What is a possible substitution to this ODE?

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I have to solve this ODE:

$$
(y-2x)fracdydx=3y-6x+1
$$

I have to make it in a form of separable variable equation and for that I have tried the following substitutions:
$$
u=y-2x
$$
and
$$
u=3y-6x+1
$$

But none made it "solvable", for example in the first I get to a point where I have this:
$$
frace^y-2x=x+K
$$
where K is any constant. At this point I am unable to solve for y.

Can anyone help me solve this?

differential-equations

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asked 2 hours ago

Duartejfs

234

add a comment | 

up vote
3
down vote

favorite

1

I have to solve this ODE:

$$
(y-2x)fracdydx=3y-6x+1
$$

I have to make it in a form of separable variable equation and for that I have tried the following substitutions:
$$
u=y-2x
$$
and
$$
u=3y-6x+1
$$

But none made it "solvable", for example in the first I get to a point where I have this:
$$
frace^y-2x=x+K
$$
where K is any constant. At this point I am unable to solve for y.

Can anyone help me solve this?

differential-equations

share|cite|improve this question

asked 2 hours ago

Duartejfs

234

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

I have to solve this ODE:

$$
(y-2x)fracdydx=3y-6x+1
$$

I have to make it in a form of separable variable equation and for that I have tried the following substitutions:
$$
u=y-2x
$$
and
$$
u=3y-6x+1
$$

But none made it "solvable", for example in the first I get to a point where I have this:
$$
frace^y-2x=x+K
$$
where K is any constant. At this point I am unable to solve for y.

Can anyone help me solve this?

differential-equations

share|cite|improve this question

asked 2 hours ago

Duartejfs

234

I have to solve this ODE:

$$
(y-2x)fracdydx=3y-6x+1
$$

I have to make it in a form of separable variable equation and for that I have tried the following substitutions:
$$
u=y-2x
$$
and
$$
u=3y-6x+1
$$

But none made it "solvable", for example in the first I get to a point where I have this:
$$
frace^y-2x=x+K
$$
where K is any constant. At this point I am unable to solve for y.

Can anyone help me solve this?

differential-equations

differential-equations

share|cite|improve this question

asked 2 hours ago

Duartejfs

234

share|cite|improve this question

asked 2 hours ago

Duartejfs

234

share|cite|improve this question

share|cite|improve this question

asked 2 hours ago

Duartejfs

234

asked 2 hours ago

Duartejfs

234

asked 2 hours ago

Duartejfs

234

234

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

$$(y-2x)fracdydx=3y-6x+1$$
Substitute $y-2x=z implies y'-2=z'$
$$z(z'+2)=3z+1$$
$$zz'=z+1$$
$$int frac zdzz+1=x+K$$
$$z-ln (z+1)=x+K$$
$$y -ln(y-2x+1)=3x+K$$
Implicit form is correct too. So you can keep it this way

Or consider x as a function of y
$$x(y)=frac 13(y-ln(y-2x+1))+c$$

share|cite|improve this answer

answered 2 hours ago

Isham

11.2k3929

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I had gotten to, but I didn't know I could leave it in the implicit form. By the way, if I was given the condition $$y(x_0)=y_0$$ how could I find the maximal solution with the implicit form?
  – Duartejfs
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If this is homework, whether you can "leave it in the implicit form" or not depends upon your teacher! Ask him or her. As for "$y(x_0)= y_0$". since you have $y- ln(y- 2x+ 1)= 3x+ K$, replace y with $y_0$ and x with $x_0$: $y_0- ln(y_0- 2x_0+ 1)= 3x_0+ K$ so that $K= y_0- 3x_0- ln(y_0- 2x_0+ 1)$ and $y- ln(y- 2x+ 1)= 3x+ y_0- 3x_0- ln(y_0- 2x_0+ 1)$.
  – user247327
  24 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Starting from Isham's answer.

Using
$$3x+c=y-ln(y-2x+1)$$ the explicit solution is given by
$$y=-Wleft(-e^-(x+c+1)right)+2 x-1$$ where appears Lambert function.

Using the condition $y(x_0)=y_0$, we get
$$c=-log left(-(2 x_0-y_0-1), e^(2 x_0-y_0-1)right)-x_0-1$$ and then
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$

share|cite|improve this answer

answered 34 mins ago

Claude Leibovici

113k1155128

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

$$(y-2x)fracdydx=3y-6x+1$$
Substitute $y-2x=z implies y'-2=z'$
$$z(z'+2)=3z+1$$
$$zz'=z+1$$
$$int frac zdzz+1=x+K$$
$$z-ln (z+1)=x+K$$
$$y -ln(y-2x+1)=3x+K$$
Implicit form is correct too. So you can keep it this way

Or consider x as a function of y
$$x(y)=frac 13(y-ln(y-2x+1))+c$$

share|cite|improve this answer

answered 2 hours ago

Isham

11.2k3929

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I had gotten to, but I didn't know I could leave it in the implicit form. By the way, if I was given the condition $$y(x_0)=y_0$$ how could I find the maximal solution with the implicit form?
  – Duartejfs
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If this is homework, whether you can "leave it in the implicit form" or not depends upon your teacher! Ask him or her. As for "$y(x_0)= y_0$". since you have $y- ln(y- 2x+ 1)= 3x+ K$, replace y with $y_0$ and x with $x_0$: $y_0- ln(y_0- 2x_0+ 1)= 3x_0+ K$ so that $K= y_0- 3x_0- ln(y_0- 2x_0+ 1)$ and $y- ln(y- 2x+ 1)= 3x+ y_0- 3x_0- ln(y_0- 2x_0+ 1)$.
  – user247327
  24 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

$$(y-2x)fracdydx=3y-6x+1$$
Substitute $y-2x=z implies y'-2=z'$
$$z(z'+2)=3z+1$$
$$zz'=z+1$$
$$int frac zdzz+1=x+K$$
$$z-ln (z+1)=x+K$$
$$y -ln(y-2x+1)=3x+K$$
Implicit form is correct too. So you can keep it this way

Or consider x as a function of y
$$x(y)=frac 13(y-ln(y-2x+1))+c$$

share|cite|improve this answer

answered 2 hours ago

Isham

11.2k3929

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I had gotten to, but I didn't know I could leave it in the implicit form. By the way, if I was given the condition $$y(x_0)=y_0$$ how could I find the maximal solution with the implicit form?
  – Duartejfs
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If this is homework, whether you can "leave it in the implicit form" or not depends upon your teacher! Ask him or her. As for "$y(x_0)= y_0$". since you have $y- ln(y- 2x+ 1)= 3x+ K$, replace y with $y_0$ and x with $x_0$: $y_0- ln(y_0- 2x_0+ 1)= 3x_0+ K$ so that $K= y_0- 3x_0- ln(y_0- 2x_0+ 1)$ and $y- ln(y- 2x+ 1)= 3x+ y_0- 3x_0- ln(y_0- 2x_0+ 1)$.
  – user247327
  24 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

$$(y-2x)fracdydx=3y-6x+1$$
Substitute $y-2x=z implies y'-2=z'$
$$z(z'+2)=3z+1$$
$$zz'=z+1$$
$$int frac zdzz+1=x+K$$
$$z-ln (z+1)=x+K$$
$$y -ln(y-2x+1)=3x+K$$
Implicit form is correct too. So you can keep it this way

Or consider x as a function of y
$$x(y)=frac 13(y-ln(y-2x+1))+c$$

share|cite|improve this answer

answered 2 hours ago

Isham

11.2k3929

$$(y-2x)fracdydx=3y-6x+1$$
Substitute $y-2x=z implies y'-2=z'$
$$z(z'+2)=3z+1$$
$$zz'=z+1$$
$$int frac zdzz+1=x+K$$
$$z-ln (z+1)=x+K$$
$$y -ln(y-2x+1)=3x+K$$
Implicit form is correct too. So you can keep it this way

Or consider x as a function of y
$$x(y)=frac 13(y-ln(y-2x+1))+c$$

share|cite|improve this answer

answered 2 hours ago

Isham

11.2k3929

share|cite|improve this answer

share|cite|improve this answer

answered 2 hours ago

Isham

11.2k3929

answered 2 hours ago

Isham

11.2k3929

answered 2 hours ago

Isham

11.2k3929

11.2k3929

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I had gotten to, but I didn't know I could leave it in the implicit form. By the way, if I was given the condition $$y(x_0)=y_0$$ how could I find the maximal solution with the implicit form?
  – Duartejfs
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If this is homework, whether you can "leave it in the implicit form" or not depends upon your teacher! Ask him or her. As for "$y(x_0)= y_0$". since you have $y- ln(y- 2x+ 1)= 3x+ K$, replace y with $y_0$ and x with $x_0$: $y_0- ln(y_0- 2x_0+ 1)= 3x_0+ K$ so that $K= y_0- 3x_0- ln(y_0- 2x_0+ 1)$ and $y- ln(y- 2x+ 1)= 3x+ y_0- 3x_0- ln(y_0- 2x_0+ 1)$.
  – user247327
  24 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I had gotten to, but I didn't know I could leave it in the implicit form. By the way, if I was given the condition $$y(x_0)=y_0$$ how could I find the maximal solution with the implicit form?
  – Duartejfs
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If this is homework, whether you can "leave it in the implicit form" or not depends upon your teacher! Ask him or her. As for "$y(x_0)= y_0$". since you have $y- ln(y- 2x+ 1)= 3x+ K$, replace y with $y_0$ and x with $x_0$: $y_0- ln(y_0- 2x_0+ 1)= 3x_0+ K$ so that $K= y_0- 3x_0- ln(y_0- 2x_0+ 1)$ and $y- ln(y- 2x+ 1)= 3x+ y_0- 3x_0- ln(y_0- 2x_0+ 1)$.
  – user247327
  24 mins ago
  

  
  
  
  

That's what I had gotten to, but I didn't know I could leave it in the implicit form. By the way, if I was given the condition $$y(x_0)=y_0$$ how could I find the maximal solution with the implicit form?
– Duartejfs
2 hours ago

That's what I had gotten to, but I didn't know I could leave it in the implicit form. By the way, if I was given the condition $$y(x_0)=y_0$$ how could I find the maximal solution with the implicit form?
– Duartejfs
2 hours ago

1

1

If this is homework, whether you can "leave it in the implicit form" or not depends upon your teacher! Ask him or her. As for "$y(x_0)= y_0$". since you have $y- ln(y- 2x+ 1)= 3x+ K$, replace y with $y_0$ and x with $x_0$: $y_0- ln(y_0- 2x_0+ 1)= 3x_0+ K$ so that $K= y_0- 3x_0- ln(y_0- 2x_0+ 1)$ and $y- ln(y- 2x+ 1)= 3x+ y_0- 3x_0- ln(y_0- 2x_0+ 1)$.
– user247327
24 mins ago

If this is homework, whether you can "leave it in the implicit form" or not depends upon your teacher! Ask him or her. As for "$y(x_0)= y_0$". since you have $y- ln(y- 2x+ 1)= 3x+ K$, replace y with $y_0$ and x with $x_0$: $y_0- ln(y_0- 2x_0+ 1)= 3x_0+ K$ so that $K= y_0- 3x_0- ln(y_0- 2x_0+ 1)$ and $y- ln(y- 2x+ 1)= 3x+ y_0- 3x_0- ln(y_0- 2x_0+ 1)$.
– user247327
24 mins ago

add a comment | 

up vote
1
down vote

Starting from Isham's answer.

Using
$$3x+c=y-ln(y-2x+1)$$ the explicit solution is given by
$$y=-Wleft(-e^-(x+c+1)right)+2 x-1$$ where appears Lambert function.

Using the condition $y(x_0)=y_0$, we get
$$c=-log left(-(2 x_0-y_0-1), e^(2 x_0-y_0-1)right)-x_0-1$$ and then
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$

share|cite|improve this answer

answered 34 mins ago

Claude Leibovici

113k1155128

add a comment | 

up vote
1
down vote

Starting from Isham's answer.

Using
$$3x+c=y-ln(y-2x+1)$$ the explicit solution is given by
$$y=-Wleft(-e^-(x+c+1)right)+2 x-1$$ where appears Lambert function.

Using the condition $y(x_0)=y_0$, we get
$$c=-log left(-(2 x_0-y_0-1), e^(2 x_0-y_0-1)right)-x_0-1$$ and then
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$

share|cite|improve this answer

answered 34 mins ago

Claude Leibovici

113k1155128

add a comment | 

up vote
1
down vote

up vote
1
down vote

Starting from Isham's answer.

Using
$$3x+c=y-ln(y-2x+1)$$ the explicit solution is given by
$$y=-Wleft(-e^-(x+c+1)right)+2 x-1$$ where appears Lambert function.

Using the condition $y(x_0)=y_0$, we get
$$c=-log left(-(2 x_0-y_0-1), e^(2 x_0-y_0-1)right)-x_0-1$$ and then
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$

share|cite|improve this answer

answered 34 mins ago

Claude Leibovici

113k1155128

Starting from Isham's answer.

Using
$$3x+c=y-ln(y-2x+1)$$ the explicit solution is given by
$$y=-Wleft(-e^-(x+c+1)right)+2 x-1$$ where appears Lambert function.

Using the condition $y(x_0)=y_0$, we get
$$c=-log left(-(2 x_0-y_0-1), e^(2 x_0-y_0-1)right)-x_0-1$$ and then
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$

share|cite|improve this answer

answered 34 mins ago

Claude Leibovici

113k1155128

share|cite|improve this answer

share|cite|improve this answer

answered 34 mins ago

Claude Leibovici

113k1155128

answered 34 mins ago

Claude Leibovici

113k1155128

answered 34 mins ago

Claude Leibovici

113k1155128

113k1155128

add a comment | 

add a comment | 

 

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