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











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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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            一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
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            Setup Asymptote in Texstudio

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            0 Good time of the day! I am trying to compile the following file: documentclass[12pt]{article} usepackage[inline]{asymptote} begin{document} Here is a venn diagram produced with Asymptote, drawn to width 4cm: defA{A} defB{V{B}} begin{figure} begin{center} begin{asy} size(4cm,0); pen colour1=red; pen colour2=green; pair z0=(0,0); pair z1=(-1,0); pair z2=(1,0); real r=1.5; path c1=circle(z1,r); path c2=circle(z2,r); fill(c1,colour1); fill(c2,colour2); picture intersection=new picture; fill(intersection,c1,colour1+colour2); clip(intersection,c2); add(intersection); draw(c1); draw(c2); //draw("$A$",box,z1)
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