Solution to Cauchy Problem












2












$begingroup$


I am trying to solve the following Cauchy Problem:



$y'(t) = A(t)y(t),
A=begin{pmatrix}
t &-1 \
1 &t
end{pmatrix}, y(0)=y_0$



What I did:



I know that $ forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$



So the solution would be:



$y(t)= e^{ begin{pmatrix}
t^2/2 &-t \
t &t^2/2
end{pmatrix}} y_0$



Is that a good reasonning? Please note that I don't have more information regarding the domain of study of this Cauchy Problem so I don't know if that plays any decisive role in the solution found above.



Many thanks!






cauchy-problem







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I am trying to solve the following Cauchy Problem:



    $y'(t) = A(t)y(t),
    A=begin{pmatrix}
    t &-1 \
    1 &t
    end{pmatrix}, y(0)=y_0$



    What I did:



    I know that $ forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$



    So the solution would be:



    $y(t)= e^{ begin{pmatrix}
    t^2/2 &-t \
    t &t^2/2
    end{pmatrix}} y_0$



    Is that a good reasonning? Please note that I don't have more information regarding the domain of study of this Cauchy Problem so I don't know if that plays any decisive role in the solution found above.



    Many thanks!






    cauchy-problem







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I am trying to solve the following Cauchy Problem:



      $y'(t) = A(t)y(t),
      A=begin{pmatrix}
      t &-1 \
      1 &t
      end{pmatrix}, y(0)=y_0$



      What I did:



      I know that $ forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$



      So the solution would be:



      $y(t)= e^{ begin{pmatrix}
      t^2/2 &-t \
      t &t^2/2
      end{pmatrix}} y_0$



      Is that a good reasonning? Please note that I don't have more information regarding the domain of study of this Cauchy Problem so I don't know if that plays any decisive role in the solution found above.



      Many thanks!






      cauchy-problem







      share|cite|improve this question











      $endgroup$




      I am trying to solve the following Cauchy Problem:



      $y'(t) = A(t)y(t),
      A=begin{pmatrix}
      t &-1 \
      1 &t
      end{pmatrix}, y(0)=y_0$



      What I did:



      I know that $ forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$



      So the solution would be:



      $y(t)= e^{ begin{pmatrix}
      t^2/2 &-t \
      t &t^2/2
      end{pmatrix}} y_0$



      Is that a good reasonning? Please note that I don't have more information regarding the domain of study of this Cauchy Problem so I don't know if that plays any decisive role in the solution found above.



      Many thanks!






      cauchy-problem




      cauchy-problem






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 15:21







      PerelMan

















      asked Feb 9 '18 at 13:13









      PerelManPerelMan

      654313




      654313






















          1 Answer
          1






          active

          oldest

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          2












          $begingroup$

          You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
            $endgroup$
            – PerelMan
            Feb 9 '18 at 14:11






          • 1




            $begingroup$
            Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
            $endgroup$
            – mathcounterexamples.net
            Feb 9 '18 at 15:01











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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
            $endgroup$
            – PerelMan
            Feb 9 '18 at 14:11






          • 1




            $begingroup$
            Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
            $endgroup$
            – mathcounterexamples.net
            Feb 9 '18 at 15:01
















          2












          $begingroup$

          You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
            $endgroup$
            – PerelMan
            Feb 9 '18 at 14:11






          • 1




            $begingroup$
            Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
            $endgroup$
            – mathcounterexamples.net
            Feb 9 '18 at 15:01














          2












          2








          2





          $begingroup$

          You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.






          share|cite|improve this answer









          $endgroup$



          You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 9 '18 at 13:54









          mathcounterexamples.netmathcounterexamples.net

          27k22157




          27k22157












          • $begingroup$
            Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
            $endgroup$
            – PerelMan
            Feb 9 '18 at 14:11






          • 1




            $begingroup$
            Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
            $endgroup$
            – mathcounterexamples.net
            Feb 9 '18 at 15:01


















          • $begingroup$
            Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
            $endgroup$
            – PerelMan
            Feb 9 '18 at 14:11






          • 1




            $begingroup$
            Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
            $endgroup$
            – mathcounterexamples.net
            Feb 9 '18 at 15:01
















          $begingroup$
          Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
          $endgroup$
          – PerelMan
          Feb 9 '18 at 14:11




          $begingroup$
          Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
          $endgroup$
          – PerelMan
          Feb 9 '18 at 14:11




          1




          1




          $begingroup$
          Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
          $endgroup$
          – mathcounterexamples.net
          Feb 9 '18 at 15:01




          $begingroup$
          Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
          $endgroup$
          – mathcounterexamples.net
          Feb 9 '18 at 15:01


















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