Solve $x^2frac{d^2y}{dx^2}+frac{dy}{dx}+x^2y=0$












2












$begingroup$


Solve the differential equation $$x^2frac{d^2y}{dx^2}+frac{dy}{dx}+x^2y=0$$



My try:



$$x^2 frac{frac{d^2y}{dx^2}}{y}=-x^2-frac{frac{dy}{dx}}{y}$$



RHS is a perfect differential, but LHS is not



any clue?






calculus algebra-precalculus ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What are y, y_1, y_2? My guess: y a function from R to R, y_1 is its derivative, and y_2 is its second derivative.
    $endgroup$
    – John B
    Dec 5 '18 at 14:47






  • 1




    $begingroup$
    Do you mean $y_2=y''$ etc ?
    $endgroup$
    – gammatester
    Dec 5 '18 at 14:47












  • $begingroup$
    yes i edited accordingly
    $endgroup$
    – Umesh shankar
    Dec 5 '18 at 15:03










  • $begingroup$
    Why do you think this should have a closed form ?
    $endgroup$
    – user120527
    Dec 5 '18 at 15:20










  • $begingroup$
    What exactly is here understood under "solution"? A power series, asymptotic behaviour? Reduction to a Bessel or similar equation?
    $endgroup$
    – LutzL
    Dec 5 '18 at 15:21
















2












$begingroup$


Solve the differential equation $$x^2frac{d^2y}{dx^2}+frac{dy}{dx}+x^2y=0$$



My try:



$$x^2 frac{frac{d^2y}{dx^2}}{y}=-x^2-frac{frac{dy}{dx}}{y}$$



RHS is a perfect differential, but LHS is not



any clue?






calculus algebra-precalculus ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What are y, y_1, y_2? My guess: y a function from R to R, y_1 is its derivative, and y_2 is its second derivative.
    $endgroup$
    – John B
    Dec 5 '18 at 14:47






  • 1




    $begingroup$
    Do you mean $y_2=y''$ etc ?
    $endgroup$
    – gammatester
    Dec 5 '18 at 14:47












  • $begingroup$
    yes i edited accordingly
    $endgroup$
    – Umesh shankar
    Dec 5 '18 at 15:03










  • $begingroup$
    Why do you think this should have a closed form ?
    $endgroup$
    – user120527
    Dec 5 '18 at 15:20










  • $begingroup$
    What exactly is here understood under "solution"? A power series, asymptotic behaviour? Reduction to a Bessel or similar equation?
    $endgroup$
    – LutzL
    Dec 5 '18 at 15:21














2












2








2


2



$begingroup$


Solve the differential equation $$x^2frac{d^2y}{dx^2}+frac{dy}{dx}+x^2y=0$$



My try:



$$x^2 frac{frac{d^2y}{dx^2}}{y}=-x^2-frac{frac{dy}{dx}}{y}$$



RHS is a perfect differential, but LHS is not



any clue?






calculus algebra-precalculus ordinary-differential-equations







share|cite|improve this question











$endgroup$




Solve the differential equation $$x^2frac{d^2y}{dx^2}+frac{dy}{dx}+x^2y=0$$



My try:



$$x^2 frac{frac{d^2y}{dx^2}}{y}=-x^2-frac{frac{dy}{dx}}{y}$$



RHS is a perfect differential, but LHS is not



any clue?






calculus algebra-precalculus ordinary-differential-equations




calculus algebra-precalculus ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 15:03







Umesh shankar

















asked Dec 5 '18 at 14:40









Umesh shankarUmesh shankar

2,85031220




2,85031220








  • 1




    $begingroup$
    What are y, y_1, y_2? My guess: y a function from R to R, y_1 is its derivative, and y_2 is its second derivative.
    $endgroup$
    – John B
    Dec 5 '18 at 14:47






  • 1




    $begingroup$
    Do you mean $y_2=y''$ etc ?
    $endgroup$
    – gammatester
    Dec 5 '18 at 14:47












  • $begingroup$
    yes i edited accordingly
    $endgroup$
    – Umesh shankar
    Dec 5 '18 at 15:03










  • $begingroup$
    Why do you think this should have a closed form ?
    $endgroup$
    – user120527
    Dec 5 '18 at 15:20










  • $begingroup$
    What exactly is here understood under "solution"? A power series, asymptotic behaviour? Reduction to a Bessel or similar equation?
    $endgroup$
    – LutzL
    Dec 5 '18 at 15:21














  • 1




    $begingroup$
    What are y, y_1, y_2? My guess: y a function from R to R, y_1 is its derivative, and y_2 is its second derivative.
    $endgroup$
    – John B
    Dec 5 '18 at 14:47






  • 1




    $begingroup$
    Do you mean $y_2=y''$ etc ?
    $endgroup$
    – gammatester
    Dec 5 '18 at 14:47












  • $begingroup$
    yes i edited accordingly
    $endgroup$
    – Umesh shankar
    Dec 5 '18 at 15:03










  • $begingroup$
    Why do you think this should have a closed form ?
    $endgroup$
    – user120527
    Dec 5 '18 at 15:20










  • $begingroup$
    What exactly is here understood under "solution"? A power series, asymptotic behaviour? Reduction to a Bessel or similar equation?
    $endgroup$
    – LutzL
    Dec 5 '18 at 15:21








1




1




$begingroup$
What are y, y_1, y_2? My guess: y a function from R to R, y_1 is its derivative, and y_2 is its second derivative.
$endgroup$
– John B
Dec 5 '18 at 14:47




$begingroup$
What are y, y_1, y_2? My guess: y a function from R to R, y_1 is its derivative, and y_2 is its second derivative.
$endgroup$
– John B
Dec 5 '18 at 14:47




1




1




$begingroup$
Do you mean $y_2=y''$ etc ?
$endgroup$
– gammatester
Dec 5 '18 at 14:47






$begingroup$
Do you mean $y_2=y''$ etc ?
$endgroup$
– gammatester
Dec 5 '18 at 14:47














$begingroup$
yes i edited accordingly
$endgroup$
– Umesh shankar
Dec 5 '18 at 15:03




$begingroup$
yes i edited accordingly
$endgroup$
– Umesh shankar
Dec 5 '18 at 15:03












$begingroup$
Why do you think this should have a closed form ?
$endgroup$
– user120527
Dec 5 '18 at 15:20




$begingroup$
Why do you think this should have a closed form ?
$endgroup$
– user120527
Dec 5 '18 at 15:20












$begingroup$
What exactly is here understood under "solution"? A power series, asymptotic behaviour? Reduction to a Bessel or similar equation?
$endgroup$
– LutzL
Dec 5 '18 at 15:21




$begingroup$
What exactly is here understood under "solution"? A power series, asymptotic behaviour? Reduction to a Bessel or similar equation?
$endgroup$
– LutzL
Dec 5 '18 at 15:21










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint:



Let $y=e^{ix}u$ ,



Then $dfrac{dy}{dx}=e^{ix}dfrac{du}{dx}+ie^{ix}u$



$dfrac{d^2y}{dx^2}=e^{ix}dfrac{d^2u}{dx^2}+ie^{ix}dfrac{du}{dx}+ie^{ix}dfrac{du}{dx}-e^{ix}u=e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}u$



$therefore x^2left(e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}uright)+e^{ix}dfrac{du}{dx}+ie^{ix}u+x^2e^{ix}u=0$



$x^2left(dfrac{d^2u}{dx^2}+2idfrac{du}{dx}-uright)+dfrac{du}{dx}+iu+x^2u=0$



$x^2dfrac{d^2u}{dx^2}+(2ix^2+1)dfrac{du}{dx}+iu=0$



Which relates to Heun's Doubly-Confluent Equation.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    It is a Sturm Liouville equation, write
    $$frac{d}{dx}left(e^{-1/x}y'(x)right)+e^{-1/x}y(x)=0$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You are on right track, but some clarification could be good for the untrained student.
      $endgroup$
      – mathreadler
      Dec 6 '18 at 14:11













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint:



    Let $y=e^{ix}u$ ,



    Then $dfrac{dy}{dx}=e^{ix}dfrac{du}{dx}+ie^{ix}u$



    $dfrac{d^2y}{dx^2}=e^{ix}dfrac{d^2u}{dx^2}+ie^{ix}dfrac{du}{dx}+ie^{ix}dfrac{du}{dx}-e^{ix}u=e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}u$



    $therefore x^2left(e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}uright)+e^{ix}dfrac{du}{dx}+ie^{ix}u+x^2e^{ix}u=0$



    $x^2left(dfrac{d^2u}{dx^2}+2idfrac{du}{dx}-uright)+dfrac{du}{dx}+iu+x^2u=0$



    $x^2dfrac{d^2u}{dx^2}+(2ix^2+1)dfrac{du}{dx}+iu=0$



    Which relates to Heun's Doubly-Confluent Equation.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint:



      Let $y=e^{ix}u$ ,



      Then $dfrac{dy}{dx}=e^{ix}dfrac{du}{dx}+ie^{ix}u$



      $dfrac{d^2y}{dx^2}=e^{ix}dfrac{d^2u}{dx^2}+ie^{ix}dfrac{du}{dx}+ie^{ix}dfrac{du}{dx}-e^{ix}u=e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}u$



      $therefore x^2left(e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}uright)+e^{ix}dfrac{du}{dx}+ie^{ix}u+x^2e^{ix}u=0$



      $x^2left(dfrac{d^2u}{dx^2}+2idfrac{du}{dx}-uright)+dfrac{du}{dx}+iu+x^2u=0$



      $x^2dfrac{d^2u}{dx^2}+(2ix^2+1)dfrac{du}{dx}+iu=0$



      Which relates to Heun's Doubly-Confluent Equation.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint:



        Let $y=e^{ix}u$ ,



        Then $dfrac{dy}{dx}=e^{ix}dfrac{du}{dx}+ie^{ix}u$



        $dfrac{d^2y}{dx^2}=e^{ix}dfrac{d^2u}{dx^2}+ie^{ix}dfrac{du}{dx}+ie^{ix}dfrac{du}{dx}-e^{ix}u=e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}u$



        $therefore x^2left(e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}uright)+e^{ix}dfrac{du}{dx}+ie^{ix}u+x^2e^{ix}u=0$



        $x^2left(dfrac{d^2u}{dx^2}+2idfrac{du}{dx}-uright)+dfrac{du}{dx}+iu+x^2u=0$



        $x^2dfrac{d^2u}{dx^2}+(2ix^2+1)dfrac{du}{dx}+iu=0$



        Which relates to Heun's Doubly-Confluent Equation.






        share|cite|improve this answer









        $endgroup$



        Hint:



        Let $y=e^{ix}u$ ,



        Then $dfrac{dy}{dx}=e^{ix}dfrac{du}{dx}+ie^{ix}u$



        $dfrac{d^2y}{dx^2}=e^{ix}dfrac{d^2u}{dx^2}+ie^{ix}dfrac{du}{dx}+ie^{ix}dfrac{du}{dx}-e^{ix}u=e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}u$



        $therefore x^2left(e^{ix}dfrac{d^2u}{dx^2}+2ie^{ix}dfrac{du}{dx}-e^{ix}uright)+e^{ix}dfrac{du}{dx}+ie^{ix}u+x^2e^{ix}u=0$



        $x^2left(dfrac{d^2u}{dx^2}+2idfrac{du}{dx}-uright)+dfrac{du}{dx}+iu+x^2u=0$



        $x^2dfrac{d^2u}{dx^2}+(2ix^2+1)dfrac{du}{dx}+iu=0$



        Which relates to Heun's Doubly-Confluent Equation.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 12:53









        doraemonpauldoraemonpaul

        12.6k31660




        12.6k31660























            0












            $begingroup$

            It is a Sturm Liouville equation, write
            $$frac{d}{dx}left(e^{-1/x}y'(x)right)+e^{-1/x}y(x)=0$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You are on right track, but some clarification could be good for the untrained student.
              $endgroup$
              – mathreadler
              Dec 6 '18 at 14:11


















            0












            $begingroup$

            It is a Sturm Liouville equation, write
            $$frac{d}{dx}left(e^{-1/x}y'(x)right)+e^{-1/x}y(x)=0$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You are on right track, but some clarification could be good for the untrained student.
              $endgroup$
              – mathreadler
              Dec 6 '18 at 14:11
















            0












            0








            0





            $begingroup$

            It is a Sturm Liouville equation, write
            $$frac{d}{dx}left(e^{-1/x}y'(x)right)+e^{-1/x}y(x)=0$$






            share|cite|improve this answer









            $endgroup$



            It is a Sturm Liouville equation, write
            $$frac{d}{dx}left(e^{-1/x}y'(x)right)+e^{-1/x}y(x)=0$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 '18 at 15:23









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            76.7k42866




            76.7k42866












            • $begingroup$
              You are on right track, but some clarification could be good for the untrained student.
              $endgroup$
              – mathreadler
              Dec 6 '18 at 14:11




















            • $begingroup$
              You are on right track, but some clarification could be good for the untrained student.
              $endgroup$
              – mathreadler
              Dec 6 '18 at 14:11


















            $begingroup$
            You are on right track, but some clarification could be good for the untrained student.
            $endgroup$
            – mathreadler
            Dec 6 '18 at 14:11






            $begingroup$
            You are on right track, but some clarification could be good for the untrained student.
            $endgroup$
            – mathreadler
            Dec 6 '18 at 14:11




















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