Frobenius Method to solve $x(1 - x)y'' - 3xy' - y = 0$












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$begingroup$


So, Im trying to self-learn method of frobenius, and I would like to ask if someone can explain to me how can we solve the following DE about $ x = 0$ using this method.



$$ x(1 - x)y'' - 3xy' - y = 0 $$






ordinary-differential-equations power-series frobenius-method







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    So, Im trying to self-learn method of frobenius, and I would like to ask if someone can explain to me how can we solve the following DE about $ x = 0$ using this method.



    $$ x(1 - x)y'' - 3xy' - y = 0 $$






    ordinary-differential-equations power-series frobenius-method







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      So, Im trying to self-learn method of frobenius, and I would like to ask if someone can explain to me how can we solve the following DE about $ x = 0$ using this method.



      $$ x(1 - x)y'' - 3xy' - y = 0 $$






      ordinary-differential-equations power-series frobenius-method







      share|cite|improve this question











      $endgroup$




      So, Im trying to self-learn method of frobenius, and I would like to ask if someone can explain to me how can we solve the following DE about $ x = 0$ using this method.



      $$ x(1 - x)y'' - 3xy' - y = 0 $$






      ordinary-differential-equations power-series frobenius-method




      ordinary-differential-equations power-series frobenius-method






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Nov 30 '16 at 0:17







      user137731

















      asked Oct 28 '12 at 5:04









      ILoveMathILoveMath

      1




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          2 Answers
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          $begingroup$

          The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.



          Let $y(x) = displaystyle sum_{n=0}^{infty} a_n x^n$. Then we get that $$y'(x) = sum_{n=0}^{infty} na_n x^{n-1}$$
          $$3xy'(x) = sum_{n=0}^{infty} 3na_n x^{n}$$
          $$y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-2}$$
          $$xy''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-1} = sum_{n=0}^{infty} n(n+1)a_{n+1} x^{n}$$
          $$x^2y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n}$$
          The ODE is $$xy'' - x^2 y'' -3xy' - y = 0$$
          Plugging in the appropriate series expansions, we get that
          $$sum_{n=0}^{infty} left(n(n+1)a_{n+1} - n(n-1)a_n - 3na_n - a_nright)x^n = 0$$
          Hence, we get that
          $$n(n+1)a_{n+1} = (n(n-1) +3n+1)a_n = (n+1)^2 a_n implies a_{n+1} = dfrac{n+1}{n}a_n$$
          First note that $a_0 = 0$. Choose $a_1$ arbitrarily. Then we get that $a_2 = 2a_1$, $a_3 = 3a_1$, $a_4 = 4a_1$ and in general, $a_{n} = na_1$.
          Hence, the solution is given by
          $$y(x) = a_1 left(x+2x^2 + 3x^3 + cdotsright)$$
          This power series is valid only within $vert x vert <1$. In this region, we can simplify the power series to get
          begin{align}
          y(x) & = a_1 x left(1 + 2x + 3x^2 + cdots right)\
          & = a_1 x dfrac{d}{dx} left(x + x^2 + x^3 + cdots right)\
          & = a_1 x dfrac{d}{dx} left(dfrac{x}{1-x}right)\
          & = a_1 dfrac{x}{(1-x)^2}
          end{align}






          share|cite|improve this answer









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            0












            $begingroup$

            The order reduction method seeks a second basis solution in the form $y=y_1u$, where $y_1(x)=frac{x}{(1-x)^2}$ is the already found basis solution.
            $$
            x(1-x)[y_1u''+2y_1'u']-3x[y_1u']=0
            implies frac{u''}{u'}=frac{3y_1-2(1-x)y_1'}{(1-x)y_1}
            $$
            Insert $y_1(x)=dfrac1{(1-x)^2}-dfrac1{1-x}$, $y_1'=dfrac2{(1-x)^3}-dfrac1{(1-x)^2}$, $y_1''=dfrac{6}{(1-x)^4}-dfrac{2}{(1-x)^3}$
            into that formula to find
            begin{align}
            frac{u''}{u'}&=frac{-frac1{(1-x)^2}-frac1{1-x}}{frac{x}{1-x}}=-frac{2-x}{x(1-x)}=-frac{2}x+frac1{1-x}
            \
            implies u'&=frac1{x^2(1-x)}=frac{1+x}{x^2}-frac1{1-x}
            \
            implies u&=-frac1x+ln|x(1-x)|
            end{align}
            so that the second basis solution is
            $$
            y_2=frac{xln|x(1-x)|-1}{(1-x)^2}
            $$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              active

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              2 Answers
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              active

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              4












              $begingroup$

              The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.



              Let $y(x) = displaystyle sum_{n=0}^{infty} a_n x^n$. Then we get that $$y'(x) = sum_{n=0}^{infty} na_n x^{n-1}$$
              $$3xy'(x) = sum_{n=0}^{infty} 3na_n x^{n}$$
              $$y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-2}$$
              $$xy''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-1} = sum_{n=0}^{infty} n(n+1)a_{n+1} x^{n}$$
              $$x^2y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n}$$
              The ODE is $$xy'' - x^2 y'' -3xy' - y = 0$$
              Plugging in the appropriate series expansions, we get that
              $$sum_{n=0}^{infty} left(n(n+1)a_{n+1} - n(n-1)a_n - 3na_n - a_nright)x^n = 0$$
              Hence, we get that
              $$n(n+1)a_{n+1} = (n(n-1) +3n+1)a_n = (n+1)^2 a_n implies a_{n+1} = dfrac{n+1}{n}a_n$$
              First note that $a_0 = 0$. Choose $a_1$ arbitrarily. Then we get that $a_2 = 2a_1$, $a_3 = 3a_1$, $a_4 = 4a_1$ and in general, $a_{n} = na_1$.
              Hence, the solution is given by
              $$y(x) = a_1 left(x+2x^2 + 3x^3 + cdotsright)$$
              This power series is valid only within $vert x vert <1$. In this region, we can simplify the power series to get
              begin{align}
              y(x) & = a_1 x left(1 + 2x + 3x^2 + cdots right)\
              & = a_1 x dfrac{d}{dx} left(x + x^2 + x^3 + cdots right)\
              & = a_1 x dfrac{d}{dx} left(dfrac{x}{1-x}right)\
              & = a_1 dfrac{x}{(1-x)^2}
              end{align}






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.



                Let $y(x) = displaystyle sum_{n=0}^{infty} a_n x^n$. Then we get that $$y'(x) = sum_{n=0}^{infty} na_n x^{n-1}$$
                $$3xy'(x) = sum_{n=0}^{infty} 3na_n x^{n}$$
                $$y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-2}$$
                $$xy''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-1} = sum_{n=0}^{infty} n(n+1)a_{n+1} x^{n}$$
                $$x^2y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n}$$
                The ODE is $$xy'' - x^2 y'' -3xy' - y = 0$$
                Plugging in the appropriate series expansions, we get that
                $$sum_{n=0}^{infty} left(n(n+1)a_{n+1} - n(n-1)a_n - 3na_n - a_nright)x^n = 0$$
                Hence, we get that
                $$n(n+1)a_{n+1} = (n(n-1) +3n+1)a_n = (n+1)^2 a_n implies a_{n+1} = dfrac{n+1}{n}a_n$$
                First note that $a_0 = 0$. Choose $a_1$ arbitrarily. Then we get that $a_2 = 2a_1$, $a_3 = 3a_1$, $a_4 = 4a_1$ and in general, $a_{n} = na_1$.
                Hence, the solution is given by
                $$y(x) = a_1 left(x+2x^2 + 3x^3 + cdotsright)$$
                This power series is valid only within $vert x vert <1$. In this region, we can simplify the power series to get
                begin{align}
                y(x) & = a_1 x left(1 + 2x + 3x^2 + cdots right)\
                & = a_1 x dfrac{d}{dx} left(x + x^2 + x^3 + cdots right)\
                & = a_1 x dfrac{d}{dx} left(dfrac{x}{1-x}right)\
                & = a_1 dfrac{x}{(1-x)^2}
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.



                  Let $y(x) = displaystyle sum_{n=0}^{infty} a_n x^n$. Then we get that $$y'(x) = sum_{n=0}^{infty} na_n x^{n-1}$$
                  $$3xy'(x) = sum_{n=0}^{infty} 3na_n x^{n}$$
                  $$y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-2}$$
                  $$xy''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-1} = sum_{n=0}^{infty} n(n+1)a_{n+1} x^{n}$$
                  $$x^2y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n}$$
                  The ODE is $$xy'' - x^2 y'' -3xy' - y = 0$$
                  Plugging in the appropriate series expansions, we get that
                  $$sum_{n=0}^{infty} left(n(n+1)a_{n+1} - n(n-1)a_n - 3na_n - a_nright)x^n = 0$$
                  Hence, we get that
                  $$n(n+1)a_{n+1} = (n(n-1) +3n+1)a_n = (n+1)^2 a_n implies a_{n+1} = dfrac{n+1}{n}a_n$$
                  First note that $a_0 = 0$. Choose $a_1$ arbitrarily. Then we get that $a_2 = 2a_1$, $a_3 = 3a_1$, $a_4 = 4a_1$ and in general, $a_{n} = na_1$.
                  Hence, the solution is given by
                  $$y(x) = a_1 left(x+2x^2 + 3x^3 + cdotsright)$$
                  This power series is valid only within $vert x vert <1$. In this region, we can simplify the power series to get
                  begin{align}
                  y(x) & = a_1 x left(1 + 2x + 3x^2 + cdots right)\
                  & = a_1 x dfrac{d}{dx} left(x + x^2 + x^3 + cdots right)\
                  & = a_1 x dfrac{d}{dx} left(dfrac{x}{1-x}right)\
                  & = a_1 dfrac{x}{(1-x)^2}
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.



                  Let $y(x) = displaystyle sum_{n=0}^{infty} a_n x^n$. Then we get that $$y'(x) = sum_{n=0}^{infty} na_n x^{n-1}$$
                  $$3xy'(x) = sum_{n=0}^{infty} 3na_n x^{n}$$
                  $$y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-2}$$
                  $$xy''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n-1} = sum_{n=0}^{infty} n(n+1)a_{n+1} x^{n}$$
                  $$x^2y''(x) = sum_{n=0}^{infty} n(n-1)a_n x^{n}$$
                  The ODE is $$xy'' - x^2 y'' -3xy' - y = 0$$
                  Plugging in the appropriate series expansions, we get that
                  $$sum_{n=0}^{infty} left(n(n+1)a_{n+1} - n(n-1)a_n - 3na_n - a_nright)x^n = 0$$
                  Hence, we get that
                  $$n(n+1)a_{n+1} = (n(n-1) +3n+1)a_n = (n+1)^2 a_n implies a_{n+1} = dfrac{n+1}{n}a_n$$
                  First note that $a_0 = 0$. Choose $a_1$ arbitrarily. Then we get that $a_2 = 2a_1$, $a_3 = 3a_1$, $a_4 = 4a_1$ and in general, $a_{n} = na_1$.
                  Hence, the solution is given by
                  $$y(x) = a_1 left(x+2x^2 + 3x^3 + cdotsright)$$
                  This power series is valid only within $vert x vert <1$. In this region, we can simplify the power series to get
                  begin{align}
                  y(x) & = a_1 x left(1 + 2x + 3x^2 + cdots right)\
                  & = a_1 x dfrac{d}{dx} left(x + x^2 + x^3 + cdots right)\
                  & = a_1 x dfrac{d}{dx} left(dfrac{x}{1-x}right)\
                  & = a_1 dfrac{x}{(1-x)^2}
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 28 '12 at 5:23







                  user17762






























                      0












                      $begingroup$

                      The order reduction method seeks a second basis solution in the form $y=y_1u$, where $y_1(x)=frac{x}{(1-x)^2}$ is the already found basis solution.
                      $$
                      x(1-x)[y_1u''+2y_1'u']-3x[y_1u']=0
                      implies frac{u''}{u'}=frac{3y_1-2(1-x)y_1'}{(1-x)y_1}
                      $$
                      Insert $y_1(x)=dfrac1{(1-x)^2}-dfrac1{1-x}$, $y_1'=dfrac2{(1-x)^3}-dfrac1{(1-x)^2}$, $y_1''=dfrac{6}{(1-x)^4}-dfrac{2}{(1-x)^3}$
                      into that formula to find
                      begin{align}
                      frac{u''}{u'}&=frac{-frac1{(1-x)^2}-frac1{1-x}}{frac{x}{1-x}}=-frac{2-x}{x(1-x)}=-frac{2}x+frac1{1-x}
                      \
                      implies u'&=frac1{x^2(1-x)}=frac{1+x}{x^2}-frac1{1-x}
                      \
                      implies u&=-frac1x+ln|x(1-x)|
                      end{align}
                      so that the second basis solution is
                      $$
                      y_2=frac{xln|x(1-x)|-1}{(1-x)^2}
                      $$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        The order reduction method seeks a second basis solution in the form $y=y_1u$, where $y_1(x)=frac{x}{(1-x)^2}$ is the already found basis solution.
                        $$
                        x(1-x)[y_1u''+2y_1'u']-3x[y_1u']=0
                        implies frac{u''}{u'}=frac{3y_1-2(1-x)y_1'}{(1-x)y_1}
                        $$
                        Insert $y_1(x)=dfrac1{(1-x)^2}-dfrac1{1-x}$, $y_1'=dfrac2{(1-x)^3}-dfrac1{(1-x)^2}$, $y_1''=dfrac{6}{(1-x)^4}-dfrac{2}{(1-x)^3}$
                        into that formula to find
                        begin{align}
                        frac{u''}{u'}&=frac{-frac1{(1-x)^2}-frac1{1-x}}{frac{x}{1-x}}=-frac{2-x}{x(1-x)}=-frac{2}x+frac1{1-x}
                        \
                        implies u'&=frac1{x^2(1-x)}=frac{1+x}{x^2}-frac1{1-x}
                        \
                        implies u&=-frac1x+ln|x(1-x)|
                        end{align}
                        so that the second basis solution is
                        $$
                        y_2=frac{xln|x(1-x)|-1}{(1-x)^2}
                        $$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          The order reduction method seeks a second basis solution in the form $y=y_1u$, where $y_1(x)=frac{x}{(1-x)^2}$ is the already found basis solution.
                          $$
                          x(1-x)[y_1u''+2y_1'u']-3x[y_1u']=0
                          implies frac{u''}{u'}=frac{3y_1-2(1-x)y_1'}{(1-x)y_1}
                          $$
                          Insert $y_1(x)=dfrac1{(1-x)^2}-dfrac1{1-x}$, $y_1'=dfrac2{(1-x)^3}-dfrac1{(1-x)^2}$, $y_1''=dfrac{6}{(1-x)^4}-dfrac{2}{(1-x)^3}$
                          into that formula to find
                          begin{align}
                          frac{u''}{u'}&=frac{-frac1{(1-x)^2}-frac1{1-x}}{frac{x}{1-x}}=-frac{2-x}{x(1-x)}=-frac{2}x+frac1{1-x}
                          \
                          implies u'&=frac1{x^2(1-x)}=frac{1+x}{x^2}-frac1{1-x}
                          \
                          implies u&=-frac1x+ln|x(1-x)|
                          end{align}
                          so that the second basis solution is
                          $$
                          y_2=frac{xln|x(1-x)|-1}{(1-x)^2}
                          $$






                          share|cite|improve this answer









                          $endgroup$



                          The order reduction method seeks a second basis solution in the form $y=y_1u$, where $y_1(x)=frac{x}{(1-x)^2}$ is the already found basis solution.
                          $$
                          x(1-x)[y_1u''+2y_1'u']-3x[y_1u']=0
                          implies frac{u''}{u'}=frac{3y_1-2(1-x)y_1'}{(1-x)y_1}
                          $$
                          Insert $y_1(x)=dfrac1{(1-x)^2}-dfrac1{1-x}$, $y_1'=dfrac2{(1-x)^3}-dfrac1{(1-x)^2}$, $y_1''=dfrac{6}{(1-x)^4}-dfrac{2}{(1-x)^3}$
                          into that formula to find
                          begin{align}
                          frac{u''}{u'}&=frac{-frac1{(1-x)^2}-frac1{1-x}}{frac{x}{1-x}}=-frac{2-x}{x(1-x)}=-frac{2}x+frac1{1-x}
                          \
                          implies u'&=frac1{x^2(1-x)}=frac{1+x}{x^2}-frac1{1-x}
                          \
                          implies u&=-frac1x+ln|x(1-x)|
                          end{align}
                          so that the second basis solution is
                          $$
                          y_2=frac{xln|x(1-x)|-1}{(1-x)^2}
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 20 '18 at 18:42









                          LutzLLutzL

                          59.8k42057




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                              Ida-Boy-Ed-Garten

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                              Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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