^3$
$wdfracdwdy=2w^2+(6y-2y^2)w-4y^3$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=dfrac1z$ ,
Then $dfracdwdy=-dfrac1z^2dfracdzdy$
$therefore-dfrac1z^3dfracdzdy=dfrac2z^2+dfrac6y-2y^2z-4y^3$
$dfracdzdy=4y^3z^3+(2y^2-6y)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered May 10 '14 at 8:51
doraemonpaul
12.1k31660
$dfracdydx=dfrac1sqrtx^2+y^2$
$dfracdxdy=sqrtx^2+y^2$
Apply the Euler substitution:
Let $u=x+sqrtx^2+y^2$ ,
Then $x=dfracu2-dfracy^22u$
$dfracdxdy=left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu$
$thereforeleft(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=u-left(dfracu2-dfracy^22uright)$
$left(dfrac12+dfracy^22u^2right)dfracdudy-dfracyu=dfracu2+dfracy^22u$
$left(dfrac12+dfracy^22u^2right)dfracdudy=dfracu2+dfracy^2+2y2u$
$(u^2+y^2)dfracdudy=u^3+(y^2+2y)u$
Let $v=u^2$ ,
Then $dfracdvdy=2udfracdudy$
$thereforedfracu^2+y^22udfracdvdy=u^3+(y^2+2y)u$
$(u^2+y^2)dfracdvdy=2u^4+(2y^2+4y)u^2$
$(v+y^2)dfracdvdy=2v^2+(2y^2+4y)v$
Let $w=v+y^2$ ,
Then $v=w-y^2$
$dfracdvdy=dfracdwdy-2y$
$therefore wleft(dfracdwdy-2yright)=2(w-y^2)^2+(2y^2+4y)(w-y^2)$
$wdfracdwdy-2yw=2w^2+(4y-2y^2)w-4y^3$
$wdfracdwdy=2w^2+(6y-2y^2)w-4y^3$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=dfrac1z$ ,
Then $dfracdwdy=-dfrac1z^2dfracdzdy$
$therefore-dfrac1z^3dfracdzdy=dfrac2z^2+dfrac6y-2y^2z-4y^3$
$dfracdzdy=4y^3z^3+(2y^2-6y)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered May 10 '14 at 8:51
doraemonpaul
12.1k31660
edited Feb 11 '16 at 20:48
answered May 10 '14 at 8:51
doraemonpaul
12.1k31660
answered May 10 '14 at 8:51
doraemonpaul
12.1k31660
answered May 10 '14 at 8:51
doraemonpaul
12.1k31660
12.1k31660
4
Sorry, I don't understand the conclusion. Have you managed to find the general solution? If yes, could you please write it down explicitly?
– Start wearing purple
May 11 '14 at 22:09
I think OP should be a warned that before trying the method (which in this case will be very complicated and messy) that it is very unlikely that the Abel equation you end up with has a nice (or any) elementary solution. One indication of this (but does not prove anything) is that none of the best mathematical software packages (Mathematica, Maple) we have today is unable to solve it.
– Winther
Feb 6 '15 at 16:28
Consider a substitution such as $x=y tan(u)$ as suggested by Symbolab web site.
– NoChance
Aug 19 at 10:31
add a comment |Â
4
Sorry, I don't understand the conclusion. Have you managed to find the general solution? If yes, could you please write it down explicitly?
– Start wearing purple
May 11 '14 at 22:09
I think OP should be a warned that before trying the method (which in this case will be very complicated and messy) that it is very unlikely that the Abel equation you end up with has a nice (or any) elementary solution. One indication of this (but does not prove anything) is that none of the best mathematical software packages (Mathematica, Maple) we have today is unable to solve it.
– Winther
Feb 6 '15 at 16:28
Consider a substitution such as $x=y tan(u)$ as suggested by Symbolab web site.
– NoChance
Aug 19 at 10:31
4
4
Sorry, I don't understand the conclusion. Have you managed to find the general solution? If yes, could you please write it down explicitly?
– Start wearing purple
May 11 '14 at 22:09
Sorry, I don't understand the conclusion. Have you managed to find the general solution? If yes, could you please write it down explicitly?
– Start wearing purple
May 11 '14 at 22:09
I think OP should be a warned that before trying the method (which in this case will be very complicated and messy) that it is very unlikely that the Abel equation you end up with has a nice (or any) elementary solution. One indication of this (but does not prove anything) is that none of the best mathematical software packages (Mathematica, Maple) we have today is unable to solve it.
– Winther
Feb 6 '15 at 16:28
I think OP should be a warned that before trying the method (which in this case will be very complicated and messy) that it is very unlikely that the Abel equation you end up with has a nice (or any) elementary solution. One indication of this (but does not prove anything) is that none of the best mathematical software packages (Mathematica, Maple) we have today is unable to solve it.
– Winther
Feb 6 '15 at 16:28
Consider a substitution such as $x=y tan(u)$ as suggested by Symbolab web site.
– NoChance
Aug 19 at 10:31
Consider a substitution such as $x=y tan(u)$ as suggested by Symbolab web site.
– NoChance
Aug 19 at 10:31
add a comment |Â
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Where did you take this problem ? I am just curious : so simple differential equation leading (to me, at least) to a real nightmare !
– Claude Leibovici
May 8 '14 at 11:48
The solution has to be approximated numerically I guess. Wolfram Differential Equations does not provide a closed form.
– Nicky Hekster
May 8 '14 at 11:52
2
This is equivalent to $x'(y)=sqrtx^2+y^2$.
– Lucian
May 8 '14 at 13:03
1
$x(y)$ grows at least as fast as exponential, since distance to origin is no less than distance from axis. Eventually the y value will be so much less than the x value that asymptotically it should behave like $x = e^y$ or so, no?
– DanielV
May 8 '14 at 13:40
@ClaudeLeibovici Every truly nonlinear ODE is a real nightmare. Let me suggest you one more example: $y''=y^2+x$. If you find even one particular solution analytically, this would be an extremely significant mathematical result.
– Start wearing purple
May 9 '14 at 12:55