solve a second order ODE











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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.










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  • 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






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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






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0
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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).
$$left(
begin{array}{ccc}
m & n & y(x) \
1 & 2 & , _0F_1left(;frac{1}{3};frac{4
x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
x^{3/2}}{9}right)\
1 & 3 & , _0F_1left(;frac{2}{5};frac{9
x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
x^{5/3}}{25}right)\
1 & 4 & , _0F_1left(;frac{3}{7};frac{16
x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
x^{7/4}}{49}right)\
1 & 5 & , _0F_1left(;frac{4}{9};frac{25
x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
x^{9/5}}{81}right)\
1 & 6 & , _0F_1left(;frac{5}{11};frac{36
x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
x^{11/6}}{121}right)\
1 & 7 & , _0F_1left(;frac{6}{13};frac{49
x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
x^{13/7}}{169}right) \
2 & 3 & , _0F_1left(;frac{1}{4};frac{9
x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
x^{4/3}}{16}right)\
2 & 5 & , _0F_1left(;frac{3}{8};frac{25
x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
x^{8/5}}{64}right)\
2 & 7 & , _0F_1left(;frac{5}{12};frac{49
x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
x^{12/7}}{144}right)\
3 & 4 & , _0F_1left(;frac{1}{5};frac{16
x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
x^{5/4}}{25}right)\
3 & 5 & , _0F_1left(;frac{2}{7};frac{25
x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
x^{7/5}}{49}right)\
3 & 7 & , _0F_1left(;frac{4}{11};frac{49
x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
x^{11/7}}{121}right)\
3 & 8 & , _0F_1left(;frac{5}{13};frac{64
x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
x^{13/8}}{169}right)\
4 & 5 & , _0F_1left(;frac{1}{6};frac{25
x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
x^{6/5}}{36}right)\
4 & 7 & , _0F_1left(;frac{3}{10};frac{49
x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
x^{10/7}}{100}right)\
4 & 9 & , _0F_1left(;frac{5}{14};frac{81
x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
x^{14/9}}{196}right)
end{array}
right) $$






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

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    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























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        down vote










        up vote
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        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






























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