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:
$$
frace^y-2xy-2x=x+K
$$

where K is any constant. At this point I am unable to solve for y.



Can anyone help me solve this?










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    up vote
    5
    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:
    $$
    frace^y-2xy-2x=x+K
    $$

    where K is any constant. At this point I am unable to solve for y.



    Can anyone help me solve this?










    share|cite|improve this question























      up vote
      5
      down vote

      favorite
      1









      up vote
      5
      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:
      $$
      frace^y-2xy-2x=x+K
      $$

      where K is any constant. At this point I am unable to solve for y.



      Can anyone help me solve this?










      share|cite|improve this question













      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:
      $$
      frace^y-2xy-2x=x+K
      $$

      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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      share|cite|improve this question











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      share|cite|improve this question










      asked Sep 22 at 11:03









      Duartejfs

      484




      484




















          2 Answers
          2






          active

          oldest

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          up vote
          5
          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




















          • 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
            Sep 22 at 11:20







          • 3




            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
            Sep 22 at 13:34

















          up vote
          4
          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




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            5
            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




















            • 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
              Sep 22 at 11:20







            • 3




              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
              Sep 22 at 13:34














            up vote
            5
            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




















            • 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
              Sep 22 at 11:20







            • 3




              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
              Sep 22 at 13:34












            up vote
            5
            down vote



            accepted







            up vote
            5
            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












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












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 22 at 11:13









            Isham

            11.5k3929




            11.5k3929











            • 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
              Sep 22 at 11:20







            • 3




              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
              Sep 22 at 13:34
















            • 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
              Sep 22 at 11:20







            • 3




              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
              Sep 22 at 13:34















            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
            Sep 22 at 11:20





            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
            Sep 22 at 11:20





            3




            3




            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
            Sep 22 at 13:34




            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
            Sep 22 at 13:34










            up vote
            4
            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
























              up vote
              4
              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






















                up vote
                4
                down vote










                up vote
                4
                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












                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












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 22 at 13:23









                Claude Leibovici

                114k1155129




                114k1155129



























                     

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