what went wrong in solving this singular ODE?












1












$begingroup$


I want to solve $$x(2-x) y'' +(1+x)y' -frac{3}{x} y = 0$$ with singular-regular point $x_0 =0$, the general solution by Frobenius is of the form $y = x^r sum limits_{n=0}^{infty} a_n x^n$. i found that $r = -1, frac{3}{2}$. substituting the value $r=-1$ and the standard value $a_0 =1$ gave me that $$a_{n+1} = frac{n^2-4n+3}{2n^2-n-3}a_n$$ which means that $a_1 = a_2 = cdots =0$



So in the end i have one solution which is $y = frac{1}{x}(1+0 x^1 +0 x^2+cdots) = frac{1}{x}$ but the correct solution is $y = frac{1}{x}-1$, what went wrong in my solution ?






ordinary-differential-equations power-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why is this tagged as a PDE ?
    $endgroup$
    – Rebellos
    Dec 22 '18 at 10:29










  • $begingroup$
    $a_1=-1$ and $a_2=a_3=...=0$ using the recurrence law you wrote. The expected solution follows.
    $endgroup$
    – Rafa Budría
    Dec 22 '18 at 10:35












  • $begingroup$
    @RafaBudría i see that, missed it, thanks.
    $endgroup$
    – Ahmad
    Dec 22 '18 at 10:41
















1












$begingroup$


I want to solve $$x(2-x) y'' +(1+x)y' -frac{3}{x} y = 0$$ with singular-regular point $x_0 =0$, the general solution by Frobenius is of the form $y = x^r sum limits_{n=0}^{infty} a_n x^n$. i found that $r = -1, frac{3}{2}$. substituting the value $r=-1$ and the standard value $a_0 =1$ gave me that $$a_{n+1} = frac{n^2-4n+3}{2n^2-n-3}a_n$$ which means that $a_1 = a_2 = cdots =0$



So in the end i have one solution which is $y = frac{1}{x}(1+0 x^1 +0 x^2+cdots) = frac{1}{x}$ but the correct solution is $y = frac{1}{x}-1$, what went wrong in my solution ?






ordinary-differential-equations power-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why is this tagged as a PDE ?
    $endgroup$
    – Rebellos
    Dec 22 '18 at 10:29










  • $begingroup$
    $a_1=-1$ and $a_2=a_3=...=0$ using the recurrence law you wrote. The expected solution follows.
    $endgroup$
    – Rafa Budría
    Dec 22 '18 at 10:35












  • $begingroup$
    @RafaBudría i see that, missed it, thanks.
    $endgroup$
    – Ahmad
    Dec 22 '18 at 10:41














1












1








1





$begingroup$


I want to solve $$x(2-x) y'' +(1+x)y' -frac{3}{x} y = 0$$ with singular-regular point $x_0 =0$, the general solution by Frobenius is of the form $y = x^r sum limits_{n=0}^{infty} a_n x^n$. i found that $r = -1, frac{3}{2}$. substituting the value $r=-1$ and the standard value $a_0 =1$ gave me that $$a_{n+1} = frac{n^2-4n+3}{2n^2-n-3}a_n$$ which means that $a_1 = a_2 = cdots =0$



So in the end i have one solution which is $y = frac{1}{x}(1+0 x^1 +0 x^2+cdots) = frac{1}{x}$ but the correct solution is $y = frac{1}{x}-1$, what went wrong in my solution ?






ordinary-differential-equations power-series







share|cite|improve this question











$endgroup$




I want to solve $$x(2-x) y'' +(1+x)y' -frac{3}{x} y = 0$$ with singular-regular point $x_0 =0$, the general solution by Frobenius is of the form $y = x^r sum limits_{n=0}^{infty} a_n x^n$. i found that $r = -1, frac{3}{2}$. substituting the value $r=-1$ and the standard value $a_0 =1$ gave me that $$a_{n+1} = frac{n^2-4n+3}{2n^2-n-3}a_n$$ which means that $a_1 = a_2 = cdots =0$



So in the end i have one solution which is $y = frac{1}{x}(1+0 x^1 +0 x^2+cdots) = frac{1}{x}$ but the correct solution is $y = frac{1}{x}-1$, what went wrong in my solution ?






ordinary-differential-equations power-series




ordinary-differential-equations power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 11:18









LutzL

60k42057




60k42057










asked Dec 22 '18 at 10:22









AhmadAhmad

2,6151725




2,6151725








  • 1




    $begingroup$
    Why is this tagged as a PDE ?
    $endgroup$
    – Rebellos
    Dec 22 '18 at 10:29










  • $begingroup$
    $a_1=-1$ and $a_2=a_3=...=0$ using the recurrence law you wrote. The expected solution follows.
    $endgroup$
    – Rafa Budría
    Dec 22 '18 at 10:35












  • $begingroup$
    @RafaBudría i see that, missed it, thanks.
    $endgroup$
    – Ahmad
    Dec 22 '18 at 10:41














  • 1




    $begingroup$
    Why is this tagged as a PDE ?
    $endgroup$
    – Rebellos
    Dec 22 '18 at 10:29










  • $begingroup$
    $a_1=-1$ and $a_2=a_3=...=0$ using the recurrence law you wrote. The expected solution follows.
    $endgroup$
    – Rafa Budría
    Dec 22 '18 at 10:35












  • $begingroup$
    @RafaBudría i see that, missed it, thanks.
    $endgroup$
    – Ahmad
    Dec 22 '18 at 10:41








1




1




$begingroup$
Why is this tagged as a PDE ?
$endgroup$
– Rebellos
Dec 22 '18 at 10:29




$begingroup$
Why is this tagged as a PDE ?
$endgroup$
– Rebellos
Dec 22 '18 at 10:29












$begingroup$
$a_1=-1$ and $a_2=a_3=...=0$ using the recurrence law you wrote. The expected solution follows.
$endgroup$
– Rafa Budría
Dec 22 '18 at 10:35






$begingroup$
$a_1=-1$ and $a_2=a_3=...=0$ using the recurrence law you wrote. The expected solution follows.
$endgroup$
– Rafa Budría
Dec 22 '18 at 10:35














$begingroup$
@RafaBudría i see that, missed it, thanks.
$endgroup$
– Ahmad
Dec 22 '18 at 10:41




$begingroup$
@RafaBudría i see that, missed it, thanks.
$endgroup$
– Ahmad
Dec 22 '18 at 10:41










1 Answer
1






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$begingroup$

You have



$$
a_{n + 1} = a_nfrac{n^2 - 4n + 3}{2n^2 - n - 3}
$$



Using $a_0 = color{blue}{1}$ you get



$$
a_1 = 1 frac{3}{(-3)} = color{red}{-1}
$$



and



$$
a_2 = 0 = a_3 = cdots
$$



So one of the solutions is



$$
y_1(x) = x^{-1}left(a_0 + a_1 x + a_2 x^2 + cdots right) = x^{-1}left[color{blue}{1} + (color{red}{-1})xright] = -1 + 1/x
$$






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    1 Answer
    1






    active

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    1 Answer
    1






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    1












    $begingroup$

    You have



    $$
    a_{n + 1} = a_nfrac{n^2 - 4n + 3}{2n^2 - n - 3}
    $$



    Using $a_0 = color{blue}{1}$ you get



    $$
    a_1 = 1 frac{3}{(-3)} = color{red}{-1}
    $$



    and



    $$
    a_2 = 0 = a_3 = cdots
    $$



    So one of the solutions is



    $$
    y_1(x) = x^{-1}left(a_0 + a_1 x + a_2 x^2 + cdots right) = x^{-1}left[color{blue}{1} + (color{red}{-1})xright] = -1 + 1/x
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You have



      $$
      a_{n + 1} = a_nfrac{n^2 - 4n + 3}{2n^2 - n - 3}
      $$



      Using $a_0 = color{blue}{1}$ you get



      $$
      a_1 = 1 frac{3}{(-3)} = color{red}{-1}
      $$



      and



      $$
      a_2 = 0 = a_3 = cdots
      $$



      So one of the solutions is



      $$
      y_1(x) = x^{-1}left(a_0 + a_1 x + a_2 x^2 + cdots right) = x^{-1}left[color{blue}{1} + (color{red}{-1})xright] = -1 + 1/x
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You have



        $$
        a_{n + 1} = a_nfrac{n^2 - 4n + 3}{2n^2 - n - 3}
        $$



        Using $a_0 = color{blue}{1}$ you get



        $$
        a_1 = 1 frac{3}{(-3)} = color{red}{-1}
        $$



        and



        $$
        a_2 = 0 = a_3 = cdots
        $$



        So one of the solutions is



        $$
        y_1(x) = x^{-1}left(a_0 + a_1 x + a_2 x^2 + cdots right) = x^{-1}left[color{blue}{1} + (color{red}{-1})xright] = -1 + 1/x
        $$






        share|cite|improve this answer









        $endgroup$



        You have



        $$
        a_{n + 1} = a_nfrac{n^2 - 4n + 3}{2n^2 - n - 3}
        $$



        Using $a_0 = color{blue}{1}$ you get



        $$
        a_1 = 1 frac{3}{(-3)} = color{red}{-1}
        $$



        and



        $$
        a_2 = 0 = a_3 = cdots
        $$



        So one of the solutions is



        $$
        y_1(x) = x^{-1}left(a_0 + a_1 x + a_2 x^2 + cdots right) = x^{-1}left[color{blue}{1} + (color{red}{-1})xright] = -1 + 1/x
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 10:45









        caveraccaverac

        14.8k31130




        14.8k31130






























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