solve a second order ODE











up vote
0
down vote

favorite












Can we solve analytically (find a closed form solution) the second order ODE



$$x^{alpha}y^{primeprime}=y,quad x>0$$
where $alphain,]0,1[$.
Consider the conditions
$$y(0)=1,quad y^{prime}(0)=1$$
This equation appears in a fractional model that describes viscoelasticity properties in certain materials.










share|cite|improve this question
























  • Fourier transform?
    – Rushabh Mehta
    Nov 22 at 3:19










  • Could you precise the boundary conditions ?
    – Claude Leibovici
    Nov 22 at 3:23










  • @ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
    – Medo
    Nov 22 at 3:28










  • Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
    – Medo
    Nov 22 at 3:29










  • @ Mattos. That at best would give a special function. Can we go around that?
    – Medo
    Nov 22 at 3:32















up vote
0
down vote

favorite












Can we solve analytically (find a closed form solution) the second order ODE



$$x^{alpha}y^{primeprime}=y,quad x>0$$
where $alphain,]0,1[$.
Consider the conditions
$$y(0)=1,quad y^{prime}(0)=1$$
This equation appears in a fractional model that describes viscoelasticity properties in certain materials.










share|cite|improve this question
























  • Fourier transform?
    – Rushabh Mehta
    Nov 22 at 3:19










  • Could you precise the boundary conditions ?
    – Claude Leibovici
    Nov 22 at 3:23










  • @ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
    – Medo
    Nov 22 at 3:28










  • Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
    – Medo
    Nov 22 at 3:29










  • @ Mattos. That at best would give a special function. Can we go around that?
    – Medo
    Nov 22 at 3:32













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Can we solve analytically (find a closed form solution) the second order ODE



$$x^{alpha}y^{primeprime}=y,quad x>0$$
where $alphain,]0,1[$.
Consider the conditions
$$y(0)=1,quad y^{prime}(0)=1$$
This equation appears in a fractional model that describes viscoelasticity properties in certain materials.










share|cite|improve this question















Can we solve analytically (find a closed form solution) the second order ODE



$$x^{alpha}y^{primeprime}=y,quad x>0$$
where $alphain,]0,1[$.
Consider the conditions
$$y(0)=1,quad y^{prime}(0)=1$$
This equation appears in a fractional model that describes viscoelasticity properties in certain materials.







differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 3:31

























asked Nov 22 at 3:00









Medo

611213




611213












  • Fourier transform?
    – Rushabh Mehta
    Nov 22 at 3:19










  • Could you precise the boundary conditions ?
    – Claude Leibovici
    Nov 22 at 3:23










  • @ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
    – Medo
    Nov 22 at 3:28










  • Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
    – Medo
    Nov 22 at 3:29










  • @ Mattos. That at best would give a special function. Can we go around that?
    – Medo
    Nov 22 at 3:32


















  • Fourier transform?
    – Rushabh Mehta
    Nov 22 at 3:19










  • Could you precise the boundary conditions ?
    – Claude Leibovici
    Nov 22 at 3:23










  • @ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
    – Medo
    Nov 22 at 3:28










  • Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
    – Medo
    Nov 22 at 3:29










  • @ Mattos. That at best would give a special function. Can we go around that?
    – Medo
    Nov 22 at 3:32
















Fourier transform?
– Rushabh Mehta
Nov 22 at 3:19




Fourier transform?
– Rushabh Mehta
Nov 22 at 3:19












Could you precise the boundary conditions ?
– Claude Leibovici
Nov 22 at 3:23




Could you precise the boundary conditions ?
– Claude Leibovici
Nov 22 at 3:23












@ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
– Medo
Nov 22 at 3:28




@ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
– Medo
Nov 22 at 3:28












Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
– Medo
Nov 22 at 3:29




Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
– Medo
Nov 22 at 3:29












@ Mattos. That at best would give a special function. Can we go around that?
– Medo
Nov 22 at 3:32




@ Mattos. That at best would give a special function. Can we go around that?
– Medo
Nov 22 at 3:32










2 Answers
2






active

oldest

votes

















up vote
0
down vote













Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html



enter image description here



Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$






share|cite|improve this answer





















  • This does not seem to take into account the boundary conditions.
    – Claude Leibovici
    Nov 22 at 4:01












  • Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
    – Medo
    Nov 22 at 7:34












  • What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
    – Claude Leibovici
    Nov 22 at 7:44




















up vote
0
down vote













Too long for a comment.



As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.



Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
$$left(
begin{array}{ccc}
m & n & y(x) \
1 & 2 & , _0F_1left(;frac{1}{3};frac{4
x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
x^{3/2}}{9}right)\
1 & 3 & , _0F_1left(;frac{2}{5};frac{9
x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
x^{5/3}}{25}right)\
1 & 4 & , _0F_1left(;frac{3}{7};frac{16
x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
x^{7/4}}{49}right)\
1 & 5 & , _0F_1left(;frac{4}{9};frac{25
x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
x^{9/5}}{81}right)\
1 & 6 & , _0F_1left(;frac{5}{11};frac{36
x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
x^{11/6}}{121}right)\
1 & 7 & , _0F_1left(;frac{6}{13};frac{49
x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
x^{13/7}}{169}right) \
2 & 3 & , _0F_1left(;frac{1}{4};frac{9
x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
x^{4/3}}{16}right)\
2 & 5 & , _0F_1left(;frac{3}{8};frac{25
x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
x^{8/5}}{64}right)\
2 & 7 & , _0F_1left(;frac{5}{12};frac{49
x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
x^{12/7}}{144}right)\
3 & 4 & , _0F_1left(;frac{1}{5};frac{16
x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
x^{5/4}}{25}right)\
3 & 5 & , _0F_1left(;frac{2}{7};frac{25
x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
x^{7/5}}{49}right)\
3 & 7 & , _0F_1left(;frac{4}{11};frac{49
x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
x^{11/7}}{121}right)\
3 & 8 & , _0F_1left(;frac{5}{13};frac{64
x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
x^{13/8}}{169}right)\
4 & 5 & , _0F_1left(;frac{1}{6};frac{25
x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
x^{6/5}}{36}right)\
4 & 7 & , _0F_1left(;frac{3}{10};frac{49
x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
x^{10/7}}{100}right)\
4 & 9 & , _0F_1left(;frac{5}{14};frac{81
x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
x^{14/9}}{196}right)
end{array}
right) $$






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008696%2fsolve-a-second-order-ode%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
    http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html



    enter image description here



    Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$






    share|cite|improve this answer





















    • This does not seem to take into account the boundary conditions.
      – Claude Leibovici
      Nov 22 at 4:01












    • Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
      – Medo
      Nov 22 at 7:34












    • What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
      – Claude Leibovici
      Nov 22 at 7:44

















    up vote
    0
    down vote













    Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
    http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html



    enter image description here



    Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$






    share|cite|improve this answer





















    • This does not seem to take into account the boundary conditions.
      – Claude Leibovici
      Nov 22 at 4:01












    • Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
      – Medo
      Nov 22 at 7:34












    • What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
      – Claude Leibovici
      Nov 22 at 7:44















    up vote
    0
    down vote










    up vote
    0
    down vote









    Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
    http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html



    enter image description here



    Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$






    share|cite|improve this answer












    Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
    http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html



    enter image description here



    Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 22 at 3:58









    Medo

    611213




    611213












    • This does not seem to take into account the boundary conditions.
      – Claude Leibovici
      Nov 22 at 4:01












    • Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
      – Medo
      Nov 22 at 7:34












    • What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
      – Claude Leibovici
      Nov 22 at 7:44




















    • This does not seem to take into account the boundary conditions.
      – Claude Leibovici
      Nov 22 at 4:01












    • Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
      – Medo
      Nov 22 at 7:34












    • What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
      – Claude Leibovici
      Nov 22 at 7:44


















    This does not seem to take into account the boundary conditions.
    – Claude Leibovici
    Nov 22 at 4:01






    This does not seem to take into account the boundary conditions.
    – Claude Leibovici
    Nov 22 at 4:01














    Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
    – Medo
    Nov 22 at 7:34






    Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
    – Medo
    Nov 22 at 7:34














    What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
    – Claude Leibovici
    Nov 22 at 7:44






    What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
    – Claude Leibovici
    Nov 22 at 7:44












    up vote
    0
    down vote













    Too long for a comment.



    As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.



    Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
    $$left(
    begin{array}{ccc}
    m & n & y(x) \
    1 & 2 & , _0F_1left(;frac{1}{3};frac{4
    x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
    x^{3/2}}{9}right)\
    1 & 3 & , _0F_1left(;frac{2}{5};frac{9
    x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
    x^{5/3}}{25}right)\
    1 & 4 & , _0F_1left(;frac{3}{7};frac{16
    x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
    x^{7/4}}{49}right)\
    1 & 5 & , _0F_1left(;frac{4}{9};frac{25
    x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
    x^{9/5}}{81}right)\
    1 & 6 & , _0F_1left(;frac{5}{11};frac{36
    x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
    x^{11/6}}{121}right)\
    1 & 7 & , _0F_1left(;frac{6}{13};frac{49
    x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
    x^{13/7}}{169}right) \
    2 & 3 & , _0F_1left(;frac{1}{4};frac{9
    x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
    x^{4/3}}{16}right)\
    2 & 5 & , _0F_1left(;frac{3}{8};frac{25
    x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
    x^{8/5}}{64}right)\
    2 & 7 & , _0F_1left(;frac{5}{12};frac{49
    x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
    x^{12/7}}{144}right)\
    3 & 4 & , _0F_1left(;frac{1}{5};frac{16
    x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
    x^{5/4}}{25}right)\
    3 & 5 & , _0F_1left(;frac{2}{7};frac{25
    x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
    x^{7/5}}{49}right)\
    3 & 7 & , _0F_1left(;frac{4}{11};frac{49
    x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
    x^{11/7}}{121}right)\
    3 & 8 & , _0F_1left(;frac{5}{13};frac{64
    x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
    x^{13/8}}{169}right)\
    4 & 5 & , _0F_1left(;frac{1}{6};frac{25
    x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
    x^{6/5}}{36}right)\
    4 & 7 & , _0F_1left(;frac{3}{10};frac{49
    x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
    x^{10/7}}{100}right)\
    4 & 9 & , _0F_1left(;frac{5}{14};frac{81
    x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
    x^{14/9}}{196}right)
    end{array}
    right) $$






    share|cite|improve this answer

























      up vote
      0
      down vote













      Too long for a comment.



      As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.



      Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
      $$left(
      begin{array}{ccc}
      m & n & y(x) \
      1 & 2 & , _0F_1left(;frac{1}{3};frac{4
      x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
      x^{3/2}}{9}right)\
      1 & 3 & , _0F_1left(;frac{2}{5};frac{9
      x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
      x^{5/3}}{25}right)\
      1 & 4 & , _0F_1left(;frac{3}{7};frac{16
      x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
      x^{7/4}}{49}right)\
      1 & 5 & , _0F_1left(;frac{4}{9};frac{25
      x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
      x^{9/5}}{81}right)\
      1 & 6 & , _0F_1left(;frac{5}{11};frac{36
      x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
      x^{11/6}}{121}right)\
      1 & 7 & , _0F_1left(;frac{6}{13};frac{49
      x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
      x^{13/7}}{169}right) \
      2 & 3 & , _0F_1left(;frac{1}{4};frac{9
      x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
      x^{4/3}}{16}right)\
      2 & 5 & , _0F_1left(;frac{3}{8};frac{25
      x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
      x^{8/5}}{64}right)\
      2 & 7 & , _0F_1left(;frac{5}{12};frac{49
      x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
      x^{12/7}}{144}right)\
      3 & 4 & , _0F_1left(;frac{1}{5};frac{16
      x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
      x^{5/4}}{25}right)\
      3 & 5 & , _0F_1left(;frac{2}{7};frac{25
      x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
      x^{7/5}}{49}right)\
      3 & 7 & , _0F_1left(;frac{4}{11};frac{49
      x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
      x^{11/7}}{121}right)\
      3 & 8 & , _0F_1left(;frac{5}{13};frac{64
      x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
      x^{13/8}}{169}right)\
      4 & 5 & , _0F_1left(;frac{1}{6};frac{25
      x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
      x^{6/5}}{36}right)\
      4 & 7 & , _0F_1left(;frac{3}{10};frac{49
      x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
      x^{10/7}}{100}right)\
      4 & 9 & , _0F_1left(;frac{5}{14};frac{81
      x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
      x^{14/9}}{196}right)
      end{array}
      right) $$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Too long for a comment.



        As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.



        Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
        $$left(
        begin{array}{ccc}
        m & n & y(x) \
        1 & 2 & , _0F_1left(;frac{1}{3};frac{4
        x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
        x^{3/2}}{9}right)\
        1 & 3 & , _0F_1left(;frac{2}{5};frac{9
        x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
        x^{5/3}}{25}right)\
        1 & 4 & , _0F_1left(;frac{3}{7};frac{16
        x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
        x^{7/4}}{49}right)\
        1 & 5 & , _0F_1left(;frac{4}{9};frac{25
        x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
        x^{9/5}}{81}right)\
        1 & 6 & , _0F_1left(;frac{5}{11};frac{36
        x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
        x^{11/6}}{121}right)\
        1 & 7 & , _0F_1left(;frac{6}{13};frac{49
        x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
        x^{13/7}}{169}right) \
        2 & 3 & , _0F_1left(;frac{1}{4};frac{9
        x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
        x^{4/3}}{16}right)\
        2 & 5 & , _0F_1left(;frac{3}{8};frac{25
        x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
        x^{8/5}}{64}right)\
        2 & 7 & , _0F_1left(;frac{5}{12};frac{49
        x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
        x^{12/7}}{144}right)\
        3 & 4 & , _0F_1left(;frac{1}{5};frac{16
        x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
        x^{5/4}}{25}right)\
        3 & 5 & , _0F_1left(;frac{2}{7};frac{25
        x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
        x^{7/5}}{49}right)\
        3 & 7 & , _0F_1left(;frac{4}{11};frac{49
        x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
        x^{11/7}}{121}right)\
        3 & 8 & , _0F_1left(;frac{5}{13};frac{64
        x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
        x^{13/8}}{169}right)\
        4 & 5 & , _0F_1left(;frac{1}{6};frac{25
        x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
        x^{6/5}}{36}right)\
        4 & 7 & , _0F_1left(;frac{3}{10};frac{49
        x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
        x^{10/7}}{100}right)\
        4 & 9 & , _0F_1left(;frac{5}{14};frac{81
        x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
        x^{14/9}}{196}right)
        end{array}
        right) $$






        share|cite|improve this answer












        Too long for a comment.



        As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.



        Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
        $$left(
        begin{array}{ccc}
        m & n & y(x) \
        1 & 2 & , _0F_1left(;frac{1}{3};frac{4
        x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
        x^{3/2}}{9}right)\
        1 & 3 & , _0F_1left(;frac{2}{5};frac{9
        x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
        x^{5/3}}{25}right)\
        1 & 4 & , _0F_1left(;frac{3}{7};frac{16
        x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
        x^{7/4}}{49}right)\
        1 & 5 & , _0F_1left(;frac{4}{9};frac{25
        x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
        x^{9/5}}{81}right)\
        1 & 6 & , _0F_1left(;frac{5}{11};frac{36
        x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
        x^{11/6}}{121}right)\
        1 & 7 & , _0F_1left(;frac{6}{13};frac{49
        x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
        x^{13/7}}{169}right) \
        2 & 3 & , _0F_1left(;frac{1}{4};frac{9
        x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
        x^{4/3}}{16}right)\
        2 & 5 & , _0F_1left(;frac{3}{8};frac{25
        x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
        x^{8/5}}{64}right)\
        2 & 7 & , _0F_1left(;frac{5}{12};frac{49
        x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
        x^{12/7}}{144}right)\
        3 & 4 & , _0F_1left(;frac{1}{5};frac{16
        x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
        x^{5/4}}{25}right)\
        3 & 5 & , _0F_1left(;frac{2}{7};frac{25
        x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
        x^{7/5}}{49}right)\
        3 & 7 & , _0F_1left(;frac{4}{11};frac{49
        x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
        x^{11/7}}{121}right)\
        3 & 8 & , _0F_1left(;frac{5}{13};frac{64
        x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
        x^{13/8}}{169}right)\
        4 & 5 & , _0F_1left(;frac{1}{6};frac{25
        x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
        x^{6/5}}{36}right)\
        4 & 7 & , _0F_1left(;frac{3}{10};frac{49
        x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
        x^{10/7}}{100}right)\
        4 & 9 & , _0F_1left(;frac{5}{14};frac{81
        x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
        x^{14/9}}{196}right)
        end{array}
        right) $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 5:11









        Claude Leibovici

        117k1156131




        117k1156131






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008696%2fsolve-a-second-order-ode%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Ellipse (mathématiques)

            Quarter-circle Tiles

            Mont Emei