Independent solution means no repeated root.












0












$begingroup$


If $y_1 $ and $y_2 $ are independent solution of homogeneous second order linear differential equation.Then




  1. neither $y_1$ and nor $y_2$ has repeated zeros ?


  2. They don't have common point of extremum.



I don't see why its true. I tried write it as linear combination in non trivial way but have no clue how to use hypothesis ( i will add more details if i found any) .please give me a hint to start with






ordinary-differential-equations







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    If $y_1 $ and $y_2 $ are independent solution of homogeneous second order linear differential equation.Then




    1. neither $y_1$ and nor $y_2$ has repeated zeros ?


    2. They don't have common point of extremum.



    I don't see why its true. I tried write it as linear combination in non trivial way but have no clue how to use hypothesis ( i will add more details if i found any) .please give me a hint to start with






    ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      If $y_1 $ and $y_2 $ are independent solution of homogeneous second order linear differential equation.Then




      1. neither $y_1$ and nor $y_2$ has repeated zeros ?


      2. They don't have common point of extremum.



      I don't see why its true. I tried write it as linear combination in non trivial way but have no clue how to use hypothesis ( i will add more details if i found any) .please give me a hint to start with






      ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      If $y_1 $ and $y_2 $ are independent solution of homogeneous second order linear differential equation.Then




      1. neither $y_1$ and nor $y_2$ has repeated zeros ?


      2. They don't have common point of extremum.



      I don't see why its true. I tried write it as linear combination in non trivial way but have no clue how to use hypothesis ( i will add more details if i found any) .please give me a hint to start with






      ordinary-differential-equations




      ordinary-differential-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 29 at 18:55









      Cloud JRCloud JR

      910518




      910518






















          1 Answer
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          $begingroup$

          If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.






          share|cite|improve this answer









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          • $begingroup$
            for 2) how to use uniqueness, itsn't obvious for me
            $endgroup$
            – Cloud JR
            Jan 30 at 13:55












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          $begingroup$

          If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            for 2) how to use uniqueness, itsn't obvious for me
            $endgroup$
            – Cloud JR
            Jan 30 at 13:55
















          1












          $begingroup$

          If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            for 2) how to use uniqueness, itsn't obvious for me
            $endgroup$
            – Cloud JR
            Jan 30 at 13:55














          1












          1








          1





          $begingroup$

          If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.






          share|cite|improve this answer









          $endgroup$



          If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 29 at 19:08









          David C. UllrichDavid C. Ullrich

          61.6k43995




          61.6k43995












          • $begingroup$
            for 2) how to use uniqueness, itsn't obvious for me
            $endgroup$
            – Cloud JR
            Jan 30 at 13:55


















          • $begingroup$
            for 2) how to use uniqueness, itsn't obvious for me
            $endgroup$
            – Cloud JR
            Jan 30 at 13:55
















          $begingroup$
          for 2) how to use uniqueness, itsn't obvious for me
          $endgroup$
          – Cloud JR
          Jan 30 at 13:55




          $begingroup$
          for 2) how to use uniqueness, itsn't obvious for me
          $endgroup$
          – Cloud JR
          Jan 30 at 13:55


















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