How to solve ode of this form












6












$begingroup$


$
a_n(y')^n + a_{n-1}(y')^{n-1} + cdots + a_0 y' =0
$



I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before.






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
    $endgroup$
    – Seth
    1 hour ago






  • 1




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    1 hour ago










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    1 hour ago


















6












$begingroup$


$
a_n(y')^n + a_{n-1}(y')^{n-1} + cdots + a_0 y' =0
$



I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before.






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
    $endgroup$
    – Seth
    1 hour ago






  • 1




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    1 hour ago










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    1 hour ago
















6












6








6


2



$begingroup$


$
a_n(y')^n + a_{n-1}(y')^{n-1} + cdots + a_0 y' =0
$



I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before.






ordinary-differential-equations







share|cite|improve this question









$endgroup$




$
a_n(y')^n + a_{n-1}(y')^{n-1} + cdots + a_0 y' =0
$



I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before.






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









user47475user47475

634




634








  • 1




    $begingroup$
    I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
    $endgroup$
    – Seth
    1 hour ago






  • 1




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    1 hour ago










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    1 hour ago
















  • 1




    $begingroup$
    I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
    $endgroup$
    – Seth
    1 hour ago






  • 1




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    1 hour ago










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    1 hour ago










1




1




$begingroup$
I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
$endgroup$
– Seth
1 hour ago




$begingroup$
I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
$endgroup$
– Seth
1 hour ago




1




1




$begingroup$
Do you mean $cdots + a_1 y^prime + a_0$ at the end?
$endgroup$
– parsiad
1 hour ago




$begingroup$
Do you mean $cdots + a_1 y^prime + a_0$ at the end?
$endgroup$
– parsiad
1 hour ago












$begingroup$
Would make sense @parsiad
$endgroup$
– Neo Darwin
1 hour ago




$begingroup$
Would make sense @parsiad
$endgroup$
– Neo Darwin
1 hour ago




1




1




$begingroup$
@NeoDarwin: encouraging OP to edit the post.
$endgroup$
– parsiad
1 hour ago






$begingroup$
@NeoDarwin: encouraging OP to edit the post.
$endgroup$
– parsiad
1 hour ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Well. As pointed out in the comments by Seth.



Let us consider an example



begin{align}
(y')^2+2y'+1=0
end{align}
then it follows $y' = -1$ so $y = -t+C$.



So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What if it does not have real roots?
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
    $endgroup$
    – Jacky Chong
    1 hour ago












  • $begingroup$
    Thanks. Good answer and logical.
    $endgroup$
    – Neo Darwin
    1 hour ago



















1












$begingroup$


TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
$$
a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
$$
By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
$$
a_{n}r^{n}+cdots+a_{1}r+a_{0}
$$
has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
$$
left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
$$
By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Well. As pointed out in the comments by Seth.



    Let us consider an example



    begin{align}
    (y')^2+2y'+1=0
    end{align}
    then it follows $y' = -1$ so $y = -t+C$.



    So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What if it does not have real roots?
      $endgroup$
      – Neo Darwin
      1 hour ago






    • 1




      $begingroup$
      Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
      $endgroup$
      – Jacky Chong
      1 hour ago












    • $begingroup$
      Thanks. Good answer and logical.
      $endgroup$
      – Neo Darwin
      1 hour ago
















    3












    $begingroup$

    Well. As pointed out in the comments by Seth.



    Let us consider an example



    begin{align}
    (y')^2+2y'+1=0
    end{align}
    then it follows $y' = -1$ so $y = -t+C$.



    So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What if it does not have real roots?
      $endgroup$
      – Neo Darwin
      1 hour ago






    • 1




      $begingroup$
      Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
      $endgroup$
      – Jacky Chong
      1 hour ago












    • $begingroup$
      Thanks. Good answer and logical.
      $endgroup$
      – Neo Darwin
      1 hour ago














    3












    3








    3





    $begingroup$

    Well. As pointed out in the comments by Seth.



    Let us consider an example



    begin{align}
    (y')^2+2y'+1=0
    end{align}
    then it follows $y' = -1$ so $y = -t+C$.



    So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.






    share|cite|improve this answer









    $endgroup$



    Well. As pointed out in the comments by Seth.



    Let us consider an example



    begin{align}
    (y')^2+2y'+1=0
    end{align}
    then it follows $y' = -1$ so $y = -t+C$.



    So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    Jacky ChongJacky Chong

    18.6k21128




    18.6k21128












    • $begingroup$
      What if it does not have real roots?
      $endgroup$
      – Neo Darwin
      1 hour ago






    • 1




      $begingroup$
      Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
      $endgroup$
      – Jacky Chong
      1 hour ago












    • $begingroup$
      Thanks. Good answer and logical.
      $endgroup$
      – Neo Darwin
      1 hour ago


















    • $begingroup$
      What if it does not have real roots?
      $endgroup$
      – Neo Darwin
      1 hour ago






    • 1




      $begingroup$
      Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
      $endgroup$
      – Jacky Chong
      1 hour ago












    • $begingroup$
      Thanks. Good answer and logical.
      $endgroup$
      – Neo Darwin
      1 hour ago
















    $begingroup$
    What if it does not have real roots?
    $endgroup$
    – Neo Darwin
    1 hour ago




    $begingroup$
    What if it does not have real roots?
    $endgroup$
    – Neo Darwin
    1 hour ago




    1




    1




    $begingroup$
    Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
    $endgroup$
    – Jacky Chong
    1 hour ago






    $begingroup$
    Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
    $endgroup$
    – Jacky Chong
    1 hour ago














    $begingroup$
    Thanks. Good answer and logical.
    $endgroup$
    – Neo Darwin
    1 hour ago




    $begingroup$
    Thanks. Good answer and logical.
    $endgroup$
    – Neo Darwin
    1 hour ago











    1












    $begingroup$


    TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




    Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
    $$
    a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
    $$
    By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
    $$
    a_{n}r^{n}+cdots+a_{1}r+a_{0}
    $$
    has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
    $$
    left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
    $$
    By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
    In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$


      TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




      Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
      $$
      a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
      $$
      By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
      $$
      a_{n}r^{n}+cdots+a_{1}r+a_{0}
      $$
      has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
      $$
      left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
      $$
      By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
      In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$


        TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




        Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
        $$
        a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
        $$
        By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
        $$
        a_{n}r^{n}+cdots+a_{1}r+a_{0}
        $$
        has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
        $$
        left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
        $$
        By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
        In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.






        share|cite|improve this answer











        $endgroup$




        TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




        Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
        $$
        a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
        $$
        By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
        $$
        a_{n}r^{n}+cdots+a_{1}r+a_{0}
        $$
        has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
        $$
        left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
        $$
        By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
        In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        parsiadparsiad

        17.5k32353




        17.5k32353






























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