Solving a system of linear ODEs subject to specific conditions











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I know this question has been asked and answered before, but my version of the question is a bit easier. Also, I've my exam in a few hours, so please don't close this question. Thanks in advance!



$x_0'=-ax_0+bx_1$



$x_1'=ax_0-(a+b)x_1+2bx_2$



$x_2'=ax_1-2bx_2$



Given that $x_0+x_1+x_2=1$ always. Also, initially, $x_0=1$.



Usually, the previous line is not there given for questions already asked. I guess it will simplify it sufficiently. But, I was unable to solve it. Please help. Thanks






differential-equations







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    up vote
    0
    down vote

    favorite












    I know this question has been asked and answered before, but my version of the question is a bit easier. Also, I've my exam in a few hours, so please don't close this question. Thanks in advance!



    $x_0'=-ax_0+bx_1$



    $x_1'=ax_0-(a+b)x_1+2bx_2$



    $x_2'=ax_1-2bx_2$



    Given that $x_0+x_1+x_2=1$ always. Also, initially, $x_0=1$.



    Usually, the previous line is not there given for questions already asked. I guess it will simplify it sufficiently. But, I was unable to solve it. Please help. Thanks






    differential-equations







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I know this question has been asked and answered before, but my version of the question is a bit easier. Also, I've my exam in a few hours, so please don't close this question. Thanks in advance!



      $x_0'=-ax_0+bx_1$



      $x_1'=ax_0-(a+b)x_1+2bx_2$



      $x_2'=ax_1-2bx_2$



      Given that $x_0+x_1+x_2=1$ always. Also, initially, $x_0=1$.



      Usually, the previous line is not there given for questions already asked. I guess it will simplify it sufficiently. But, I was unable to solve it. Please help. Thanks






      differential-equations







      share|cite|improve this question













      I know this question has been asked and answered before, but my version of the question is a bit easier. Also, I've my exam in a few hours, so please don't close this question. Thanks in advance!



      $x_0'=-ax_0+bx_1$



      $x_1'=ax_0-(a+b)x_1+2bx_2$



      $x_2'=ax_1-2bx_2$



      Given that $x_0+x_1+x_2=1$ always. Also, initially, $x_0=1$.



      Usually, the previous line is not there given for questions already asked. I guess it will simplify it sufficiently. But, I was unable to solve it. Please help. Thanks






      differential-equations




      differential-equations






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 20 at 3:52









      Ankit Kumar

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