Indicial equation of $(x^2-1)^2y''+(x+1)y'-y=0$












1












$begingroup$


Let $alpha$ and $beta$ with $alpha>beta$ be the roots of the indicial equation of $(x^2-1)^2y''+(x+1)y'-y=0$ at $x= -1$.




Then what is the value of $alpha-4beta$ ?




I am trying to solve this by Frobenius series solution method. Assuming trial solution of the form $y=sum_{m=0}^{infty} c_m(x+1)^{m+rho}$. Then I get the indicial equation as $$rho^2-1=0.$$ So $alpha=1$ and $beta=-1$. Thus the value of $alpha-4beta$ is $5$.



But the answer is $2$. The question appears in GATE 2017. So where i am wrong. Any help is highly appreciated.






ordinary-differential-equations frobenius-method







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's y'', not y^{''}
    $endgroup$
    – egreg
    Jan 26 '18 at 18:26






  • 1




    $begingroup$
    The indicial equation is $rho^2-frac{3}{4}rho-frac{1}{4}=0$.
    $endgroup$
    – Wang
    Jan 26 '18 at 19:44










  • $begingroup$
    @Wang Sir can you explain how it arise?
    $endgroup$
    – SAHEB PAL
    Jan 26 '18 at 19:58










  • $begingroup$
    The indicial equation of the linear ODE $y''+p(x)y'+q(x)y=0$ near the regular singular point $x=x_0$ is given by $$ r^2+p_0r+q_0=0, $$ where $p_0=lim_{xto x_0}(x-x_0)p(x)$ and $q_0=lim_{xto x_0}(x-x_0)^2q(x)$.
    $endgroup$
    – Wang
    Jan 26 '18 at 20:25








  • 1




    $begingroup$
    @Wang Thanks for your valuable comment. Here the indicial equation would be $r^2+(p_0-1)r+q_0=0$ and I get the indicial equation as your previous comment.
    $endgroup$
    – SAHEB PAL
    Jan 26 '18 at 20:49
















1












$begingroup$


Let $alpha$ and $beta$ with $alpha>beta$ be the roots of the indicial equation of $(x^2-1)^2y''+(x+1)y'-y=0$ at $x= -1$.




Then what is the value of $alpha-4beta$ ?




I am trying to solve this by Frobenius series solution method. Assuming trial solution of the form $y=sum_{m=0}^{infty} c_m(x+1)^{m+rho}$. Then I get the indicial equation as $$rho^2-1=0.$$ So $alpha=1$ and $beta=-1$. Thus the value of $alpha-4beta$ is $5$.



But the answer is $2$. The question appears in GATE 2017. So where i am wrong. Any help is highly appreciated.






ordinary-differential-equations frobenius-method







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's y'', not y^{''}
    $endgroup$
    – egreg
    Jan 26 '18 at 18:26






  • 1




    $begingroup$
    The indicial equation is $rho^2-frac{3}{4}rho-frac{1}{4}=0$.
    $endgroup$
    – Wang
    Jan 26 '18 at 19:44










  • $begingroup$
    @Wang Sir can you explain how it arise?
    $endgroup$
    – SAHEB PAL
    Jan 26 '18 at 19:58










  • $begingroup$
    The indicial equation of the linear ODE $y''+p(x)y'+q(x)y=0$ near the regular singular point $x=x_0$ is given by $$ r^2+p_0r+q_0=0, $$ where $p_0=lim_{xto x_0}(x-x_0)p(x)$ and $q_0=lim_{xto x_0}(x-x_0)^2q(x)$.
    $endgroup$
    – Wang
    Jan 26 '18 at 20:25








  • 1




    $begingroup$
    @Wang Thanks for your valuable comment. Here the indicial equation would be $r^2+(p_0-1)r+q_0=0$ and I get the indicial equation as your previous comment.
    $endgroup$
    – SAHEB PAL
    Jan 26 '18 at 20:49














1












1








1


2



$begingroup$


Let $alpha$ and $beta$ with $alpha>beta$ be the roots of the indicial equation of $(x^2-1)^2y''+(x+1)y'-y=0$ at $x= -1$.




Then what is the value of $alpha-4beta$ ?




I am trying to solve this by Frobenius series solution method. Assuming trial solution of the form $y=sum_{m=0}^{infty} c_m(x+1)^{m+rho}$. Then I get the indicial equation as $$rho^2-1=0.$$ So $alpha=1$ and $beta=-1$. Thus the value of $alpha-4beta$ is $5$.



But the answer is $2$. The question appears in GATE 2017. So where i am wrong. Any help is highly appreciated.






ordinary-differential-equations frobenius-method







share|cite|improve this question











$endgroup$




Let $alpha$ and $beta$ with $alpha>beta$ be the roots of the indicial equation of $(x^2-1)^2y''+(x+1)y'-y=0$ at $x= -1$.




Then what is the value of $alpha-4beta$ ?




I am trying to solve this by Frobenius series solution method. Assuming trial solution of the form $y=sum_{m=0}^{infty} c_m(x+1)^{m+rho}$. Then I get the indicial equation as $$rho^2-1=0.$$ So $alpha=1$ and $beta=-1$. Thus the value of $alpha-4beta$ is $5$.



But the answer is $2$. The question appears in GATE 2017. So where i am wrong. Any help is highly appreciated.






ordinary-differential-equations frobenius-method




ordinary-differential-equations frobenius-method






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 '18 at 18:28









Rebellos

15.7k31250




15.7k31250










asked Jan 26 '18 at 18:21









SAHEB PALSAHEB PAL

998413




998413












  • $begingroup$
    It's y'', not y^{''}
    $endgroup$
    – egreg
    Jan 26 '18 at 18:26






  • 1




    $begingroup$
    The indicial equation is $rho^2-frac{3}{4}rho-frac{1}{4}=0$.
    $endgroup$
    – Wang
    Jan 26 '18 at 19:44










  • $begingroup$
    @Wang Sir can you explain how it arise?
    $endgroup$
    – SAHEB PAL
    Jan 26 '18 at 19:58










  • $begingroup$
    The indicial equation of the linear ODE $y''+p(x)y'+q(x)y=0$ near the regular singular point $x=x_0$ is given by $$ r^2+p_0r+q_0=0, $$ where $p_0=lim_{xto x_0}(x-x_0)p(x)$ and $q_0=lim_{xto x_0}(x-x_0)^2q(x)$.
    $endgroup$
    – Wang
    Jan 26 '18 at 20:25








  • 1




    $begingroup$
    @Wang Thanks for your valuable comment. Here the indicial equation would be $r^2+(p_0-1)r+q_0=0$ and I get the indicial equation as your previous comment.
    $endgroup$
    – SAHEB PAL
    Jan 26 '18 at 20:49


















  • $begingroup$
    It's y'', not y^{''}
    $endgroup$
    – egreg
    Jan 26 '18 at 18:26






  • 1




    $begingroup$
    The indicial equation is $rho^2-frac{3}{4}rho-frac{1}{4}=0$.
    $endgroup$
    – Wang
    Jan 26 '18 at 19:44










  • $begingroup$
    @Wang Sir can you explain how it arise?
    $endgroup$
    – SAHEB PAL
    Jan 26 '18 at 19:58










  • $begingroup$
    The indicial equation of the linear ODE $y''+p(x)y'+q(x)y=0$ near the regular singular point $x=x_0$ is given by $$ r^2+p_0r+q_0=0, $$ where $p_0=lim_{xto x_0}(x-x_0)p(x)$ and $q_0=lim_{xto x_0}(x-x_0)^2q(x)$.
    $endgroup$
    – Wang
    Jan 26 '18 at 20:25








  • 1




    $begingroup$
    @Wang Thanks for your valuable comment. Here the indicial equation would be $r^2+(p_0-1)r+q_0=0$ and I get the indicial equation as your previous comment.
    $endgroup$
    – SAHEB PAL
    Jan 26 '18 at 20:49
















$begingroup$
It's y'', not y^{''}
$endgroup$
– egreg
Jan 26 '18 at 18:26




$begingroup$
It's y'', not y^{''}
$endgroup$
– egreg
Jan 26 '18 at 18:26




1




1




$begingroup$
The indicial equation is $rho^2-frac{3}{4}rho-frac{1}{4}=0$.
$endgroup$
– Wang
Jan 26 '18 at 19:44




$begingroup$
The indicial equation is $rho^2-frac{3}{4}rho-frac{1}{4}=0$.
$endgroup$
– Wang
Jan 26 '18 at 19:44












$begingroup$
@Wang Sir can you explain how it arise?
$endgroup$
– SAHEB PAL
Jan 26 '18 at 19:58




$begingroup$
@Wang Sir can you explain how it arise?
$endgroup$
– SAHEB PAL
Jan 26 '18 at 19:58












$begingroup$
The indicial equation of the linear ODE $y''+p(x)y'+q(x)y=0$ near the regular singular point $x=x_0$ is given by $$ r^2+p_0r+q_0=0, $$ where $p_0=lim_{xto x_0}(x-x_0)p(x)$ and $q_0=lim_{xto x_0}(x-x_0)^2q(x)$.
$endgroup$
– Wang
Jan 26 '18 at 20:25






$begingroup$
The indicial equation of the linear ODE $y''+p(x)y'+q(x)y=0$ near the regular singular point $x=x_0$ is given by $$ r^2+p_0r+q_0=0, $$ where $p_0=lim_{xto x_0}(x-x_0)p(x)$ and $q_0=lim_{xto x_0}(x-x_0)^2q(x)$.
$endgroup$
– Wang
Jan 26 '18 at 20:25






1




1




$begingroup$
@Wang Thanks for your valuable comment. Here the indicial equation would be $r^2+(p_0-1)r+q_0=0$ and I get the indicial equation as your previous comment.
$endgroup$
– SAHEB PAL
Jan 26 '18 at 20:49




$begingroup$
@Wang Thanks for your valuable comment. Here the indicial equation would be $r^2+(p_0-1)r+q_0=0$ and I get the indicial equation as your previous comment.
$endgroup$
– SAHEB PAL
Jan 26 '18 at 20:49










1 Answer
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$begingroup$

$$(x^2-1)^2y''+(x+1)y'-y=0$$
At $x=-1$
We have,
$$alpha(x)(x+1)^2y''+beta(x)(x+1)y'+gamma(x)y=0$$
With$ begin{cases} alpha(x)=(x-1)^2 implies alpha(-1)=4\
beta(x)=1 implies beta(-1)=1\
gamma(x)=-1 implies gamma(-1)=-1
end{cases}
$



Then the indicial equation is



$$r^2+left(frac {beta(-1)}{alpha(-1)}-1right)r+frac {gamma(-1)}{alpha(-1)}=0$$
$$r^2-frac {3}{4}r-frac {1}{4}=0$$
$$(r-1)(r+frac {1}{4})=0 implies r=1, r=-frac 1 4$$



$$text{And }alpha -4beta=1-4frac {-1} 4=2$$






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    $begingroup$

    $$(x^2-1)^2y''+(x+1)y'-y=0$$
    At $x=-1$
    We have,
    $$alpha(x)(x+1)^2y''+beta(x)(x+1)y'+gamma(x)y=0$$
    With$ begin{cases} alpha(x)=(x-1)^2 implies alpha(-1)=4\
    beta(x)=1 implies beta(-1)=1\
    gamma(x)=-1 implies gamma(-1)=-1
    end{cases}
    $



    Then the indicial equation is



    $$r^2+left(frac {beta(-1)}{alpha(-1)}-1right)r+frac {gamma(-1)}{alpha(-1)}=0$$
    $$r^2-frac {3}{4}r-frac {1}{4}=0$$
    $$(r-1)(r+frac {1}{4})=0 implies r=1, r=-frac 1 4$$



    $$text{And }alpha -4beta=1-4frac {-1} 4=2$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $$(x^2-1)^2y''+(x+1)y'-y=0$$
      At $x=-1$
      We have,
      $$alpha(x)(x+1)^2y''+beta(x)(x+1)y'+gamma(x)y=0$$
      With$ begin{cases} alpha(x)=(x-1)^2 implies alpha(-1)=4\
      beta(x)=1 implies beta(-1)=1\
      gamma(x)=-1 implies gamma(-1)=-1
      end{cases}
      $



      Then the indicial equation is



      $$r^2+left(frac {beta(-1)}{alpha(-1)}-1right)r+frac {gamma(-1)}{alpha(-1)}=0$$
      $$r^2-frac {3}{4}r-frac {1}{4}=0$$
      $$(r-1)(r+frac {1}{4})=0 implies r=1, r=-frac 1 4$$



      $$text{And }alpha -4beta=1-4frac {-1} 4=2$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $$(x^2-1)^2y''+(x+1)y'-y=0$$
        At $x=-1$
        We have,
        $$alpha(x)(x+1)^2y''+beta(x)(x+1)y'+gamma(x)y=0$$
        With$ begin{cases} alpha(x)=(x-1)^2 implies alpha(-1)=4\
        beta(x)=1 implies beta(-1)=1\
        gamma(x)=-1 implies gamma(-1)=-1
        end{cases}
        $



        Then the indicial equation is



        $$r^2+left(frac {beta(-1)}{alpha(-1)}-1right)r+frac {gamma(-1)}{alpha(-1)}=0$$
        $$r^2-frac {3}{4}r-frac {1}{4}=0$$
        $$(r-1)(r+frac {1}{4})=0 implies r=1, r=-frac 1 4$$



        $$text{And }alpha -4beta=1-4frac {-1} 4=2$$






        share|cite|improve this answer











        $endgroup$



        $$(x^2-1)^2y''+(x+1)y'-y=0$$
        At $x=-1$
        We have,
        $$alpha(x)(x+1)^2y''+beta(x)(x+1)y'+gamma(x)y=0$$
        With$ begin{cases} alpha(x)=(x-1)^2 implies alpha(-1)=4\
        beta(x)=1 implies beta(-1)=1\
        gamma(x)=-1 implies gamma(-1)=-1
        end{cases}
        $



        Then the indicial equation is



        $$r^2+left(frac {beta(-1)}{alpha(-1)}-1right)r+frac {gamma(-1)}{alpha(-1)}=0$$
        $$r^2-frac {3}{4}r-frac {1}{4}=0$$
        $$(r-1)(r+frac {1}{4})=0 implies r=1, r=-frac 1 4$$



        $$text{And }alpha -4beta=1-4frac {-1} 4=2$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 27 '18 at 1:46

























        answered Jan 27 '18 at 1:30









        IshamIsham

        12.7k3929




        12.7k3929






























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