Unable to solve differential… $yy'=y+1$
up vote
0
down vote
favorite
The differential equation is:
$$yy'=y+1$$I've been trying to solve this problem all evening, but I cannot figure it out, and none of the online calculators show me the steps on how to find the solution. Could anybody help me out?
calculus differential-equations
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 18 at 20:42
Indre
32
add a comment |
up vote
0
down vote
favorite
The differential equation is:
$$yy'=y+1$$I've been trying to solve this problem all evening, but I cannot figure it out, and none of the online calculators show me the steps on how to find the solution. Could anybody help me out?
calculus differential-equations
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 18 at 20:42
Indre
32
It's separable, no? I mean the variable $x$ doesn't even appear explicitly.
– lulu
Nov 18 at 20:43
It is separable, yes
– Indre
Nov 18 at 20:45
So...just isolate the terms involving $y$ on one side and $dx$ on the other.
– lulu
Nov 18 at 20:46
It is both separable and automonous.
– Yves Daoust
Nov 18 at 21:10
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The differential equation is:
$$yy'=y+1$$I've been trying to solve this problem all evening, but I cannot figure it out, and none of the online calculators show me the steps on how to find the solution. Could anybody help me out?
calculus differential-equations
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 18 at 20:42
Indre
32
The differential equation is:
$$yy'=y+1$$I've been trying to solve this problem all evening, but I cannot figure it out, and none of the online calculators show me the steps on how to find the solution. Could anybody help me out?
calculus differential-equations
calculus differential-equations
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 18 at 20:42
Indre
32
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 18 at 20:42
Indre
32
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
1,6622519
asked Nov 18 at 20:42
Indre
32
asked Nov 18 at 20:42
Indre
32
asked Nov 18 at 20:42
Indre
32
32
It's separable, no? I mean the variable $x$ doesn't even appear explicitly.
– lulu
Nov 18 at 20:43
It is separable, yes
– Indre
Nov 18 at 20:45
So...just isolate the terms involving $y$ on one side and $dx$ on the other.
– lulu
Nov 18 at 20:46
It is both separable and automonous.
– Yves Daoust
Nov 18 at 21:10
add a comment |
It's separable, no? I mean the variable $x$ doesn't even appear explicitly.
– lulu
Nov 18 at 20:43
It is separable, yes
– Indre
Nov 18 at 20:45
So...just isolate the terms involving $y$ on one side and $dx$ on the other.
– lulu
Nov 18 at 20:46
It is both separable and automonous.
– Yves Daoust
Nov 18 at 21:10
It's separable, no? I mean the variable $x$ doesn't even appear explicitly.
– lulu
Nov 18 at 20:43
It's separable, no? I mean the variable $x$ doesn't even appear explicitly.
– lulu
Nov 18 at 20:43
It is separable, yes
– Indre
Nov 18 at 20:45
It is separable, yes
– Indre
Nov 18 at 20:45
So...just isolate the terms involving $y$ on one side and $dx$ on the other.
– lulu
Nov 18 at 20:46
So...just isolate the terms involving $y$ on one side and $dx$ on the other.
– lulu
Nov 18 at 20:46
It is both separable and automonous.
– Yves Daoust
Nov 18 at 21:10
It is both separable and automonous.
– Yves Daoust
Nov 18 at 21:10
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
This can be reduced to just some calculus, by writing $dy/dx = frac{y+1}{y}$. This can now be separated, giving
$$frac{y dy}{y+1} = dx$$
Which can be integrated by making a substitution to give
$$y - log(|y+1|) + C = x$$
I don't think you will be able to solve this explicitly for $y$ though.
answered Nov 18 at 20:50
Alfred Yerger
10.1k2146
I think it may be it. Thanks for the help!
– Indre
Nov 18 at 20:53
$-(y+1)e^{-(y+1)}=c,e^{-x}$ can be solved as $y=-1-W(c,e^{-x})$ using the Lambert-W function.
– LutzL
Nov 18 at 21:55
add a comment |
up vote
1
down vote
$$frac{y,dy}{y+1}=dx.$$
By integration from $x_0$,
$$y-y_0-logfrac{y+1}{y_0+1}=x-x_0.$$
answered Nov 18 at 21:12
Yves Daoust
122k668217
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This can be reduced to just some calculus, by writing $dy/dx = frac{y+1}{y}$. This can now be separated, giving
$$frac{y dy}{y+1} = dx$$
Which can be integrated by making a substitution to give
$$y - log(|y+1|) + C = x$$
I don't think you will be able to solve this explicitly for $y$ though.
answered Nov 18 at 20:50
Alfred Yerger
10.1k2146
I think it may be it. Thanks for the help!
– Indre
Nov 18 at 20:53
$-(y+1)e^{-(y+1)}=c,e^{-x}$ can be solved as $y=-1-W(c,e^{-x})$ using the Lambert-W function.
– LutzL
Nov 18 at 21:55
add a comment |
up vote
1
down vote
accepted
This can be reduced to just some calculus, by writing $dy/dx = frac{y+1}{y}$. This can now be separated, giving
$$frac{y dy}{y+1} = dx$$
Which can be integrated by making a substitution to give
$$y - log(|y+1|) + C = x$$
I don't think you will be able to solve this explicitly for $y$ though.
answered Nov 18 at 20:50
Alfred Yerger
10.1k2146
I think it may be it. Thanks for the help!
– Indre
Nov 18 at 20:53
$-(y+1)e^{-(y+1)}=c,e^{-x}$ can be solved as $y=-1-W(c,e^{-x})$ using the Lambert-W function.
– LutzL
Nov 18 at 21:55
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This can be reduced to just some calculus, by writing $dy/dx = frac{y+1}{y}$. This can now be separated, giving
$$frac{y dy}{y+1} = dx$$
Which can be integrated by making a substitution to give
$$y - log(|y+1|) + C = x$$
I don't think you will be able to solve this explicitly for $y$ though.
answered Nov 18 at 20:50
Alfred Yerger
10.1k2146
This can be reduced to just some calculus, by writing $dy/dx = frac{y+1}{y}$. This can now be separated, giving
$$frac{y dy}{y+1} = dx$$
Which can be integrated by making a substitution to give
$$y - log(|y+1|) + C = x$$
I don't think you will be able to solve this explicitly for $y$ though.
answered Nov 18 at 20:50
Alfred Yerger
10.1k2146
answered Nov 18 at 20:50
Alfred Yerger
10.1k2146
answered Nov 18 at 20:50
Alfred Yerger
10.1k2146
answered Nov 18 at 20:50
Alfred Yerger
10.1k2146
10.1k2146
I think it may be it. Thanks for the help!
– Indre
Nov 18 at 20:53
$-(y+1)e^{-(y+1)}=c,e^{-x}$ can be solved as $y=-1-W(c,e^{-x})$ using the Lambert-W function.
– LutzL
Nov 18 at 21:55
add a comment |
I think it may be it. Thanks for the help!
– Indre
Nov 18 at 20:53
$-(y+1)e^{-(y+1)}=c,e^{-x}$ can be solved as $y=-1-W(c,e^{-x})$ using the Lambert-W function.
– LutzL
Nov 18 at 21:55
I think it may be it. Thanks for the help!
– Indre
Nov 18 at 20:53
I think it may be it. Thanks for the help!
– Indre
Nov 18 at 20:53
$-(y+1)e^{-(y+1)}=c,e^{-x}$ can be solved as $y=-1-W(c,e^{-x})$ using the Lambert-W function.
– LutzL
Nov 18 at 21:55
$-(y+1)e^{-(y+1)}=c,e^{-x}$ can be solved as $y=-1-W(c,e^{-x})$ using the Lambert-W function.
– LutzL
Nov 18 at 21:55
add a comment |
up vote
1
down vote
$$frac{y,dy}{y+1}=dx.$$
By integration from $x_0$,
$$y-y_0-logfrac{y+1}{y_0+1}=x-x_0.$$
answered Nov 18 at 21:12
Yves Daoust
122k668217
add a comment |
up vote
1
down vote
$$frac{y,dy}{y+1}=dx.$$
By integration from $x_0$,
$$y-y_0-logfrac{y+1}{y_0+1}=x-x_0.$$
answered Nov 18 at 21:12
Yves Daoust
122k668217
add a comment |
up vote
1
down vote
up vote
1
down vote
$$frac{y,dy}{y+1}=dx.$$
By integration from $x_0$,
$$y-y_0-logfrac{y+1}{y_0+1}=x-x_0.$$
answered Nov 18 at 21:12
Yves Daoust
122k668217
$$frac{y,dy}{y+1}=dx.$$
By integration from $x_0$,
$$y-y_0-logfrac{y+1}{y_0+1}=x-x_0.$$
answered Nov 18 at 21:12
Yves Daoust
122k668217
answered Nov 18 at 21:12
Yves Daoust
122k668217
answered Nov 18 at 21:12
Yves Daoust
122k668217
answered Nov 18 at 21:12
Yves Daoust
122k668217
122k668217
add a comment |
add a comment |
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It's separable, no? I mean the variable $x$ doesn't even appear explicitly.
– lulu
Nov 18 at 20:43
It is separable, yes
– Indre
Nov 18 at 20:45
So...just isolate the terms involving $y$ on one side and $dx$ on the other.
– lulu
Nov 18 at 20:46
It is both separable and automonous.
– Yves Daoust
Nov 18 at 21:10