How to solve this pair of differential equations?












1












$begingroup$


The equations are



$$y' = y + 3z$$$$z' = -3y + z$$I vaguely remember I need to differentiate one to get the form of the other, but by differentiating the first and subbing in the second I get$$y'' = y' + 3z'=> y''- y' + 9y = 3z$$ I don't know what to do from here.



Edit:
From here I got to $y''-2y'+10y = 0$, so from $m^2-2m+10=0$ we get$$y=e^x(Acos(3x) + Bsin(3x))$$Where would I go from here?






calculus ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $z = (y'-y)/3.$
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 20:54










  • $begingroup$
    Small correction above. $y'' - y' + 9y = 3z$
    $endgroup$
    – Doug M
    Dec 4 '18 at 21:21










  • $begingroup$
    @MathPhys Thee equations needed to solve three variables $(x,y,z)$
    $endgroup$
    – Narasimham
    Dec 4 '18 at 21:37










  • $begingroup$
    Hint: $(ypm zi)' = (1mp 3i)(ypm zi)$
    $endgroup$
    – achille hui
    Dec 5 '18 at 20:37
















1












$begingroup$


The equations are



$$y' = y + 3z$$$$z' = -3y + z$$I vaguely remember I need to differentiate one to get the form of the other, but by differentiating the first and subbing in the second I get$$y'' = y' + 3z'=> y''- y' + 9y = 3z$$ I don't know what to do from here.



Edit:
From here I got to $y''-2y'+10y = 0$, so from $m^2-2m+10=0$ we get$$y=e^x(Acos(3x) + Bsin(3x))$$Where would I go from here?






calculus ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $z = (y'-y)/3.$
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 20:54










  • $begingroup$
    Small correction above. $y'' - y' + 9y = 3z$
    $endgroup$
    – Doug M
    Dec 4 '18 at 21:21










  • $begingroup$
    @MathPhys Thee equations needed to solve three variables $(x,y,z)$
    $endgroup$
    – Narasimham
    Dec 4 '18 at 21:37










  • $begingroup$
    Hint: $(ypm zi)' = (1mp 3i)(ypm zi)$
    $endgroup$
    – achille hui
    Dec 5 '18 at 20:37














1












1








1





$begingroup$


The equations are



$$y' = y + 3z$$$$z' = -3y + z$$I vaguely remember I need to differentiate one to get the form of the other, but by differentiating the first and subbing in the second I get$$y'' = y' + 3z'=> y''- y' + 9y = 3z$$ I don't know what to do from here.



Edit:
From here I got to $y''-2y'+10y = 0$, so from $m^2-2m+10=0$ we get$$y=e^x(Acos(3x) + Bsin(3x))$$Where would I go from here?






calculus ordinary-differential-equations







share|cite|improve this question











$endgroup$




The equations are



$$y' = y + 3z$$$$z' = -3y + z$$I vaguely remember I need to differentiate one to get the form of the other, but by differentiating the first and subbing in the second I get$$y'' = y' + 3z'=> y''- y' + 9y = 3z$$ I don't know what to do from here.



Edit:
From here I got to $y''-2y'+10y = 0$, so from $m^2-2m+10=0$ we get$$y=e^x(Acos(3x) + Bsin(3x))$$Where would I go from here?






calculus ordinary-differential-equations




calculus ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 21:42







mathPhys

















asked Dec 4 '18 at 20:53









mathPhysmathPhys

114




114








  • 1




    $begingroup$
    $z = (y'-y)/3.$
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 20:54










  • $begingroup$
    Small correction above. $y'' - y' + 9y = 3z$
    $endgroup$
    – Doug M
    Dec 4 '18 at 21:21










  • $begingroup$
    @MathPhys Thee equations needed to solve three variables $(x,y,z)$
    $endgroup$
    – Narasimham
    Dec 4 '18 at 21:37










  • $begingroup$
    Hint: $(ypm zi)' = (1mp 3i)(ypm zi)$
    $endgroup$
    – achille hui
    Dec 5 '18 at 20:37














  • 1




    $begingroup$
    $z = (y'-y)/3.$
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 20:54










  • $begingroup$
    Small correction above. $y'' - y' + 9y = 3z$
    $endgroup$
    – Doug M
    Dec 4 '18 at 21:21










  • $begingroup$
    @MathPhys Thee equations needed to solve three variables $(x,y,z)$
    $endgroup$
    – Narasimham
    Dec 4 '18 at 21:37










  • $begingroup$
    Hint: $(ypm zi)' = (1mp 3i)(ypm zi)$
    $endgroup$
    – achille hui
    Dec 5 '18 at 20:37








1




1




$begingroup$
$z = (y'-y)/3.$
$endgroup$
– T. Bongers
Dec 4 '18 at 20:54




$begingroup$
$z = (y'-y)/3.$
$endgroup$
– T. Bongers
Dec 4 '18 at 20:54












$begingroup$
Small correction above. $y'' - y' + 9y = 3z$
$endgroup$
– Doug M
Dec 4 '18 at 21:21




$begingroup$
Small correction above. $y'' - y' + 9y = 3z$
$endgroup$
– Doug M
Dec 4 '18 at 21:21












$begingroup$
@MathPhys Thee equations needed to solve three variables $(x,y,z)$
$endgroup$
– Narasimham
Dec 4 '18 at 21:37




$begingroup$
@MathPhys Thee equations needed to solve three variables $(x,y,z)$
$endgroup$
– Narasimham
Dec 4 '18 at 21:37












$begingroup$
Hint: $(ypm zi)' = (1mp 3i)(ypm zi)$
$endgroup$
– achille hui
Dec 5 '18 at 20:37




$begingroup$
Hint: $(ypm zi)' = (1mp 3i)(ypm zi)$
$endgroup$
– achille hui
Dec 5 '18 at 20:37










3 Answers
3






active

oldest

votes


















1












$begingroup$

From the first equation, $z = frac{1}{3}(y'-y)$ and $y'' = y' + 3z'$. From the second eqution, $z' = -3y+z = -3y+frac{1}{3}(y'-y)$. So, $$y'' = y'+ 3(-3y+frac{1}{3}(y'-y)) \ Rightarrow y'' - 2y' + 10y = 0 \ Rightarrow y = (Acos(3x)+Bsin(3x))exp(x) $$



Recalling that $z = frac{1}{3}(y'-y)$, we first calculate $$y' = y + (Bcos(3x)-Asin(3x))exp(x) $$



and so



$$ z = left(frac{B}{3}cos(3x)-frac{A}{3}sin(3x)right)exp(x) $$






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    Your equation is
    $$x'(t)=Mx(t)$$
    Where
    $$M=begin{bmatrix}1&3\-3&1 end{bmatrix}$$
    $$x(t)=begin{bmatrix}y(t)\z(t)end{bmatrix}$$
    Which has the solution
    $$x(t)=e^{Mt}x(0)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What is the reason behind the downvote?
      $endgroup$
      – Botond
      Dec 5 '18 at 23:27










    • $begingroup$
      I think people don't really like matrix exponentiation around here. Just have a look at the similar fate of my answer
      $endgroup$
      – Federico
      Dec 12 '18 at 17:31










    • $begingroup$
      @Federico Yes, sadly. But it's more elegant :)
      $endgroup$
      – Botond
      Dec 12 '18 at 19:34










    • $begingroup$
      Honestly, it's also objectively useless for the OP, though
      $endgroup$
      – Federico
      Dec 12 '18 at 19:37



















    -2












    $begingroup$

    $(y,z)'=Acdot(y,z)$ where $A=begin{pmatrix}1&3\-3&1end{pmatrix}$. Hence
    $$
    (y,z)(t) = exp(tA)cdot(y,z)(0)
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why the downvote? The answer is correct. Maybe go read about matrix exponentiation instead of downvoting
      $endgroup$
      – Federico
      Dec 6 '18 at 13:24






    • 1




      $begingroup$
      Maybe wonder whether your post can possibly be useful to the OP, from what one can guess of their background, rather than fulminating against voters. (BTW, I didn't vote but I am tempted to.)
      $endgroup$
      – Did
      Dec 12 '18 at 17:18










    • $begingroup$
      @Did It might be useful, if for nothing else, at least as an invitation to learn about matrix exponentiation.
      $endgroup$
      – Federico
      Dec 12 '18 at 17:20










    • $begingroup$
      @Did Also, I've never downvoted correct answers. At least leave them alone if you don't find them particularly useful. To me a downvote is to signal that something is wrong, misleading, very badly presented, etc...
      $endgroup$
      – Federico
      Dec 12 '18 at 17:22












    • $begingroup$
      You might want to try holding this rather fragile position -- but then please do not be surprised when others do not follow you and act accordingly.
      $endgroup$
      – Did
      Dec 12 '18 at 17:22











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    From the first equation, $z = frac{1}{3}(y'-y)$ and $y'' = y' + 3z'$. From the second eqution, $z' = -3y+z = -3y+frac{1}{3}(y'-y)$. So, $$y'' = y'+ 3(-3y+frac{1}{3}(y'-y)) \ Rightarrow y'' - 2y' + 10y = 0 \ Rightarrow y = (Acos(3x)+Bsin(3x))exp(x) $$



    Recalling that $z = frac{1}{3}(y'-y)$, we first calculate $$y' = y + (Bcos(3x)-Asin(3x))exp(x) $$



    and so



    $$ z = left(frac{B}{3}cos(3x)-frac{A}{3}sin(3x)right)exp(x) $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      From the first equation, $z = frac{1}{3}(y'-y)$ and $y'' = y' + 3z'$. From the second eqution, $z' = -3y+z = -3y+frac{1}{3}(y'-y)$. So, $$y'' = y'+ 3(-3y+frac{1}{3}(y'-y)) \ Rightarrow y'' - 2y' + 10y = 0 \ Rightarrow y = (Acos(3x)+Bsin(3x))exp(x) $$



      Recalling that $z = frac{1}{3}(y'-y)$, we first calculate $$y' = y + (Bcos(3x)-Asin(3x))exp(x) $$



      and so



      $$ z = left(frac{B}{3}cos(3x)-frac{A}{3}sin(3x)right)exp(x) $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        From the first equation, $z = frac{1}{3}(y'-y)$ and $y'' = y' + 3z'$. From the second eqution, $z' = -3y+z = -3y+frac{1}{3}(y'-y)$. So, $$y'' = y'+ 3(-3y+frac{1}{3}(y'-y)) \ Rightarrow y'' - 2y' + 10y = 0 \ Rightarrow y = (Acos(3x)+Bsin(3x))exp(x) $$



        Recalling that $z = frac{1}{3}(y'-y)$, we first calculate $$y' = y + (Bcos(3x)-Asin(3x))exp(x) $$



        and so



        $$ z = left(frac{B}{3}cos(3x)-frac{A}{3}sin(3x)right)exp(x) $$






        share|cite|improve this answer









        $endgroup$



        From the first equation, $z = frac{1}{3}(y'-y)$ and $y'' = y' + 3z'$. From the second eqution, $z' = -3y+z = -3y+frac{1}{3}(y'-y)$. So, $$y'' = y'+ 3(-3y+frac{1}{3}(y'-y)) \ Rightarrow y'' - 2y' + 10y = 0 \ Rightarrow y = (Acos(3x)+Bsin(3x))exp(x) $$



        Recalling that $z = frac{1}{3}(y'-y)$, we first calculate $$y' = y + (Bcos(3x)-Asin(3x))exp(x) $$



        and so



        $$ z = left(frac{B}{3}cos(3x)-frac{A}{3}sin(3x)right)exp(x) $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 21:59









        AlexanderJ93AlexanderJ93

        6,158823




        6,158823























            -1












            $begingroup$

            Your equation is
            $$x'(t)=Mx(t)$$
            Where
            $$M=begin{bmatrix}1&3\-3&1 end{bmatrix}$$
            $$x(t)=begin{bmatrix}y(t)\z(t)end{bmatrix}$$
            Which has the solution
            $$x(t)=e^{Mt}x(0)$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What is the reason behind the downvote?
              $endgroup$
              – Botond
              Dec 5 '18 at 23:27










            • $begingroup$
              I think people don't really like matrix exponentiation around here. Just have a look at the similar fate of my answer
              $endgroup$
              – Federico
              Dec 12 '18 at 17:31










            • $begingroup$
              @Federico Yes, sadly. But it's more elegant :)
              $endgroup$
              – Botond
              Dec 12 '18 at 19:34










            • $begingroup$
              Honestly, it's also objectively useless for the OP, though
              $endgroup$
              – Federico
              Dec 12 '18 at 19:37
















            -1












            $begingroup$

            Your equation is
            $$x'(t)=Mx(t)$$
            Where
            $$M=begin{bmatrix}1&3\-3&1 end{bmatrix}$$
            $$x(t)=begin{bmatrix}y(t)\z(t)end{bmatrix}$$
            Which has the solution
            $$x(t)=e^{Mt}x(0)$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What is the reason behind the downvote?
              $endgroup$
              – Botond
              Dec 5 '18 at 23:27










            • $begingroup$
              I think people don't really like matrix exponentiation around here. Just have a look at the similar fate of my answer
              $endgroup$
              – Federico
              Dec 12 '18 at 17:31










            • $begingroup$
              @Federico Yes, sadly. But it's more elegant :)
              $endgroup$
              – Botond
              Dec 12 '18 at 19:34










            • $begingroup$
              Honestly, it's also objectively useless for the OP, though
              $endgroup$
              – Federico
              Dec 12 '18 at 19:37














            -1












            -1








            -1





            $begingroup$

            Your equation is
            $$x'(t)=Mx(t)$$
            Where
            $$M=begin{bmatrix}1&3\-3&1 end{bmatrix}$$
            $$x(t)=begin{bmatrix}y(t)\z(t)end{bmatrix}$$
            Which has the solution
            $$x(t)=e^{Mt}x(0)$$






            share|cite|improve this answer









            $endgroup$



            Your equation is
            $$x'(t)=Mx(t)$$
            Where
            $$M=begin{bmatrix}1&3\-3&1 end{bmatrix}$$
            $$x(t)=begin{bmatrix}y(t)\z(t)end{bmatrix}$$
            Which has the solution
            $$x(t)=e^{Mt}x(0)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 21:11









            BotondBotond

            5,6822732




            5,6822732












            • $begingroup$
              What is the reason behind the downvote?
              $endgroup$
              – Botond
              Dec 5 '18 at 23:27










            • $begingroup$
              I think people don't really like matrix exponentiation around here. Just have a look at the similar fate of my answer
              $endgroup$
              – Federico
              Dec 12 '18 at 17:31










            • $begingroup$
              @Federico Yes, sadly. But it's more elegant :)
              $endgroup$
              – Botond
              Dec 12 '18 at 19:34










            • $begingroup$
              Honestly, it's also objectively useless for the OP, though
              $endgroup$
              – Federico
              Dec 12 '18 at 19:37


















            • $begingroup$
              What is the reason behind the downvote?
              $endgroup$
              – Botond
              Dec 5 '18 at 23:27










            • $begingroup$
              I think people don't really like matrix exponentiation around here. Just have a look at the similar fate of my answer
              $endgroup$
              – Federico
              Dec 12 '18 at 17:31










            • $begingroup$
              @Federico Yes, sadly. But it's more elegant :)
              $endgroup$
              – Botond
              Dec 12 '18 at 19:34










            • $begingroup$
              Honestly, it's also objectively useless for the OP, though
              $endgroup$
              – Federico
              Dec 12 '18 at 19:37
















            $begingroup$
            What is the reason behind the downvote?
            $endgroup$
            – Botond
            Dec 5 '18 at 23:27




            $begingroup$
            What is the reason behind the downvote?
            $endgroup$
            – Botond
            Dec 5 '18 at 23:27












            $begingroup$
            I think people don't really like matrix exponentiation around here. Just have a look at the similar fate of my answer
            $endgroup$
            – Federico
            Dec 12 '18 at 17:31




            $begingroup$
            I think people don't really like matrix exponentiation around here. Just have a look at the similar fate of my answer
            $endgroup$
            – Federico
            Dec 12 '18 at 17:31












            $begingroup$
            @Federico Yes, sadly. But it's more elegant :)
            $endgroup$
            – Botond
            Dec 12 '18 at 19:34




            $begingroup$
            @Federico Yes, sadly. But it's more elegant :)
            $endgroup$
            – Botond
            Dec 12 '18 at 19:34












            $begingroup$
            Honestly, it's also objectively useless for the OP, though
            $endgroup$
            – Federico
            Dec 12 '18 at 19:37




            $begingroup$
            Honestly, it's also objectively useless for the OP, though
            $endgroup$
            – Federico
            Dec 12 '18 at 19:37











            -2












            $begingroup$

            $(y,z)'=Acdot(y,z)$ where $A=begin{pmatrix}1&3\-3&1end{pmatrix}$. Hence
            $$
            (y,z)(t) = exp(tA)cdot(y,z)(0)
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why the downvote? The answer is correct. Maybe go read about matrix exponentiation instead of downvoting
              $endgroup$
              – Federico
              Dec 6 '18 at 13:24






            • 1




              $begingroup$
              Maybe wonder whether your post can possibly be useful to the OP, from what one can guess of their background, rather than fulminating against voters. (BTW, I didn't vote but I am tempted to.)
              $endgroup$
              – Did
              Dec 12 '18 at 17:18










            • $begingroup$
              @Did It might be useful, if for nothing else, at least as an invitation to learn about matrix exponentiation.
              $endgroup$
              – Federico
              Dec 12 '18 at 17:20










            • $begingroup$
              @Did Also, I've never downvoted correct answers. At least leave them alone if you don't find them particularly useful. To me a downvote is to signal that something is wrong, misleading, very badly presented, etc...
              $endgroup$
              – Federico
              Dec 12 '18 at 17:22












            • $begingroup$
              You might want to try holding this rather fragile position -- but then please do not be surprised when others do not follow you and act accordingly.
              $endgroup$
              – Did
              Dec 12 '18 at 17:22
















            -2












            $begingroup$

            $(y,z)'=Acdot(y,z)$ where $A=begin{pmatrix}1&3\-3&1end{pmatrix}$. Hence
            $$
            (y,z)(t) = exp(tA)cdot(y,z)(0)
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why the downvote? The answer is correct. Maybe go read about matrix exponentiation instead of downvoting
              $endgroup$
              – Federico
              Dec 6 '18 at 13:24






            • 1




              $begingroup$
              Maybe wonder whether your post can possibly be useful to the OP, from what one can guess of their background, rather than fulminating against voters. (BTW, I didn't vote but I am tempted to.)
              $endgroup$
              – Did
              Dec 12 '18 at 17:18










            • $begingroup$
              @Did It might be useful, if for nothing else, at least as an invitation to learn about matrix exponentiation.
              $endgroup$
              – Federico
              Dec 12 '18 at 17:20










            • $begingroup$
              @Did Also, I've never downvoted correct answers. At least leave them alone if you don't find them particularly useful. To me a downvote is to signal that something is wrong, misleading, very badly presented, etc...
              $endgroup$
              – Federico
              Dec 12 '18 at 17:22












            • $begingroup$
              You might want to try holding this rather fragile position -- but then please do not be surprised when others do not follow you and act accordingly.
              $endgroup$
              – Did
              Dec 12 '18 at 17:22














            -2












            -2








            -2





            $begingroup$

            $(y,z)'=Acdot(y,z)$ where $A=begin{pmatrix}1&3\-3&1end{pmatrix}$. Hence
            $$
            (y,z)(t) = exp(tA)cdot(y,z)(0)
            $$






            share|cite|improve this answer









            $endgroup$



            $(y,z)'=Acdot(y,z)$ where $A=begin{pmatrix}1&3\-3&1end{pmatrix}$. Hence
            $$
            (y,z)(t) = exp(tA)cdot(y,z)(0)
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 21:10









            FedericoFederico

            4,984514




            4,984514












            • $begingroup$
              Why the downvote? The answer is correct. Maybe go read about matrix exponentiation instead of downvoting
              $endgroup$
              – Federico
              Dec 6 '18 at 13:24






            • 1




              $begingroup$
              Maybe wonder whether your post can possibly be useful to the OP, from what one can guess of their background, rather than fulminating against voters. (BTW, I didn't vote but I am tempted to.)
              $endgroup$
              – Did
              Dec 12 '18 at 17:18










            • $begingroup$
              @Did It might be useful, if for nothing else, at least as an invitation to learn about matrix exponentiation.
              $endgroup$
              – Federico
              Dec 12 '18 at 17:20










            • $begingroup$
              @Did Also, I've never downvoted correct answers. At least leave them alone if you don't find them particularly useful. To me a downvote is to signal that something is wrong, misleading, very badly presented, etc...
              $endgroup$
              – Federico
              Dec 12 '18 at 17:22












            • $begingroup$
              You might want to try holding this rather fragile position -- but then please do not be surprised when others do not follow you and act accordingly.
              $endgroup$
              – Did
              Dec 12 '18 at 17:22


















            • $begingroup$
              Why the downvote? The answer is correct. Maybe go read about matrix exponentiation instead of downvoting
              $endgroup$
              – Federico
              Dec 6 '18 at 13:24






            • 1




              $begingroup$
              Maybe wonder whether your post can possibly be useful to the OP, from what one can guess of their background, rather than fulminating against voters. (BTW, I didn't vote but I am tempted to.)
              $endgroup$
              – Did
              Dec 12 '18 at 17:18










            • $begingroup$
              @Did It might be useful, if for nothing else, at least as an invitation to learn about matrix exponentiation.
              $endgroup$
              – Federico
              Dec 12 '18 at 17:20










            • $begingroup$
              @Did Also, I've never downvoted correct answers. At least leave them alone if you don't find them particularly useful. To me a downvote is to signal that something is wrong, misleading, very badly presented, etc...
              $endgroup$
              – Federico
              Dec 12 '18 at 17:22












            • $begingroup$
              You might want to try holding this rather fragile position -- but then please do not be surprised when others do not follow you and act accordingly.
              $endgroup$
              – Did
              Dec 12 '18 at 17:22
















            $begingroup$
            Why the downvote? The answer is correct. Maybe go read about matrix exponentiation instead of downvoting
            $endgroup$
            – Federico
            Dec 6 '18 at 13:24




            $begingroup$
            Why the downvote? The answer is correct. Maybe go read about matrix exponentiation instead of downvoting
            $endgroup$
            – Federico
            Dec 6 '18 at 13:24




            1




            1




            $begingroup$
            Maybe wonder whether your post can possibly be useful to the OP, from what one can guess of their background, rather than fulminating against voters. (BTW, I didn't vote but I am tempted to.)
            $endgroup$
            – Did
            Dec 12 '18 at 17:18




            $begingroup$
            Maybe wonder whether your post can possibly be useful to the OP, from what one can guess of their background, rather than fulminating against voters. (BTW, I didn't vote but I am tempted to.)
            $endgroup$
            – Did
            Dec 12 '18 at 17:18












            $begingroup$
            @Did It might be useful, if for nothing else, at least as an invitation to learn about matrix exponentiation.
            $endgroup$
            – Federico
            Dec 12 '18 at 17:20




            $begingroup$
            @Did It might be useful, if for nothing else, at least as an invitation to learn about matrix exponentiation.
            $endgroup$
            – Federico
            Dec 12 '18 at 17:20












            $begingroup$
            @Did Also, I've never downvoted correct answers. At least leave them alone if you don't find them particularly useful. To me a downvote is to signal that something is wrong, misleading, very badly presented, etc...
            $endgroup$
            – Federico
            Dec 12 '18 at 17:22






            $begingroup$
            @Did Also, I've never downvoted correct answers. At least leave them alone if you don't find them particularly useful. To me a downvote is to signal that something is wrong, misleading, very badly presented, etc...
            $endgroup$
            – Federico
            Dec 12 '18 at 17:22














            $begingroup$
            You might want to try holding this rather fragile position -- but then please do not be surprised when others do not follow you and act accordingly.
            $endgroup$
            – Did
            Dec 12 '18 at 17:22




            $begingroup$
            You might want to try holding this rather fragile position -- but then please do not be surprised when others do not follow you and act accordingly.
            $endgroup$
            – Did
            Dec 12 '18 at 17:22


















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