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What is a possible substitution to this ODE?
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3
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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
asked 2 hours ago
Duartejfs
234
add a comment |Â
up vote
3
down vote
favorite
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
asked 2 hours ago
Duartejfs
234
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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
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
asked 2 hours ago
Duartejfs
234
asked 2 hours ago
Duartejfs
234
asked 2 hours ago
Duartejfs
234
asked 2 hours ago
Duartejfs
234
asked 2 hours ago
Duartejfs
234
234
add a comment |Â
add a comment |Â
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
$$x(y)=frac 13(y-ln(y-2x+1))+c$$
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 |Â
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
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$
answered 34 mins ago
Claude Leibovici
113k1155128
add a comment |Â
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
$$x(y)=frac 13(y-ln(y-2x+1))+c$$
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 |Â
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
$$x(y)=frac 13(y-ln(y-2x+1))+c$$
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 |Â
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
$$x(y)=frac 13(y-ln(y-2x+1))+c$$
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
$$x(y)=frac 13(y-ln(y-2x+1))+c$$
answered 2 hours ago
Isham
11.2k3929
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 |Â
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 |Â
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
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$
answered 34 mins ago
Claude Leibovici
113k1155128
add a comment |Â
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
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$
answered 34 mins ago
Claude Leibovici
113k1155128
add a comment |Â
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
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$
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
$$y=-Wleft((2 x_0-y_0-1) e^(3 x_0-y_0-x-1)right)+2 x-1$$
answered 34 mins ago
Claude Leibovici
113k1155128
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 |Â
add a comment |Â
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