Find an explicit solution for $xy'+(x+frac 13)y=e^-2xy^-2$
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Beginning with the differential equation:
$$xy'+(x+frac 13)y=e^-2xy^-2$$
Divide by $x$: $$y'+y+frac y3x=e^-2xy^-2x^-1$$
Dividing by $y^-2$ gives: $$y^2y'+y^2y+frac 13xy=e^-2xx^-1$$
Normally, for Bernoulli equations, one would substitute such that $v=y^2$ and $v'=2yy'$ but I seem to have simplified poorly.
differential-equations
asked Aug 20 at 5:36
Alessandro Verniani
406
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up vote
1
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Beginning with the differential equation:
$$xy'+(x+frac 13)y=e^-2xy^-2$$
Divide by $x$: $$y'+y+frac y3x=e^-2xy^-2x^-1$$
Dividing by $y^-2$ gives: $$y^2y'+y^2y+frac 13xy=e^-2xx^-1$$
Normally, for Bernoulli equations, one would substitute such that $v=y^2$ and $v'=2yy'$ but I seem to have simplified poorly.
differential-equations
asked Aug 20 at 5:36
Alessandro Verniani
406
Except in this case, it's $v=y^1-(-2)=y^3$.
– John Wayland Bales
Aug 20 at 5:45
1
Bernoulli equation $$fracdydx+P(x)y=Q(x)y^n$$ can be reduced to a linear equation by using $v=y^1-n$. Then, for the given equation, use $v=y^1-(-2)=y^3$.
– Ãngel Mario Gallegos
Aug 20 at 5:49
your third step is wrong there should be y^3
– James
Aug 20 at 5:51
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Beginning with the differential equation:
$$xy'+(x+frac 13)y=e^-2xy^-2$$
Divide by $x$: $$y'+y+frac y3x=e^-2xy^-2x^-1$$
Dividing by $y^-2$ gives: $$y^2y'+y^2y+frac 13xy=e^-2xx^-1$$
Normally, for Bernoulli equations, one would substitute such that $v=y^2$ and $v'=2yy'$ but I seem to have simplified poorly.
differential-equations
asked Aug 20 at 5:36
Alessandro Verniani
406
Beginning with the differential equation:
$$xy'+(x+frac 13)y=e^-2xy^-2$$
Divide by $x$: $$y'+y+frac y3x=e^-2xy^-2x^-1$$
Dividing by $y^-2$ gives: $$y^2y'+y^2y+frac 13xy=e^-2xx^-1$$
Normally, for Bernoulli equations, one would substitute such that $v=y^2$ and $v'=2yy'$ but I seem to have simplified poorly.
differential-equations
asked Aug 20 at 5:36
Alessandro Verniani
406
asked Aug 20 at 5:36
Alessandro Verniani
406
asked Aug 20 at 5:36
Alessandro Verniani
406
asked Aug 20 at 5:36
Alessandro Verniani
406
406
Except in this case, it's $v=y^1-(-2)=y^3$.
– John Wayland Bales
Aug 20 at 5:45
1
Bernoulli equation $$fracdydx+P(x)y=Q(x)y^n$$ can be reduced to a linear equation by using $v=y^1-n$. Then, for the given equation, use $v=y^1-(-2)=y^3$.
– Ãngel Mario Gallegos
Aug 20 at 5:49
your third step is wrong there should be y^3
– James
Aug 20 at 5:51
add a comment |Â
Except in this case, it's $v=y^1-(-2)=y^3$.
– John Wayland Bales
Aug 20 at 5:45
1
Bernoulli equation $$fracdydx+P(x)y=Q(x)y^n$$ can be reduced to a linear equation by using $v=y^1-n$. Then, for the given equation, use $v=y^1-(-2)=y^3$.
– Ãngel Mario Gallegos
Aug 20 at 5:49
your third step is wrong there should be y^3
– James
Aug 20 at 5:51
Except in this case, it's $v=y^1-(-2)=y^3$.
– John Wayland Bales
Aug 20 at 5:45
Except in this case, it's $v=y^1-(-2)=y^3$.
– John Wayland Bales
Aug 20 at 5:45
1
1
Bernoulli equation $$fracdydx+P(x)y=Q(x)y^n$$ can be reduced to a linear equation by using $v=y^1-n$. Then, for the given equation, use $v=y^1-(-2)=y^3$.
– Ãngel Mario Gallegos
Aug 20 at 5:49
Bernoulli equation $$fracdydx+P(x)y=Q(x)y^n$$ can be reduced to a linear equation by using $v=y^1-n$. Then, for the given equation, use $v=y^1-(-2)=y^3$.
– Ãngel Mario Gallegos
Aug 20 at 5:49
your third step is wrong there should be y^3
– James
Aug 20 at 5:51
your third step is wrong there should be y^3
– James
Aug 20 at 5:51
add a comment |Â
2 Answers
2
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oldest
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up vote
1
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accepted
$$xy'+(x+frac 13)y=e^-2xy^-2$$
Multiply by $3y^2$:
$$3xy^2y'+(3x+1)y^3=3e^-2x$$
Let $quad Y=y^3$ :
$$xY'+(3x+1)Y=3e^-2x$$
This is a linear ODE easy to solve :
$$Y=fraccxe^3x+frac3xe^2x$$
$$y(x)=left(fraccxe^3x+frac3xe^2x right)^1/3$$
answered Aug 20 at 6:36
JJacquelin
40.6k21650
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The ODE given is
$$xy'+ left(x+dfrac 13right)y=e^-2xy^-2$$
I shall also divide it by $x$. So, the ODE becomes:
$$y'+left(1+dfrac13xright)y =dfrace^-2xxy^-2tag 1$$
This is Bernoulli's ODE.
We substitute $v=y^1-(-2)=y^3$.
So, $dfracmathrm dymathrm dx=dfrac13v^(2/3)dfracmathrm dvmathrm dx$
$$beginalign(1)&implies dfrac13v^(2/3)dfracmathrm dvmathrm dx+left(1+dfrac13xright)v^1/3=dfrace^-2xxv^-2/3\ &implies dfracmathrm dvmathrm dx+ 3left(1+dfrac13xright)v=3dfrace^-2xxendalign$$
I think you can take it from here.
answered Aug 20 at 6:13
DarkKnight
4731211
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$xy'+(x+frac 13)y=e^-2xy^-2$$
Multiply by $3y^2$:
$$3xy^2y'+(3x+1)y^3=3e^-2x$$
Let $quad Y=y^3$ :
$$xY'+(3x+1)Y=3e^-2x$$
This is a linear ODE easy to solve :
$$Y=fraccxe^3x+frac3xe^2x$$
$$y(x)=left(fraccxe^3x+frac3xe^2x right)^1/3$$
answered Aug 20 at 6:36
JJacquelin
40.6k21650
add a comment |Â
up vote
1
down vote
accepted
$$xy'+(x+frac 13)y=e^-2xy^-2$$
Multiply by $3y^2$:
$$3xy^2y'+(3x+1)y^3=3e^-2x$$
Let $quad Y=y^3$ :
$$xY'+(3x+1)Y=3e^-2x$$
This is a linear ODE easy to solve :
$$Y=fraccxe^3x+frac3xe^2x$$
$$y(x)=left(fraccxe^3x+frac3xe^2x right)^1/3$$
answered Aug 20 at 6:36
JJacquelin
40.6k21650
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$xy'+(x+frac 13)y=e^-2xy^-2$$
Multiply by $3y^2$:
$$3xy^2y'+(3x+1)y^3=3e^-2x$$
Let $quad Y=y^3$ :
$$xY'+(3x+1)Y=3e^-2x$$
This is a linear ODE easy to solve :
$$Y=fraccxe^3x+frac3xe^2x$$
$$y(x)=left(fraccxe^3x+frac3xe^2x right)^1/3$$
answered Aug 20 at 6:36
JJacquelin
40.6k21650
$$xy'+(x+frac 13)y=e^-2xy^-2$$
Multiply by $3y^2$:
$$3xy^2y'+(3x+1)y^3=3e^-2x$$
Let $quad Y=y^3$ :
$$xY'+(3x+1)Y=3e^-2x$$
This is a linear ODE easy to solve :
$$Y=fraccxe^3x+frac3xe^2x$$
$$y(x)=left(fraccxe^3x+frac3xe^2x right)^1/3$$
answered Aug 20 at 6:36
JJacquelin
40.6k21650
answered Aug 20 at 6:36
JJacquelin
40.6k21650
answered Aug 20 at 6:36
JJacquelin
40.6k21650
answered Aug 20 at 6:36
JJacquelin
40.6k21650
40.6k21650
add a comment |Â
add a comment |Â
up vote
3
down vote
The ODE given is
$$xy'+ left(x+dfrac 13right)y=e^-2xy^-2$$
I shall also divide it by $x$. So, the ODE becomes:
$$y'+left(1+dfrac13xright)y =dfrace^-2xxy^-2tag 1$$
This is Bernoulli's ODE.
We substitute $v=y^1-(-2)=y^3$.
So, $dfracmathrm dymathrm dx=dfrac13v^(2/3)dfracmathrm dvmathrm dx$
$$beginalign(1)&implies dfrac13v^(2/3)dfracmathrm dvmathrm dx+left(1+dfrac13xright)v^1/3=dfrace^-2xxv^-2/3\ &implies dfracmathrm dvmathrm dx+ 3left(1+dfrac13xright)v=3dfrace^-2xxendalign$$
I think you can take it from here.
answered Aug 20 at 6:13
DarkKnight
4731211
add a comment |Â
up vote
3
down vote
The ODE given is
$$xy'+ left(x+dfrac 13right)y=e^-2xy^-2$$
I shall also divide it by $x$. So, the ODE becomes:
$$y'+left(1+dfrac13xright)y =dfrace^-2xxy^-2tag 1$$
This is Bernoulli's ODE.
We substitute $v=y^1-(-2)=y^3$.
So, $dfracmathrm dymathrm dx=dfrac13v^(2/3)dfracmathrm dvmathrm dx$
$$beginalign(1)&implies dfrac13v^(2/3)dfracmathrm dvmathrm dx+left(1+dfrac13xright)v^1/3=dfrace^-2xxv^-2/3\ &implies dfracmathrm dvmathrm dx+ 3left(1+dfrac13xright)v=3dfrace^-2xxendalign$$
I think you can take it from here.
answered Aug 20 at 6:13
DarkKnight
4731211
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The ODE given is
$$xy'+ left(x+dfrac 13right)y=e^-2xy^-2$$
I shall also divide it by $x$. So, the ODE becomes:
$$y'+left(1+dfrac13xright)y =dfrace^-2xxy^-2tag 1$$
This is Bernoulli's ODE.
We substitute $v=y^1-(-2)=y^3$.
So, $dfracmathrm dymathrm dx=dfrac13v^(2/3)dfracmathrm dvmathrm dx$
$$beginalign(1)&implies dfrac13v^(2/3)dfracmathrm dvmathrm dx+left(1+dfrac13xright)v^1/3=dfrace^-2xxv^-2/3\ &implies dfracmathrm dvmathrm dx+ 3left(1+dfrac13xright)v=3dfrace^-2xxendalign$$
I think you can take it from here.
answered Aug 20 at 6:13
DarkKnight
4731211
The ODE given is
$$xy'+ left(x+dfrac 13right)y=e^-2xy^-2$$
I shall also divide it by $x$. So, the ODE becomes:
$$y'+left(1+dfrac13xright)y =dfrace^-2xxy^-2tag 1$$
This is Bernoulli's ODE.
We substitute $v=y^1-(-2)=y^3$.
So, $dfracmathrm dymathrm dx=dfrac13v^(2/3)dfracmathrm dvmathrm dx$
$$beginalign(1)&implies dfrac13v^(2/3)dfracmathrm dvmathrm dx+left(1+dfrac13xright)v^1/3=dfrace^-2xxv^-2/3\ &implies dfracmathrm dvmathrm dx+ 3left(1+dfrac13xright)v=3dfrace^-2xxendalign$$
I think you can take it from here.
answered Aug 20 at 6:13
DarkKnight
4731211
edited Aug 20 at 6:23
answered Aug 20 at 6:13
DarkKnight
4731211
answered Aug 20 at 6:13
DarkKnight
4731211
answered Aug 20 at 6:13
DarkKnight
4731211
4731211
add a comment |Â
add a comment |Â
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Except in this case, it's $v=y^1-(-2)=y^3$.
– John Wayland Bales
Aug 20 at 5:45
1
Bernoulli equation $$fracdydx+P(x)y=Q(x)y^n$$ can be reduced to a linear equation by using $v=y^1-n$. Then, for the given equation, use $v=y^1-(-2)=y^3$.
– Ãngel Mario Gallegos
Aug 20 at 5:49
your third step is wrong there should be y^3
– James
Aug 20 at 5:51