How to solve ode of the form $ a_n(y')^n + a_n-1(y')^n-1 + cdots + a_1 y' + a_0 =0 $

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6












$begingroup$


$
a_n(y')^n + a_n-1(y')^n-1 + cdots + a_1 y'+a_0 =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







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$endgroup$







  • 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
    Feb 14 at 4:07






  • 3




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    Feb 14 at 4:19










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    Feb 14 at 4:22






  • 2




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    Feb 14 at 4:23
















6












$begingroup$


$
a_n(y')^n + a_n-1(y')^n-1 + cdots + a_1 y'+a_0 =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
    Feb 14 at 4:07






  • 3




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    Feb 14 at 4:19










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    Feb 14 at 4:22






  • 2




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    Feb 14 at 4:23














6












6








6


2



$begingroup$


$
a_n(y')^n + a_n-1(y')^n-1 + cdots + a_1 y'+a_0 =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_1 y'+a_0 =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








edited Feb 14 at 16:02







user47475

















asked Feb 14 at 4:05









user47475user47475

656




656







  • 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
    Feb 14 at 4:07






  • 3




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    Feb 14 at 4:19










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    Feb 14 at 4:22






  • 2




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    Feb 14 at 4:23













  • 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
    Feb 14 at 4:07






  • 3




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    Feb 14 at 4:19










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    Feb 14 at 4:22






  • 2




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    Feb 14 at 4:23








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
Feb 14 at 4:07




$begingroup$
I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
$endgroup$
– Seth
Feb 14 at 4:07




3




3




$begingroup$
Do you mean $cdots + a_1 y^prime + a_0$ at the end?
$endgroup$
– parsiad
Feb 14 at 4:19




$begingroup$
Do you mean $cdots + a_1 y^prime + a_0$ at the end?
$endgroup$
– parsiad
Feb 14 at 4:19












$begingroup$
Would make sense @parsiad
$endgroup$
– Neo Darwin
Feb 14 at 4:22




$begingroup$
Would make sense @parsiad
$endgroup$
– Neo Darwin
Feb 14 at 4:22




2




2




$begingroup$
@NeoDarwin: encouraging OP to edit the post.
$endgroup$
– parsiad
Feb 14 at 4:23





$begingroup$
@NeoDarwin: encouraging OP to edit the post.
$endgroup$
– parsiad
Feb 14 at 4:23











2 Answers
2






active

oldest

votes


















3












$begingroup$


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 $y:[0,T)rightarrowmathbbR$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
$$
a_n(y^prime(t))^n+cdots+a_1y^prime(t)+a_0=0.
$$
By the fundamental theorem of algebra, if $a_nneq0$, the corresponding polynomial
$$
a_nr^n+cdots+a_1r+a_0
$$
has $n$ complex roots, call them $z_1,ldots,z_n$. We can re-express the ODE as
$$
left(y^prime(t)-z_1right)cdotsleft(y^prime(t)-z_nright)=0.
$$
By the above, $y^prime(t)=z_k(t)$ where $k(t)in1,ldots,n$ 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











$endgroup$




















    5












    $begingroup$

    Well. As pointed out in the comments by Seth.



    Let us consider an example



    beginalign
    (y')^2+2y'+1=0
    endalign
    then it follows $y' = -1$ so $y = -t+C$.



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






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      What if it does not have real roots?
      $endgroup$
      – Neo Darwin
      Feb 14 at 4:32






    • 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
      Feb 14 at 4:33











    • $begingroup$
      Thanks. Good answer and logical.
      $endgroup$
      – Neo Darwin
      Feb 14 at 4:43










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$


    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 $y:[0,T)rightarrowmathbbR$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
    $$
    a_n(y^prime(t))^n+cdots+a_1y^prime(t)+a_0=0.
    $$
    By the fundamental theorem of algebra, if $a_nneq0$, the corresponding polynomial
    $$
    a_nr^n+cdots+a_1r+a_0
    $$
    has $n$ complex roots, call them $z_1,ldots,z_n$. We can re-express the ODE as
    $$
    left(y^prime(t)-z_1right)cdotsleft(y^prime(t)-z_nright)=0.
    $$
    By the above, $y^prime(t)=z_k(t)$ where $k(t)in1,ldots,n$ 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











    $endgroup$

















      3












      $begingroup$


      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 $y:[0,T)rightarrowmathbbR$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
      $$
      a_n(y^prime(t))^n+cdots+a_1y^prime(t)+a_0=0.
      $$
      By the fundamental theorem of algebra, if $a_nneq0$, the corresponding polynomial
      $$
      a_nr^n+cdots+a_1r+a_0
      $$
      has $n$ complex roots, call them $z_1,ldots,z_n$. We can re-express the ODE as
      $$
      left(y^prime(t)-z_1right)cdotsleft(y^prime(t)-z_nright)=0.
      $$
      By the above, $y^prime(t)=z_k(t)$ where $k(t)in1,ldots,n$ 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











      $endgroup$















        3












        3








        3





        $begingroup$


        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 $y:[0,T)rightarrowmathbbR$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
        $$
        a_n(y^prime(t))^n+cdots+a_1y^prime(t)+a_0=0.
        $$
        By the fundamental theorem of algebra, if $a_nneq0$, the corresponding polynomial
        $$
        a_nr^n+cdots+a_1r+a_0
        $$
        has $n$ complex roots, call them $z_1,ldots,z_n$. We can re-express the ODE as
        $$
        left(y^prime(t)-z_1right)cdotsleft(y^prime(t)-z_nright)=0.
        $$
        By the above, $y^prime(t)=z_k(t)$ where $k(t)in1,ldots,n$ 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 $y:[0,T)rightarrowmathbbR$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
        $$
        a_n(y^prime(t))^n+cdots+a_1y^prime(t)+a_0=0.
        $$
        By the fundamental theorem of algebra, if $a_nneq0$, the corresponding polynomial
        $$
        a_nr^n+cdots+a_1r+a_0
        $$
        has $n$ complex roots, call them $z_1,ldots,z_n$. We can re-express the ODE as
        $$
        left(y^prime(t)-z_1right)cdotsleft(y^prime(t)-z_nright)=0.
        $$
        By the above, $y^prime(t)=z_k(t)$ where $k(t)in1,ldots,n$ 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 Feb 14 at 4:50

























        answered Feb 14 at 4:44









        parsiadparsiad

        18.1k32453




        18.1k32453





















            5












            $begingroup$

            Well. As pointed out in the comments by Seth.



            Let us consider an example



            beginalign
            (y')^2+2y'+1=0
            endalign
            then it follows $y' = -1$ so $y = -t+C$.



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






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              What if it does not have real roots?
              $endgroup$
              – Neo Darwin
              Feb 14 at 4:32






            • 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
              Feb 14 at 4:33











            • $begingroup$
              Thanks. Good answer and logical.
              $endgroup$
              – Neo Darwin
              Feb 14 at 4:43















            5












            $begingroup$

            Well. As pointed out in the comments by Seth.



            Let us consider an example



            beginalign
            (y')^2+2y'+1=0
            endalign
            then it follows $y' = -1$ so $y = -t+C$.



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






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              What if it does not have real roots?
              $endgroup$
              – Neo Darwin
              Feb 14 at 4:32






            • 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
              Feb 14 at 4:33











            • $begingroup$
              Thanks. Good answer and logical.
              $endgroup$
              – Neo Darwin
              Feb 14 at 4:43













            5












            5








            5





            $begingroup$

            Well. As pointed out in the comments by Seth.



            Let us consider an example



            beginalign
            (y')^2+2y'+1=0
            endalign
            then it follows $y' = -1$ so $y = -t+C$.



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






            share|cite|improve this answer









            $endgroup$



            Well. As pointed out in the comments by Seth.



            Let us consider an example



            beginalign
            (y')^2+2y'+1=0
            endalign
            then it follows $y' = -1$ so $y = -t+C$.



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







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 14 at 4:22









            Jacky ChongJacky Chong

            19.1k21129




            19.1k21129











            • $begingroup$
              What if it does not have real roots?
              $endgroup$
              – Neo Darwin
              Feb 14 at 4:32






            • 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
              Feb 14 at 4:33











            • $begingroup$
              Thanks. Good answer and logical.
              $endgroup$
              – Neo Darwin
              Feb 14 at 4:43
















            • $begingroup$
              What if it does not have real roots?
              $endgroup$
              – Neo Darwin
              Feb 14 at 4:32






            • 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
              Feb 14 at 4:33











            • $begingroup$
              Thanks. Good answer and logical.
              $endgroup$
              – Neo Darwin
              Feb 14 at 4:43















            $begingroup$
            What if it does not have real roots?
            $endgroup$
            – Neo Darwin
            Feb 14 at 4:32




            $begingroup$
            What if it does not have real roots?
            $endgroup$
            – Neo Darwin
            Feb 14 at 4:32




            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
            Feb 14 at 4:33





            $begingroup$
            Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
            $endgroup$
            – Jacky Chong
            Feb 14 at 4:33













            $begingroup$
            Thanks. Good answer and logical.
            $endgroup$
            – Neo Darwin
            Feb 14 at 4:43




            $begingroup$
            Thanks. Good answer and logical.
            $endgroup$
            – Neo Darwin
            Feb 14 at 4:43

















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