How to solve this pair of differential equations?












1














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 differential-equations







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  • 1




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










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










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










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
















1














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 differential-equations







share|cite|improve this question




















  • 1




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










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










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










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














1












1








1







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 differential-equations







share|cite|improve this question















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 differential-equations




calculus 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




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










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










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










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














  • 1




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










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










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










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








1




1




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




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












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




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












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




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












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




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










3 Answers
3






active

oldest

votes


















1














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





























    -1














    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





















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










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










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










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



















    -2














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






    share|cite|improve this answer





















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






    • 1




      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.)
      – Did
      Dec 12 '18 at 17:18










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










    • @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...
      – Federico
      Dec 12 '18 at 17:22












    • 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.
      – 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














    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


























      1














      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
























        1












        1








        1






        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












        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

        6,093823




        6,093823























            -1














            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





















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










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










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










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
















            -1














            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





















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










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










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










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














            -1












            -1








            -1






            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












            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

            5,5882732




            5,5882732












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










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










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










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


















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










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










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










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
















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




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












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




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












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




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












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




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











            -2














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






            share|cite|improve this answer





















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






            • 1




              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.)
              – Did
              Dec 12 '18 at 17:18










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










            • @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...
              – Federico
              Dec 12 '18 at 17:22












            • 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.
              – Did
              Dec 12 '18 at 17:22
















            -2














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






            share|cite|improve this answer





















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






            • 1




              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.)
              – Did
              Dec 12 '18 at 17:18










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










            • @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...
              – Federico
              Dec 12 '18 at 17:22












            • 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.
              – Did
              Dec 12 '18 at 17:22














            -2












            -2








            -2






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






            share|cite|improve this answer












            $(y,z)'=Acdot(y,z)$ 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

            4,829514




            4,829514












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






            • 1




              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.)
              – Did
              Dec 12 '18 at 17:18










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










            • @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...
              – Federico
              Dec 12 '18 at 17:22












            • 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.
              – Did
              Dec 12 '18 at 17:22


















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






            • 1




              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.)
              – Did
              Dec 12 '18 at 17:18










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










            • @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...
              – Federico
              Dec 12 '18 at 17:22












            • 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.
              – Did
              Dec 12 '18 at 17:22
















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




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




            1




            1




            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.)
            – Did
            Dec 12 '18 at 17:18




            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.)
            – Did
            Dec 12 '18 at 17:18












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




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












            @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...
            – Federico
            Dec 12 '18 at 17:22






            @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...
            – Federico
            Dec 12 '18 at 17:22














            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.
            – Did
            Dec 12 '18 at 17:22




            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.
            – Did
            Dec 12 '18 at 17:22


















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