$dy/dx=sqrtx^2+y^2$
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up vote
6
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favorite
$$fracdydx=sqrtx^2+y^2$$
slope=distance from origin, should be simple and interesting. May have no solution!
I have tried several approaches, best:
$(fracdydx-y)(fracdydx+y)=x^2$ multiply by $e(-x) * e(+x)$ as integrating factor. Substitute $frac12x^2=t$.
Second approach:
$y=xsinh(u)$ and $x=e(t)$ yields $fracdudt + tanh(u)=e(t)$.
Sorry, I am not yet using the proper format.
differential-equations
asked Aug 18 at 8:09
Craig Stevens
362
 |Â
show 3 more comments
up vote
6
down vote
favorite
$$fracdydx=sqrtx^2+y^2$$
slope=distance from origin, should be simple and interesting. May have no solution!
I have tried several approaches, best:
$(fracdydx-y)(fracdydx+y)=x^2$ multiply by $e(-x) * e(+x)$ as integrating factor. Substitute $frac12x^2=t$.
Second approach:
$y=xsinh(u)$ and $x=e(t)$ yields $fracdudt + tanh(u)=e(t)$.
Sorry, I am not yet using the proper format.
differential-equations
asked Aug 18 at 8:09
Craig Stevens
362
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 18 at 8:16
You should be able to get the first few terms in $y=a_0+a_1x+a_2x^2+ldots$ with all the $a_i$ functions of $a_0$.
– Empy2
Aug 18 at 8:25
1
Mathematica does not find an explicit solution. Probably there is no solution in terms of elementary (or even special) functions.
– Julián Aguirre
Aug 18 at 10:52
1
Yes, all tried so far. Polar coordinates don't help. Power series approach didn't yield result. Once y>x, then dy/dx=sqrt(2) * Y yields upward bound of y=exp(sqrt 2) * x, so no singularities. Multiplying both sides by y, then substituting y*y=v yielded a similar interesting differential equation.
– Craig Stevens
Aug 18 at 14:39
1
The substitution $t=y/x, r=sqrtx^2+y^2$ gives $dr(t^2+1)(r-t)=dt(r+tr^2)$ which at least lacks a square-root.
– Empy2
Aug 18 at 15:47
 |Â
show 3 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
$$fracdydx=sqrtx^2+y^2$$
slope=distance from origin, should be simple and interesting. May have no solution!
I have tried several approaches, best:
$(fracdydx-y)(fracdydx+y)=x^2$ multiply by $e(-x) * e(+x)$ as integrating factor. Substitute $frac12x^2=t$.
Second approach:
$y=xsinh(u)$ and $x=e(t)$ yields $fracdudt + tanh(u)=e(t)$.
Sorry, I am not yet using the proper format.
differential-equations
asked Aug 18 at 8:09
Craig Stevens
362
$$fracdydx=sqrtx^2+y^2$$
slope=distance from origin, should be simple and interesting. May have no solution!
I have tried several approaches, best:
$(fracdydx-y)(fracdydx+y)=x^2$ multiply by $e(-x) * e(+x)$ as integrating factor. Substitute $frac12x^2=t$.
Second approach:
$y=xsinh(u)$ and $x=e(t)$ yields $fracdudt + tanh(u)=e(t)$.
Sorry, I am not yet using the proper format.
differential-equations
asked Aug 18 at 8:09
Craig Stevens
362
edited Aug 20 at 12:25
asked Aug 18 at 8:09
Craig Stevens
362
asked Aug 18 at 8:09
Craig Stevens
362
asked Aug 18 at 8:09
Craig Stevens
362
362
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 18 at 8:16
You should be able to get the first few terms in $y=a_0+a_1x+a_2x^2+ldots$ with all the $a_i$ functions of $a_0$.
– Empy2
Aug 18 at 8:25
1
Mathematica does not find an explicit solution. Probably there is no solution in terms of elementary (or even special) functions.
– Julián Aguirre
Aug 18 at 10:52
1
Yes, all tried so far. Polar coordinates don't help. Power series approach didn't yield result. Once y>x, then dy/dx=sqrt(2) * Y yields upward bound of y=exp(sqrt 2) * x, so no singularities. Multiplying both sides by y, then substituting y*y=v yielded a similar interesting differential equation.
– Craig Stevens
Aug 18 at 14:39
1
The substitution $t=y/x, r=sqrtx^2+y^2$ gives $dr(t^2+1)(r-t)=dt(r+tr^2)$ which at least lacks a square-root.
– Empy2
Aug 18 at 15:47
 |Â
show 3 more comments
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 18 at 8:16
You should be able to get the first few terms in $y=a_0+a_1x+a_2x^2+ldots$ with all the $a_i$ functions of $a_0$.
– Empy2
Aug 18 at 8:25
1
Mathematica does not find an explicit solution. Probably there is no solution in terms of elementary (or even special) functions.
– Julián Aguirre
Aug 18 at 10:52
1
Yes, all tried so far. Polar coordinates don't help. Power series approach didn't yield result. Once y>x, then dy/dx=sqrt(2) * Y yields upward bound of y=exp(sqrt 2) * x, so no singularities. Multiplying both sides by y, then substituting y*y=v yielded a similar interesting differential equation.
– Craig Stevens
Aug 18 at 14:39
1
The substitution $t=y/x, r=sqrtx^2+y^2$ gives $dr(t^2+1)(r-t)=dt(r+tr^2)$ which at least lacks a square-root.
– Empy2
Aug 18 at 15:47
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 18 at 8:16
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 18 at 8:16
You should be able to get the first few terms in $y=a_0+a_1x+a_2x^2+ldots$ with all the $a_i$ functions of $a_0$.
– Empy2
Aug 18 at 8:25
You should be able to get the first few terms in $y=a_0+a_1x+a_2x^2+ldots$ with all the $a_i$ functions of $a_0$.
– Empy2
Aug 18 at 8:25
1
1
Mathematica does not find an explicit solution. Probably there is no solution in terms of elementary (or even special) functions.
– Julián Aguirre
Aug 18 at 10:52
Mathematica does not find an explicit solution. Probably there is no solution in terms of elementary (or even special) functions.
– Julián Aguirre
Aug 18 at 10:52
1
1
Yes, all tried so far. Polar coordinates don't help. Power series approach didn't yield result. Once y>x, then dy/dx=sqrt(2) * Y yields upward bound of y=exp(sqrt 2) * x, so no singularities. Multiplying both sides by y, then substituting y*y=v yielded a similar interesting differential equation.
– Craig Stevens
Aug 18 at 14:39
Yes, all tried so far. Polar coordinates don't help. Power series approach didn't yield result. Once y>x, then dy/dx=sqrt(2) * Y yields upward bound of y=exp(sqrt 2) * x, so no singularities. Multiplying both sides by y, then substituting y*y=v yielded a similar interesting differential equation.
– Craig Stevens
Aug 18 at 14:39
1
1
The substitution $t=y/x, r=sqrtx^2+y^2$ gives $dr(t^2+1)(r-t)=dt(r+tr^2)$ which at least lacks a square-root.
– Empy2
Aug 18 at 15:47
The substitution $t=y/x, r=sqrtx^2+y^2$ gives $dr(t^2+1)(r-t)=dt(r+tr^2)$ which at least lacks a square-root.
– Empy2
Aug 18 at 15:47
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
Similar to To solve $ frac dydx=frac 1sqrtx^2+y^2$:
Apply the Euler substitution:
Let $u=y+sqrtx^2+y^2$ ,
Then $y=dfracu2-dfracx^22u$
$dfracdydx=left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu$
$thereforeleft(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=u-left(dfracu2-dfracx^22uright)$
$left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=dfracu2+dfracx^22u$
$left(dfrac12+dfracx^22u^2right)dfracdudx=dfracu2+dfracx^2+2x2u$
$(u^2+x^2)dfracdudx=u^3+(x^2+2x)u$
Let $v=u^2$ ,
Then $dfracdvdx=2udfracdudx$
$thereforedfracu^2+x^22udfracdvdx=u^3+(x^2+2x)u$
$(u^2+x^2)dfracdvdx=2u^4+(2x^2+4x)u^2$
$(v+x^2)dfracdvdx=2v^2+(2x^2+4x)v$
Let $w=v+x^2$ ,
Then $v=w-x^2$
$dfracdvdx=dfracdwdx-2x$
$therefore wleft(dfracdwdx-2xright)=2(w-x^2)^2+(2x^2+4x)(w-x^2)$
$wdfracdwdx-2xw=2w^2+(4x-2x^2)w-4x^3$
$wdfracdwdx=2w^2+(6x-2x^2)w-4x^3$
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 $dfracdwdx=-dfrac1z^2dfracdzdx$
$therefore-dfrac1z^3dfracdzdx=dfrac2z^2+dfrac6x-2x^2z-4x^3$
$dfracdzdx=4x^3z^3+(2x^2-6x)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered Aug 19 at 7:42
doraemonpaul
12.1k31660
Thank you doraemonnpaul. Apparently, from your citation, this class of equation was considered unsolvable until 2011. I was a math major 40 years ago, before entering the field of medicine. Differential equations have always remained a hobby of mine, especially nonlinear. This particular equation has fascinated me for the last few years. Thank you so much for your expertise in providing the solution.
– Craig Stevens
Aug 20 at 12:11
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Similar to To solve $ frac dydx=frac 1sqrtx^2+y^2$:
Apply the Euler substitution:
Let $u=y+sqrtx^2+y^2$ ,
Then $y=dfracu2-dfracx^22u$
$dfracdydx=left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu$
$thereforeleft(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=u-left(dfracu2-dfracx^22uright)$
$left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=dfracu2+dfracx^22u$
$left(dfrac12+dfracx^22u^2right)dfracdudx=dfracu2+dfracx^2+2x2u$
$(u^2+x^2)dfracdudx=u^3+(x^2+2x)u$
Let $v=u^2$ ,
Then $dfracdvdx=2udfracdudx$
$thereforedfracu^2+x^22udfracdvdx=u^3+(x^2+2x)u$
$(u^2+x^2)dfracdvdx=2u^4+(2x^2+4x)u^2$
$(v+x^2)dfracdvdx=2v^2+(2x^2+4x)v$
Let $w=v+x^2$ ,
Then $v=w-x^2$
$dfracdvdx=dfracdwdx-2x$
$therefore wleft(dfracdwdx-2xright)=2(w-x^2)^2+(2x^2+4x)(w-x^2)$
$wdfracdwdx-2xw=2w^2+(4x-2x^2)w-4x^3$
$wdfracdwdx=2w^2+(6x-2x^2)w-4x^3$
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 $dfracdwdx=-dfrac1z^2dfracdzdx$
$therefore-dfrac1z^3dfracdzdx=dfrac2z^2+dfrac6x-2x^2z-4x^3$
$dfracdzdx=4x^3z^3+(2x^2-6x)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered Aug 19 at 7:42
doraemonpaul
12.1k31660
Thank you doraemonnpaul. Apparently, from your citation, this class of equation was considered unsolvable until 2011. I was a math major 40 years ago, before entering the field of medicine. Differential equations have always remained a hobby of mine, especially nonlinear. This particular equation has fascinated me for the last few years. Thank you so much for your expertise in providing the solution.
– Craig Stevens
Aug 20 at 12:11
add a comment |Â
up vote
1
down vote
Similar to To solve $ frac dydx=frac 1sqrtx^2+y^2$:
Apply the Euler substitution:
Let $u=y+sqrtx^2+y^2$ ,
Then $y=dfracu2-dfracx^22u$
$dfracdydx=left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu$
$thereforeleft(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=u-left(dfracu2-dfracx^22uright)$
$left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=dfracu2+dfracx^22u$
$left(dfrac12+dfracx^22u^2right)dfracdudx=dfracu2+dfracx^2+2x2u$
$(u^2+x^2)dfracdudx=u^3+(x^2+2x)u$
Let $v=u^2$ ,
Then $dfracdvdx=2udfracdudx$
$thereforedfracu^2+x^22udfracdvdx=u^3+(x^2+2x)u$
$(u^2+x^2)dfracdvdx=2u^4+(2x^2+4x)u^2$
$(v+x^2)dfracdvdx=2v^2+(2x^2+4x)v$
Let $w=v+x^2$ ,
Then $v=w-x^2$
$dfracdvdx=dfracdwdx-2x$
$therefore wleft(dfracdwdx-2xright)=2(w-x^2)^2+(2x^2+4x)(w-x^2)$
$wdfracdwdx-2xw=2w^2+(4x-2x^2)w-4x^3$
$wdfracdwdx=2w^2+(6x-2x^2)w-4x^3$
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 $dfracdwdx=-dfrac1z^2dfracdzdx$
$therefore-dfrac1z^3dfracdzdx=dfrac2z^2+dfrac6x-2x^2z-4x^3$
$dfracdzdx=4x^3z^3+(2x^2-6x)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered Aug 19 at 7:42
doraemonpaul
12.1k31660
Thank you doraemonnpaul. Apparently, from your citation, this class of equation was considered unsolvable until 2011. I was a math major 40 years ago, before entering the field of medicine. Differential equations have always remained a hobby of mine, especially nonlinear. This particular equation has fascinated me for the last few years. Thank you so much for your expertise in providing the solution.
– Craig Stevens
Aug 20 at 12:11
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Similar to To solve $ frac dydx=frac 1sqrtx^2+y^2$:
Apply the Euler substitution:
Let $u=y+sqrtx^2+y^2$ ,
Then $y=dfracu2-dfracx^22u$
$dfracdydx=left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu$
$thereforeleft(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=u-left(dfracu2-dfracx^22uright)$
$left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=dfracu2+dfracx^22u$
$left(dfrac12+dfracx^22u^2right)dfracdudx=dfracu2+dfracx^2+2x2u$
$(u^2+x^2)dfracdudx=u^3+(x^2+2x)u$
Let $v=u^2$ ,
Then $dfracdvdx=2udfracdudx$
$thereforedfracu^2+x^22udfracdvdx=u^3+(x^2+2x)u$
$(u^2+x^2)dfracdvdx=2u^4+(2x^2+4x)u^2$
$(v+x^2)dfracdvdx=2v^2+(2x^2+4x)v$
Let $w=v+x^2$ ,
Then $v=w-x^2$
$dfracdvdx=dfracdwdx-2x$
$therefore wleft(dfracdwdx-2xright)=2(w-x^2)^2+(2x^2+4x)(w-x^2)$
$wdfracdwdx-2xw=2w^2+(4x-2x^2)w-4x^3$
$wdfracdwdx=2w^2+(6x-2x^2)w-4x^3$
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 $dfracdwdx=-dfrac1z^2dfracdzdx$
$therefore-dfrac1z^3dfracdzdx=dfrac2z^2+dfrac6x-2x^2z-4x^3$
$dfracdzdx=4x^3z^3+(2x^2-6x)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered Aug 19 at 7:42
doraemonpaul
12.1k31660
Similar to To solve $ frac dydx=frac 1sqrtx^2+y^2$:
Apply the Euler substitution:
Let $u=y+sqrtx^2+y^2$ ,
Then $y=dfracu2-dfracx^22u$
$dfracdydx=left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu$
$thereforeleft(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=u-left(dfracu2-dfracx^22uright)$
$left(dfrac12+dfracx^22u^2right)dfracdudx-dfracxu=dfracu2+dfracx^22u$
$left(dfrac12+dfracx^22u^2right)dfracdudx=dfracu2+dfracx^2+2x2u$
$(u^2+x^2)dfracdudx=u^3+(x^2+2x)u$
Let $v=u^2$ ,
Then $dfracdvdx=2udfracdudx$
$thereforedfracu^2+x^22udfracdvdx=u^3+(x^2+2x)u$
$(u^2+x^2)dfracdvdx=2u^4+(2x^2+4x)u^2$
$(v+x^2)dfracdvdx=2v^2+(2x^2+4x)v$
Let $w=v+x^2$ ,
Then $v=w-x^2$
$dfracdvdx=dfracdwdx-2x$
$therefore wleft(dfracdwdx-2xright)=2(w-x^2)^2+(2x^2+4x)(w-x^2)$
$wdfracdwdx-2xw=2w^2+(4x-2x^2)w-4x^3$
$wdfracdwdx=2w^2+(6x-2x^2)w-4x^3$
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 $dfracdwdx=-dfrac1z^2dfracdzdx$
$therefore-dfrac1z^3dfracdzdx=dfrac2z^2+dfrac6x-2x^2z-4x^3$
$dfracdzdx=4x^3z^3+(2x^2-6x)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
answered Aug 19 at 7:42
doraemonpaul
12.1k31660
answered Aug 19 at 7:42
doraemonpaul
12.1k31660
answered Aug 19 at 7:42
doraemonpaul
12.1k31660
answered Aug 19 at 7:42
doraemonpaul
12.1k31660
12.1k31660
Thank you doraemonnpaul. Apparently, from your citation, this class of equation was considered unsolvable until 2011. I was a math major 40 years ago, before entering the field of medicine. Differential equations have always remained a hobby of mine, especially nonlinear. This particular equation has fascinated me for the last few years. Thank you so much for your expertise in providing the solution.
– Craig Stevens
Aug 20 at 12:11
add a comment |Â
Thank you doraemonnpaul. Apparently, from your citation, this class of equation was considered unsolvable until 2011. I was a math major 40 years ago, before entering the field of medicine. Differential equations have always remained a hobby of mine, especially nonlinear. This particular equation has fascinated me for the last few years. Thank you so much for your expertise in providing the solution.
– Craig Stevens
Aug 20 at 12:11
Thank you doraemonnpaul. Apparently, from your citation, this class of equation was considered unsolvable until 2011. I was a math major 40 years ago, before entering the field of medicine. Differential equations have always remained a hobby of mine, especially nonlinear. This particular equation has fascinated me for the last few years. Thank you so much for your expertise in providing the solution.
– Craig Stevens
Aug 20 at 12:11
Thank you doraemonnpaul. Apparently, from your citation, this class of equation was considered unsolvable until 2011. I was a math major 40 years ago, before entering the field of medicine. Differential equations have always remained a hobby of mine, especially nonlinear. This particular equation has fascinated me for the last few years. Thank you so much for your expertise in providing the solution.
– Craig Stevens
Aug 20 at 12:11
add a comment |Â
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 18 at 8:16
You should be able to get the first few terms in $y=a_0+a_1x+a_2x^2+ldots$ with all the $a_i$ functions of $a_0$.
– Empy2
Aug 18 at 8:25
1
Mathematica does not find an explicit solution. Probably there is no solution in terms of elementary (or even special) functions.
– Julián Aguirre
Aug 18 at 10:52
1
Yes, all tried so far. Polar coordinates don't help. Power series approach didn't yield result. Once y>x, then dy/dx=sqrt(2) * Y yields upward bound of y=exp(sqrt 2) * x, so no singularities. Multiplying both sides by y, then substituting y*y=v yielded a similar interesting differential equation.
– Craig Stevens
Aug 18 at 14:39
1
The substitution $t=y/x, r=sqrtx^2+y^2$ gives $dr(t^2+1)(r-t)=dt(r+tr^2)$ which at least lacks a square-root.
– Empy2
Aug 18 at 15:47