Solve $(x+y log y)y, mathrmdx=(y+x log x)x ,mathrmdy$
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Solve the differential equation $$(x+y log y)y ,mathrmdx=(y+x log x)x ,mathrmdy$$
My try: put $x=e^t$ and $y =e^w$: we get
$$(e^t+w e^w)e^w e^t ,mathrmdt=(e^w+te^t)e^we^t,mathrmdwimplies
(e^t+we^w),mathrmdt=(e^w+te^t),mathrmdw$$
Any clue here?
integration algebra-precalculus differential-equations definite-integrals indefinite-integrals
edited Sep 6 at 8:51
Gibbs
3,5521524
asked Sep 6 at 7:56
Ekaveera Kumar Sharma
5,36611126
 |Â
show 3 more comments
up vote
8
down vote
favorite
Solve the differential equation $$(x+y log y)y ,mathrmdx=(y+x log x)x ,mathrmdy$$
My try: put $x=e^t$ and $y =e^w$: we get
$$(e^t+w e^w)e^w e^t ,mathrmdt=(e^w+te^t)e^we^t,mathrmdwimplies
(e^t+we^w),mathrmdt=(e^w+te^t),mathrmdw$$
Any clue here?
integration algebra-precalculus differential-equations definite-integrals indefinite-integrals
edited Sep 6 at 8:51
Gibbs
3,5521524
asked Sep 6 at 7:56
Ekaveera Kumar Sharma
5,36611126
Is $y$ a function of $x$?
– user496634
Sep 6 at 8:03
Yes it's an implicit function
– Ekaveera Kumar Sharma
Sep 6 at 8:07
1
Just an observation: by symmetry $y(x) = x$ works.
– Gibbs
Sep 6 at 8:49
Yes but y=x only in first quadrant
– Ekaveera Kumar Sharma
Sep 6 at 8:56
Yes sure, because both $y$ and $x$ must be positive.
– Gibbs
Sep 6 at 8:57
 |Â
show 3 more comments
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Solve the differential equation $$(x+y log y)y ,mathrmdx=(y+x log x)x ,mathrmdy$$
My try: put $x=e^t$ and $y =e^w$: we get
$$(e^t+w e^w)e^w e^t ,mathrmdt=(e^w+te^t)e^we^t,mathrmdwimplies
(e^t+we^w),mathrmdt=(e^w+te^t),mathrmdw$$
Any clue here?
integration algebra-precalculus differential-equations definite-integrals indefinite-integrals
edited Sep 6 at 8:51
Gibbs
3,5521524
asked Sep 6 at 7:56
Ekaveera Kumar Sharma
5,36611126
Solve the differential equation $$(x+y log y)y ,mathrmdx=(y+x log x)x ,mathrmdy$$
My try: put $x=e^t$ and $y =e^w$: we get
$$(e^t+w e^w)e^w e^t ,mathrmdt=(e^w+te^t)e^we^t,mathrmdwimplies
(e^t+we^w),mathrmdt=(e^w+te^t),mathrmdw$$
Any clue here?
integration algebra-precalculus differential-equations definite-integrals indefinite-integrals
integration algebra-precalculus differential-equations definite-integrals indefinite-integrals
edited Sep 6 at 8:51
Gibbs
3,5521524
asked Sep 6 at 7:56
Ekaveera Kumar Sharma
5,36611126
edited Sep 6 at 8:51
Gibbs
3,5521524
asked Sep 6 at 7:56
Ekaveera Kumar Sharma
5,36611126
edited Sep 6 at 8:51
Gibbs
3,5521524
edited Sep 6 at 8:51
Gibbs
3,5521524
edited Sep 6 at 8:51
Gibbs
3,5521524
3,5521524
asked Sep 6 at 7:56
Ekaveera Kumar Sharma
5,36611126
asked Sep 6 at 7:56
Ekaveera Kumar Sharma
5,36611126
asked Sep 6 at 7:56
Ekaveera Kumar Sharma
5,36611126
5,36611126
Is $y$ a function of $x$?
– user496634
Sep 6 at 8:03
Yes it's an implicit function
– Ekaveera Kumar Sharma
Sep 6 at 8:07
1
Just an observation: by symmetry $y(x) = x$ works.
– Gibbs
Sep 6 at 8:49
Yes but y=x only in first quadrant
– Ekaveera Kumar Sharma
Sep 6 at 8:56
Yes sure, because both $y$ and $x$ must be positive.
– Gibbs
Sep 6 at 8:57
 |Â
show 3 more comments
Is $y$ a function of $x$?
– user496634
Sep 6 at 8:03
Yes it's an implicit function
– Ekaveera Kumar Sharma
Sep 6 at 8:07
1
Just an observation: by symmetry $y(x) = x$ works.
– Gibbs
Sep 6 at 8:49
Yes but y=x only in first quadrant
– Ekaveera Kumar Sharma
Sep 6 at 8:56
Yes sure, because both $y$ and $x$ must be positive.
– Gibbs
Sep 6 at 8:57
Is $y$ a function of $x$?
– user496634
Sep 6 at 8:03
Is $y$ a function of $x$?
– user496634
Sep 6 at 8:03
Yes it's an implicit function
– Ekaveera Kumar Sharma
Sep 6 at 8:07
Yes it's an implicit function
– Ekaveera Kumar Sharma
Sep 6 at 8:07
1
1
Just an observation: by symmetry $y(x) = x$ works.
– Gibbs
Sep 6 at 8:49
Just an observation: by symmetry $y(x) = x$ works.
– Gibbs
Sep 6 at 8:49
Yes but y=x only in first quadrant
– Ekaveera Kumar Sharma
Sep 6 at 8:56
Yes but y=x only in first quadrant
– Ekaveera Kumar Sharma
Sep 6 at 8:56
Yes sure, because both $y$ and $x$ must be positive.
– Gibbs
Sep 6 at 8:57
Yes sure, because both $y$ and $x$ must be positive.
– Gibbs
Sep 6 at 8:57
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
10
down vote
accepted
We have the following differential equation:
$$(x+y log y)y ,mathrmdx=(y+x log x)x ,mathrmdy$$
Rearranging the terms, we get:
$$implies xymathrmdx+y^2log(y)mathrmdx=xymathrmdy+x^2log(x)mathrmdy$$
$$implies-x^2log(x)mathrmdy+xymathrmdx=-y^2log(y)mathrmdx+xymathrmdy$$
$$implies x(-xlog(x)mathrmdy+ymathrmdx)=y(-ylog(y)mathrmdx+xmathrmdy)$$
Dividing both sides by the term $x^2y^2$, we get:
$$implies frac(-xlog(x)mathrmdy+ymathrmdx)xy^2=frac(-ylog(y)mathrmdx+xmathrmdy)x^2y$$
$$implies left(-frac1y^2mathrmdyright)log(x)+left(frac1xmathrmdxright)frac1y=left(-frac1x^2mathrmdxright)log(y)+left(frac1ymathrmdyright)frac1x$$
$$implies mathrmdleft(frac1yright)log(x)+mathrmdleft(log(x)right)frac1y=mathrmdleft(frac1xright)log(y)+mathrmdleft(log(y)right)frac1x$$
$$implies mathrmdleft(frac1ylog(x)right)=mathrmdleft(frac1xlog(y)right)$$
Integrating both sides, we get:
$$implies int mathrmdleft(frac1ylog(x)right)=int mathrmdleft(frac1xlog(y)right)$$
We get the solution as: $$colordarkbluefrac1ylog(x)=frac1xlog(y)+C$$
Where $C$ is the integration constant. This is the implicit solution.
Edit: As shown by @Claude Leibovici, the solution can be converted to an explicit one using Lambert $W$ function. The solution is as follows:
Rearranging the above equation, we get:
$$implies fracxylog(x)=log(y)+Cx$$
Taking exponential of both sides, we have:
$$implies expleft(fracxylog(x)right)=expleft(log(y)right)expleft(Cxright)$$
$$implies expleft(fracxylog(x)right)=yexpleft(Cxright)$$
Multiplying both sides by $fracxylog(x)$, we have:
$$implies fracxylog(x)expleft(fracxylog(x)right)=xexpleft(Cxright)log(x)$$
If we consider $fracxylog(x)$ as $u$, the above equation reduces to $ue^u=K$ and can be represented by Lambert $W$ function as $u=W(K)$, so we have:
$$implies fracxylog(x)=Wleft(xexpleft(Cxright)log(x)right)$$
$$implies colordarkgreeny=fracx log (x)Wleft(x e^Cx log (x)right)$$
where $W(.)$ is Lambert function. This is the explicit solution.
answered Sep 6 at 8:34
paulplusx
1,052318
3
perfect, clean and well formated answer
– Matěj Å tágl
Sep 6 at 12:09
1
Nice solution, for sure ! $to +1$
– Claude Leibovici
Sep 6 at 13:19
@ClaudeLeibovici and Matěj, Thank you :)
– paulplusx
Sep 6 at 14:55
1
May be, you could add to your good answer that the final solution is $$y=fracx log (x)Wleft(x e^c x log (x)right)$$. where $W(.)$ is Lambert function.
– Claude Leibovici
Sep 7 at 2:24
1
Try to learn about Lambert function. It allows to solve explicitly so many problems. Just search of this site Lambert function; 2277 posts ! It heps to solve a huge number of problems in mathematics, physics, chemistry, biochemistry and so on. Just for your curiosity, Euler and Lambert worked together.
– Claude Leibovici
Sep 7 at 4:06
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
We have the following differential equation:
$$(x+y log y)y ,mathrmdx=(y+x log x)x ,mathrmdy$$
Rearranging the terms, we get:
$$implies xymathrmdx+y^2log(y)mathrmdx=xymathrmdy+x^2log(x)mathrmdy$$
$$implies-x^2log(x)mathrmdy+xymathrmdx=-y^2log(y)mathrmdx+xymathrmdy$$
$$implies x(-xlog(x)mathrmdy+ymathrmdx)=y(-ylog(y)mathrmdx+xmathrmdy)$$
Dividing both sides by the term $x^2y^2$, we get:
$$implies frac(-xlog(x)mathrmdy+ymathrmdx)xy^2=frac(-ylog(y)mathrmdx+xmathrmdy)x^2y$$
$$implies left(-frac1y^2mathrmdyright)log(x)+left(frac1xmathrmdxright)frac1y=left(-frac1x^2mathrmdxright)log(y)+left(frac1ymathrmdyright)frac1x$$
$$implies mathrmdleft(frac1yright)log(x)+mathrmdleft(log(x)right)frac1y=mathrmdleft(frac1xright)log(y)+mathrmdleft(log(y)right)frac1x$$
$$implies mathrmdleft(frac1ylog(x)right)=mathrmdleft(frac1xlog(y)right)$$
Integrating both sides, we get:
$$implies int mathrmdleft(frac1ylog(x)right)=int mathrmdleft(frac1xlog(y)right)$$
We get the solution as: $$colordarkbluefrac1ylog(x)=frac1xlog(y)+C$$
Where $C$ is the integration constant. This is the implicit solution.
Edit: As shown by @Claude Leibovici, the solution can be converted to an explicit one using Lambert $W$ function. The solution is as follows:
Rearranging the above equation, we get:
$$implies fracxylog(x)=log(y)+Cx$$
Taking exponential of both sides, we have:
$$implies expleft(fracxylog(x)right)=expleft(log(y)right)expleft(Cxright)$$
$$implies expleft(fracxylog(x)right)=yexpleft(Cxright)$$
Multiplying both sides by $fracxylog(x)$, we have:
$$implies fracxylog(x)expleft(fracxylog(x)right)=xexpleft(Cxright)log(x)$$
If we consider $fracxylog(x)$ as $u$, the above equation reduces to $ue^u=K$ and can be represented by Lambert $W$ function as $u=W(K)$, so we have:
$$implies fracxylog(x)=Wleft(xexpleft(Cxright)log(x)right)$$
$$implies colordarkgreeny=fracx log (x)Wleft(x e^Cx log (x)right)$$
where $W(.)$ is Lambert function. This is the explicit solution.
answered Sep 6 at 8:34
paulplusx
1,052318
3
perfect, clean and well formated answer
– Matěj Å tágl
Sep 6 at 12:09
1
Nice solution, for sure ! $to +1$
– Claude Leibovici
Sep 6 at 13:19
@ClaudeLeibovici and Matěj, Thank you :)
– paulplusx
Sep 6 at 14:55
1
May be, you could add to your good answer that the final solution is $$y=fracx log (x)Wleft(x e^c x log (x)right)$$. where $W(.)$ is Lambert function.
– Claude Leibovici
Sep 7 at 2:24
1
Try to learn about Lambert function. It allows to solve explicitly so many problems. Just search of this site Lambert function; 2277 posts ! It heps to solve a huge number of problems in mathematics, physics, chemistry, biochemistry and so on. Just for your curiosity, Euler and Lambert worked together.
– Claude Leibovici
Sep 7 at 4:06
 |Â
show 6 more comments
up vote
10
down vote
accepted
We have the following differential equation:
$$(x+y log y)y ,mathrmdx=(y+x log x)x ,mathrmdy$$
Rearranging the terms, we get:
$$implies xymathrmdx+y^2log(y)mathrmdx=xymathrmdy+x^2log(x)mathrmdy$$
$$implies-x^2log(x)mathrmdy+xymathrmdx=-y^2log(y)mathrmdx+xymathrmdy$$
$$implies x(-xlog(x)mathrmdy+ymathrmdx)=y(-ylog(y)mathrmdx+xmathrmdy)$$
Dividing both sides by the term $x^2y^2$, we get:
$$implies frac(-xlog(x)mathrmdy+ymathrmdx)xy^2=frac(-ylog(y)mathrmdx+xmathrmdy)x^2y$$
$$implies left(-frac1y^2mathrmdyright)log(x)+left(frac1xmathrmdxright)frac1y=left(-frac1x^2mathrmdxright)log(y)+left(frac1ymathrmdyright)frac1x$$
$$implies mathrmdleft(frac1yright)log(x)+mathrmdleft(log(x)right)frac1y=mathrmdleft(frac1xright)log(y)+mathrmdleft(log(y)right)frac1x$$
$$implies mathrmdleft(frac1ylog(x)right)=mathrmdleft(frac1xlog(y)right)$$
Integrating both sides, we get:
$$implies int mathrmdleft(frac1ylog(x)right)=int mathrmdleft(frac1xlog(y)right)$$
We get the solution as: $$colordarkbluefrac1ylog(x)=frac1xlog(y)+C$$
Where $C$ is the integration constant. This is the implicit solution.
Edit: As shown by @Claude Leibovici, the solution can be converted to an explicit one using Lambert $W$ function. The solution is as follows:
Rearranging the above equation, we get:
$$implies fracxylog(x)=log(y)+Cx$$
Taking exponential of both sides, we have:
$$implies expleft(fracxylog(x)right)=expleft(log(y)right)expleft(Cxright)$$
$$implies expleft(fracxylog(x)right)=yexpleft(Cxright)$$
Multiplying both sides by $fracxylog(x)$, we have:
$$implies fracxylog(x)expleft(fracxylog(x)right)=xexpleft(Cxright)log(x)$$
If we consider $fracxylog(x)$ as $u$, the above equation reduces to $ue^u=K$ and can be represented by Lambert $W$ function as $u=W(K)$, so we have:
$$implies fracxylog(x)=Wleft(xexpleft(Cxright)log(x)right)$$
$$implies colordarkgreeny=fracx log (x)Wleft(x e^Cx log (x)right)$$
where $W(.)$ is Lambert function. This is the explicit solution.
answered Sep 6 at 8:34
paulplusx
1,052318
3
perfect, clean and well formated answer
– Matěj Å tágl
Sep 6 at 12:09
1
Nice solution, for sure ! $to +1$
– Claude Leibovici
Sep 6 at 13:19
@ClaudeLeibovici and Matěj, Thank you :)
– paulplusx
Sep 6 at 14:55
1
May be, you could add to your good answer that the final solution is $$y=fracx log (x)Wleft(x e^c x log (x)right)$$. where $W(.)$ is Lambert function.
– Claude Leibovici
Sep 7 at 2:24
1
Try to learn about Lambert function. It allows to solve explicitly so many problems. Just search of this site Lambert function; 2277 posts ! It heps to solve a huge number of problems in mathematics, physics, chemistry, biochemistry and so on. Just for your curiosity, Euler and Lambert worked together.
– Claude Leibovici
Sep 7 at 4:06
 |Â
show 6 more comments
up vote
10
down vote
accepted
up vote
10
down vote
accepted
We have the following differential equation:
$$(x+y log y)y ,mathrmdx=(y+x log x)x ,mathrmdy$$
Rearranging the terms, we get:
$$implies xymathrmdx+y^2log(y)mathrmdx=xymathrmdy+x^2log(x)mathrmdy$$
$$implies-x^2log(x)mathrmdy+xymathrmdx=-y^2log(y)mathrmdx+xymathrmdy$$
$$implies x(-xlog(x)mathrmdy+ymathrmdx)=y(-ylog(y)mathrmdx+xmathrmdy)$$
Dividing both sides by the term $x^2y^2$, we get:
$$implies frac(-xlog(x)mathrmdy+ymathrmdx)xy^2=frac(-ylog(y)mathrmdx+xmathrmdy)x^2y$$
$$implies left(-frac1y^2mathrmdyright)log(x)+left(frac1xmathrmdxright)frac1y=left(-frac1x^2mathrmdxright)log(y)+left(frac1ymathrmdyright)frac1x$$
$$implies mathrmdleft(frac1yright)log(x)+mathrmdleft(log(x)right)frac1y=mathrmdleft(frac1xright)log(y)+mathrmdleft(log(y)right)frac1x$$
$$implies mathrmdleft(frac1ylog(x)right)=mathrmdleft(frac1xlog(y)right)$$
Integrating both sides, we get:
$$implies int mathrmdleft(frac1ylog(x)right)=int mathrmdleft(frac1xlog(y)right)$$
We get the solution as: $$colordarkbluefrac1ylog(x)=frac1xlog(y)+C$$
Where $C$ is the integration constant. This is the implicit solution.
Edit: As shown by @Claude Leibovici, the solution can be converted to an explicit one using Lambert $W$ function. The solution is as follows:
Rearranging the above equation, we get:
$$implies fracxylog(x)=log(y)+Cx$$
Taking exponential of both sides, we have:
$$implies expleft(fracxylog(x)right)=expleft(log(y)right)expleft(Cxright)$$
$$implies expleft(fracxylog(x)right)=yexpleft(Cxright)$$
Multiplying both sides by $fracxylog(x)$, we have:
$$implies fracxylog(x)expleft(fracxylog(x)right)=xexpleft(Cxright)log(x)$$
If we consider $fracxylog(x)$ as $u$, the above equation reduces to $ue^u=K$ and can be represented by Lambert $W$ function as $u=W(K)$, so we have:
$$implies fracxylog(x)=Wleft(xexpleft(Cxright)log(x)right)$$
$$implies colordarkgreeny=fracx log (x)Wleft(x e^Cx log (x)right)$$
where $W(.)$ is Lambert function. This is the explicit solution.
answered Sep 6 at 8:34
paulplusx
1,052318
We have the following differential equation:
$$(x+y log y)y ,mathrmdx=(y+x log x)x ,mathrmdy$$
Rearranging the terms, we get:
$$implies xymathrmdx+y^2log(y)mathrmdx=xymathrmdy+x^2log(x)mathrmdy$$
$$implies-x^2log(x)mathrmdy+xymathrmdx=-y^2log(y)mathrmdx+xymathrmdy$$
$$implies x(-xlog(x)mathrmdy+ymathrmdx)=y(-ylog(y)mathrmdx+xmathrmdy)$$
Dividing both sides by the term $x^2y^2$, we get:
$$implies frac(-xlog(x)mathrmdy+ymathrmdx)xy^2=frac(-ylog(y)mathrmdx+xmathrmdy)x^2y$$
$$implies left(-frac1y^2mathrmdyright)log(x)+left(frac1xmathrmdxright)frac1y=left(-frac1x^2mathrmdxright)log(y)+left(frac1ymathrmdyright)frac1x$$
$$implies mathrmdleft(frac1yright)log(x)+mathrmdleft(log(x)right)frac1y=mathrmdleft(frac1xright)log(y)+mathrmdleft(log(y)right)frac1x$$
$$implies mathrmdleft(frac1ylog(x)right)=mathrmdleft(frac1xlog(y)right)$$
Integrating both sides, we get:
$$implies int mathrmdleft(frac1ylog(x)right)=int mathrmdleft(frac1xlog(y)right)$$
We get the solution as: $$colordarkbluefrac1ylog(x)=frac1xlog(y)+C$$
Where $C$ is the integration constant. This is the implicit solution.
Edit: As shown by @Claude Leibovici, the solution can be converted to an explicit one using Lambert $W$ function. The solution is as follows:
Rearranging the above equation, we get:
$$implies fracxylog(x)=log(y)+Cx$$
Taking exponential of both sides, we have:
$$implies expleft(fracxylog(x)right)=expleft(log(y)right)expleft(Cxright)$$
$$implies expleft(fracxylog(x)right)=yexpleft(Cxright)$$
Multiplying both sides by $fracxylog(x)$, we have:
$$implies fracxylog(x)expleft(fracxylog(x)right)=xexpleft(Cxright)log(x)$$
If we consider $fracxylog(x)$ as $u$, the above equation reduces to $ue^u=K$ and can be represented by Lambert $W$ function as $u=W(K)$, so we have:
$$implies fracxylog(x)=Wleft(xexpleft(Cxright)log(x)right)$$
$$implies colordarkgreeny=fracx log (x)Wleft(x e^Cx log (x)right)$$
where $W(.)$ is Lambert function. This is the explicit solution.
answered Sep 6 at 8:34
paulplusx
1,052318
edited Sep 7 at 5:04
answered Sep 6 at 8:34
paulplusx
1,052318
answered Sep 6 at 8:34
paulplusx
1,052318
answered Sep 6 at 8:34
paulplusx
1,052318
1,052318
3
perfect, clean and well formated answer
– Matěj Å tágl
Sep 6 at 12:09
1
Nice solution, for sure ! $to +1$
– Claude Leibovici
Sep 6 at 13:19
@ClaudeLeibovici and Matěj, Thank you :)
– paulplusx
Sep 6 at 14:55
1
May be, you could add to your good answer that the final solution is $$y=fracx log (x)Wleft(x e^c x log (x)right)$$. where $W(.)$ is Lambert function.
– Claude Leibovici
Sep 7 at 2:24
1
Try to learn about Lambert function. It allows to solve explicitly so many problems. Just search of this site Lambert function; 2277 posts ! It heps to solve a huge number of problems in mathematics, physics, chemistry, biochemistry and so on. Just for your curiosity, Euler and Lambert worked together.
– Claude Leibovici
Sep 7 at 4:06
 |Â
show 6 more comments
3
perfect, clean and well formated answer
– Matěj Å tágl
Sep 6 at 12:09
1
Nice solution, for sure ! $to +1$
– Claude Leibovici
Sep 6 at 13:19
@ClaudeLeibovici and Matěj, Thank you :)
– paulplusx
Sep 6 at 14:55
1
May be, you could add to your good answer that the final solution is $$y=fracx log (x)Wleft(x e^c x log (x)right)$$. where $W(.)$ is Lambert function.
– Claude Leibovici
Sep 7 at 2:24
1
Try to learn about Lambert function. It allows to solve explicitly so many problems. Just search of this site Lambert function; 2277 posts ! It heps to solve a huge number of problems in mathematics, physics, chemistry, biochemistry and so on. Just for your curiosity, Euler and Lambert worked together.
– Claude Leibovici
Sep 7 at 4:06
3
3
perfect, clean and well formated answer
– Matěj Å tágl
Sep 6 at 12:09
perfect, clean and well formated answer
– Matěj Å tágl
Sep 6 at 12:09
1
1
Nice solution, for sure ! $to +1$
– Claude Leibovici
Sep 6 at 13:19
Nice solution, for sure ! $to +1$
– Claude Leibovici
Sep 6 at 13:19
@ClaudeLeibovici and Matěj, Thank you :)
– paulplusx
Sep 6 at 14:55
@ClaudeLeibovici and Matěj, Thank you :)
– paulplusx
Sep 6 at 14:55
1
1
May be, you could add to your good answer that the final solution is $$y=fracx log (x)Wleft(x e^c x log (x)right)$$. where $W(.)$ is Lambert function.
– Claude Leibovici
Sep 7 at 2:24
May be, you could add to your good answer that the final solution is $$y=fracx log (x)Wleft(x e^c x log (x)right)$$. where $W(.)$ is Lambert function.
– Claude Leibovici
Sep 7 at 2:24
1
1
Try to learn about Lambert function. It allows to solve explicitly so many problems. Just search of this site Lambert function; 2277 posts ! It heps to solve a huge number of problems in mathematics, physics, chemistry, biochemistry and so on. Just for your curiosity, Euler and Lambert worked together.
– Claude Leibovici
Sep 7 at 4:06
Try to learn about Lambert function. It allows to solve explicitly so many problems. Just search of this site Lambert function; 2277 posts ! It heps to solve a huge number of problems in mathematics, physics, chemistry, biochemistry and so on. Just for your curiosity, Euler and Lambert worked together.
– Claude Leibovici
Sep 7 at 4:06
 |Â
show 6 more comments
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Is $y$ a function of $x$?
– user496634
Sep 6 at 8:03
Yes it's an implicit function
– Ekaveera Kumar Sharma
Sep 6 at 8:07
1
Just an observation: by symmetry $y(x) = x$ works.
– Gibbs
Sep 6 at 8:49
Yes but y=x only in first quadrant
– Ekaveera Kumar Sharma
Sep 6 at 8:56
Yes sure, because both $y$ and $x$ must be positive.
– Gibbs
Sep 6 at 8:57