Ordinary Differential Equation Problem

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Let $$g$$ and $$ h$$ be solutions of the second-order linear differential equation



$$a(x)y'' +b(x)y' + c(x)y = 0$$



on an interval where a(x) is never
zero.



Show that the Wronskian W of g and h satisfies
$$a(x)W'(x) + b(x)W(x) = 0$$



Now $$W(x) = gh' - hg$$
and
$$W'(x) = h''g - g''h$$
Then I get,
$$a(x)W'(x) + b(x)W(x) = g(a(x)h'' + b(x)h') - h(a(x)g'' + b(x)g')$$
To get this expression to equal 0 I need to show $$c(x) = 0$$ so that the h and g satisfy the ODE to bring 0 but how?







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    up vote
    -1
    down vote

    favorite












    Let $$g$$ and $$ h$$ be solutions of the second-order linear differential equation



    $$a(x)y'' +b(x)y' + c(x)y = 0$$



    on an interval where a(x) is never
    zero.



    Show that the Wronskian W of g and h satisfies
    $$a(x)W'(x) + b(x)W(x) = 0$$



    Now $$W(x) = gh' - hg$$
    and
    $$W'(x) = h''g - g''h$$
    Then I get,
    $$a(x)W'(x) + b(x)W(x) = g(a(x)h'' + b(x)h') - h(a(x)g'' + b(x)g')$$
    To get this expression to equal 0 I need to show $$c(x) = 0$$ so that the h and g satisfy the ODE to bring 0 but how?







    share|cite|improve this question
























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      Let $$g$$ and $$ h$$ be solutions of the second-order linear differential equation



      $$a(x)y'' +b(x)y' + c(x)y = 0$$



      on an interval where a(x) is never
      zero.



      Show that the Wronskian W of g and h satisfies
      $$a(x)W'(x) + b(x)W(x) = 0$$



      Now $$W(x) = gh' - hg$$
      and
      $$W'(x) = h''g - g''h$$
      Then I get,
      $$a(x)W'(x) + b(x)W(x) = g(a(x)h'' + b(x)h') - h(a(x)g'' + b(x)g')$$
      To get this expression to equal 0 I need to show $$c(x) = 0$$ so that the h and g satisfy the ODE to bring 0 but how?







      share|cite|improve this question














      Let $$g$$ and $$ h$$ be solutions of the second-order linear differential equation



      $$a(x)y'' +b(x)y' + c(x)y = 0$$



      on an interval where a(x) is never
      zero.



      Show that the Wronskian W of g and h satisfies
      $$a(x)W'(x) + b(x)W(x) = 0$$



      Now $$W(x) = gh' - hg$$
      and
      $$W'(x) = h''g - g''h$$
      Then I get,
      $$a(x)W'(x) + b(x)W(x) = g(a(x)h'' + b(x)h') - h(a(x)g'' + b(x)g')$$
      To get this expression to equal 0 I need to show $$c(x) = 0$$ so that the h and g satisfy the ODE to bring 0 but how?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 19 at 6:56

























      asked Aug 19 at 6:46









      Gordon Ramsey

      545




      545




















          1 Answer
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          You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:



          $$
          beginalign*
          a(x)h'' + b(x)h' &= -c(x)h \
          a(x)g'' + b(x)g' &= -c(x)g.
          endalign*
          $$



          Substituting into your last equation,
          $$
          beginalign*
          a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
          &= c(x)[-gh + hg] \
          &= 0.
          endalign*
          $$






          share|cite|improve this answer




















          • All good, thanks.
            – Gordon Ramsey
            Aug 19 at 7:00










          Your Answer




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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote



          accepted










          You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:



          $$
          beginalign*
          a(x)h'' + b(x)h' &= -c(x)h \
          a(x)g'' + b(x)g' &= -c(x)g.
          endalign*
          $$



          Substituting into your last equation,
          $$
          beginalign*
          a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
          &= c(x)[-gh + hg] \
          &= 0.
          endalign*
          $$






          share|cite|improve this answer




















          • All good, thanks.
            – Gordon Ramsey
            Aug 19 at 7:00














          up vote
          0
          down vote



          accepted










          You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:



          $$
          beginalign*
          a(x)h'' + b(x)h' &= -c(x)h \
          a(x)g'' + b(x)g' &= -c(x)g.
          endalign*
          $$



          Substituting into your last equation,
          $$
          beginalign*
          a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
          &= c(x)[-gh + hg] \
          &= 0.
          endalign*
          $$






          share|cite|improve this answer




















          • All good, thanks.
            – Gordon Ramsey
            Aug 19 at 7:00












          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:



          $$
          beginalign*
          a(x)h'' + b(x)h' &= -c(x)h \
          a(x)g'' + b(x)g' &= -c(x)g.
          endalign*
          $$



          Substituting into your last equation,
          $$
          beginalign*
          a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
          &= c(x)[-gh + hg] \
          &= 0.
          endalign*
          $$






          share|cite|improve this answer












          You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:



          $$
          beginalign*
          a(x)h'' + b(x)h' &= -c(x)h \
          a(x)g'' + b(x)g' &= -c(x)g.
          endalign*
          $$



          Substituting into your last equation,
          $$
          beginalign*
          a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
          &= c(x)[-gh + hg] \
          &= 0.
          endalign*
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 19 at 6:54









          youngsmasher

          35537




          35537











          • All good, thanks.
            – Gordon Ramsey
            Aug 19 at 7:00
















          • All good, thanks.
            – Gordon Ramsey
            Aug 19 at 7:00















          All good, thanks.
          – Gordon Ramsey
          Aug 19 at 7:00




          All good, thanks.
          – Gordon Ramsey
          Aug 19 at 7:00












           

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