solving a DE with Frobenius method











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I'm trying to solve:



$xy''+(5-x)y'-y=0$



using frobenius method, which I'm keep getting a wrong answer.



can someone explain how to solve this problem?



Also, are there any sources I can refer to? I really need some good instructions to followw






differential-equations frobenius-method







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    up vote
    0
    down vote

    favorite












    I'm trying to solve:



    $xy''+(5-x)y'-y=0$



    using frobenius method, which I'm keep getting a wrong answer.



    can someone explain how to solve this problem?



    Also, are there any sources I can refer to? I really need some good instructions to followw






    differential-equations frobenius-method







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm trying to solve:



      $xy''+(5-x)y'-y=0$



      using frobenius method, which I'm keep getting a wrong answer.



      can someone explain how to solve this problem?



      Also, are there any sources I can refer to? I really need some good instructions to followw






      differential-equations frobenius-method







      share|cite|improve this question













      I'm trying to solve:



      $xy''+(5-x)y'-y=0$



      using frobenius method, which I'm keep getting a wrong answer.



      can someone explain how to solve this problem?



      Also, are there any sources I can refer to? I really need some good instructions to followw






      differential-equations frobenius-method




      differential-equations frobenius-method






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      share|cite|improve this question











      share|cite|improve this question




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      asked 22 hours ago









      Subin Park

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          Hint: With writing
          $$y''+dfrac{5-x}{x}y'-dfrac{1}{x}y=0$$
          we see $x=0$ is regular point of the equation, since $p(x)=dfrac{5-x}{x}$ and $q(x)=-dfrac{1}{x}$, then
          $$p_0=lim_{xto0}xp(x)=5~~~;~~~q_0=lim_{xto0}x^2q(x)=0$$
          the characteristic equation $m(m-1)+p_0m+q_0=0$ shows we have $m=0,-4$. One may work with a solution of the form
          $$y=x^{-4}sum_{ngeqslant0}a_nx^n$$






          share|cite|improve this answer





















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            Hint: With writing
            $$y''+dfrac{5-x}{x}y'-dfrac{1}{x}y=0$$
            we see $x=0$ is regular point of the equation, since $p(x)=dfrac{5-x}{x}$ and $q(x)=-dfrac{1}{x}$, then
            $$p_0=lim_{xto0}xp(x)=5~~~;~~~q_0=lim_{xto0}x^2q(x)=0$$
            the characteristic equation $m(m-1)+p_0m+q_0=0$ shows we have $m=0,-4$. One may work with a solution of the form
            $$y=x^{-4}sum_{ngeqslant0}a_nx^n$$






            share|cite|improve this answer

























              up vote
              0
              down vote













              Hint: With writing
              $$y''+dfrac{5-x}{x}y'-dfrac{1}{x}y=0$$
              we see $x=0$ is regular point of the equation, since $p(x)=dfrac{5-x}{x}$ and $q(x)=-dfrac{1}{x}$, then
              $$p_0=lim_{xto0}xp(x)=5~~~;~~~q_0=lim_{xto0}x^2q(x)=0$$
              the characteristic equation $m(m-1)+p_0m+q_0=0$ shows we have $m=0,-4$. One may work with a solution of the form
              $$y=x^{-4}sum_{ngeqslant0}a_nx^n$$






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                Hint: With writing
                $$y''+dfrac{5-x}{x}y'-dfrac{1}{x}y=0$$
                we see $x=0$ is regular point of the equation, since $p(x)=dfrac{5-x}{x}$ and $q(x)=-dfrac{1}{x}$, then
                $$p_0=lim_{xto0}xp(x)=5~~~;~~~q_0=lim_{xto0}x^2q(x)=0$$
                the characteristic equation $m(m-1)+p_0m+q_0=0$ shows we have $m=0,-4$. One may work with a solution of the form
                $$y=x^{-4}sum_{ngeqslant0}a_nx^n$$






                share|cite|improve this answer












                Hint: With writing
                $$y''+dfrac{5-x}{x}y'-dfrac{1}{x}y=0$$
                we see $x=0$ is regular point of the equation, since $p(x)=dfrac{5-x}{x}$ and $q(x)=-dfrac{1}{x}$, then
                $$p_0=lim_{xto0}xp(x)=5~~~;~~~q_0=lim_{xto0}x^2q(x)=0$$
                the characteristic equation $m(m-1)+p_0m+q_0=0$ shows we have $m=0,-4$. One may work with a solution of the form
                $$y=x^{-4}sum_{ngeqslant0}a_nx^n$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 22 hours ago









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