if, $y=lnleft(frac{x}{1-x}right)$ prove $x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$












0












$begingroup$


If $y=lnleft(frac{x}{1-x}right)$, prove



$x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$



My Try



I was able to differentiate this expression twice without a truoble, but I'm having difficulty in obtaining 'y' in the second derivative expression. Please help.






derivatives







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$endgroup$

















    0












    $begingroup$


    If $y=lnleft(frac{x}{1-x}right)$, prove



    $x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$



    My Try



    I was able to differentiate this expression twice without a truoble, but I'm having difficulty in obtaining 'y' in the second derivative expression. Please help.






    derivatives







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      If $y=lnleft(frac{x}{1-x}right)$, prove



      $x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$



      My Try



      I was able to differentiate this expression twice without a truoble, but I'm having difficulty in obtaining 'y' in the second derivative expression. Please help.






      derivatives







      share|cite|improve this question











      $endgroup$




      If $y=lnleft(frac{x}{1-x}right)$, prove



      $x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$



      My Try



      I was able to differentiate this expression twice without a truoble, but I'm having difficulty in obtaining 'y' in the second derivative expression. Please help.






      derivatives




      derivatives






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      edited Jan 14 at 5:03









      coreyman317

      772420




      772420










      asked Jan 14 at 3:56









      emilemil

      431410




      431410






















          2 Answers
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          $begingroup$

          This is an exercise showing your given function satisfies the given differential equation!



          According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
          }=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$           $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$



          Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$



          However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            I do not think the statement to be proven is true. We clearly have
            $$y=ln x-ln(1-x),$$
            and so
            $$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
            Let $x=1/2$. Then it is easy to see that the LHS is
            $$x^3 y''=0$$
            but the RHS is
            $$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
              2






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              2












              $begingroup$

              This is an exercise showing your given function satisfies the given differential equation!



              According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
              }=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$               $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$



              Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$



              However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                This is an exercise showing your given function satisfies the given differential equation!



                According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
                }=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$                 $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$



                Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$



                However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  This is an exercise showing your given function satisfies the given differential equation!



                  According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
                  }=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$                   $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$



                  Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$



                  However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.






                  share|cite|improve this answer









                  $endgroup$



                  This is an exercise showing your given function satisfies the given differential equation!



                  According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
                  }=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$                   $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$



                  Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$



                  However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 at 4:28









                  coreyman317coreyman317

                  772420




                  772420























                      1












                      $begingroup$

                      I do not think the statement to be proven is true. We clearly have
                      $$y=ln x-ln(1-x),$$
                      and so
                      $$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
                      Let $x=1/2$. Then it is easy to see that the LHS is
                      $$x^3 y''=0$$
                      but the RHS is
                      $$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        I do not think the statement to be proven is true. We clearly have
                        $$y=ln x-ln(1-x),$$
                        and so
                        $$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
                        Let $x=1/2$. Then it is easy to see that the LHS is
                        $$x^3 y''=0$$
                        but the RHS is
                        $$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          I do not think the statement to be proven is true. We clearly have
                          $$y=ln x-ln(1-x),$$
                          and so
                          $$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
                          Let $x=1/2$. Then it is easy to see that the LHS is
                          $$x^3 y''=0$$
                          but the RHS is
                          $$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$






                          share|cite|improve this answer









                          $endgroup$



                          I do not think the statement to be proven is true. We clearly have
                          $$y=ln x-ln(1-x),$$
                          and so
                          $$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
                          Let $x=1/2$. Then it is easy to see that the LHS is
                          $$x^3 y''=0$$
                          but the RHS is
                          $$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 14 at 4:18









                          user146512user146512

                          636




                          636






























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