Systems of differential equations












1















begin{array} { c } { text { Solve the initial value problem } } \ { mathbf { x } ^ { prime } = left( begin{array} { c c } { 1 } & { - 5 } \ { 1 } & { - 3 } end{array} right) mathbf { x } , quad mathbf { x } ( 0 ) = left( begin{array} { c } { 1 } \ { 1 } end{array} right) } \ { text { and describe the behavior of the solution as } t rightarrow infty } end{array}




My answer so far is:
$$vec{x}=c_1e^{(-1+i)t}begin{pmatrix}2+i\1end{pmatrix}+c_2e^{(-1-i)t}begin{pmatrix}2-i\1end{pmatrix}$$



The textbook has the answer:
$$mathbf{x}(t)=e^{-t}begin{pmatrix}cos(t)-3sin(t)\ cos(t)-sin(t)end{pmatrix}$$



How can I simplify my solution?



UPDATE: I used the identity $e^{it} = cos t+isin t$
$$c_1e^{-t}(cos t+isin t)begin{pmatrix}2+i\1end{pmatrix}+c_2e^{-t}(cos t-isin t)begin{pmatrix}2-i\1end{pmatrix}$$






linear-algebra matrices differential-equations systems-of-equations







share|cite|improve this question




















  • 2




    $e^{(-1+i)t}=e^{-t}(cos t+isin t)$
    – Nosrati
    Dec 10 '18 at 4:56






  • 1




    $sin$ and $cos$ are bounded and $e^{-t}to0$ as $ttoinfty$.
    – Nosrati
    Dec 10 '18 at 5:01






  • 1




    At first apply initial condition to renove constants $C_1$ and $C_2$.
    – Nosrati
    Dec 10 '18 at 5:21






  • 1




    Your solution shouldn't has (have) $i$ term!
    – Nosrati
    Dec 10 '18 at 5:29








  • 1




    My method if you interest: $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$ shows $y=e^{-t}cos t$ and $y=e^{-t}sin t$
    – Nosrati
    Dec 10 '18 at 5:44


















1















begin{array} { c } { text { Solve the initial value problem } } \ { mathbf { x } ^ { prime } = left( begin{array} { c c } { 1 } & { - 5 } \ { 1 } & { - 3 } end{array} right) mathbf { x } , quad mathbf { x } ( 0 ) = left( begin{array} { c } { 1 } \ { 1 } end{array} right) } \ { text { and describe the behavior of the solution as } t rightarrow infty } end{array}




My answer so far is:
$$vec{x}=c_1e^{(-1+i)t}begin{pmatrix}2+i\1end{pmatrix}+c_2e^{(-1-i)t}begin{pmatrix}2-i\1end{pmatrix}$$



The textbook has the answer:
$$mathbf{x}(t)=e^{-t}begin{pmatrix}cos(t)-3sin(t)\ cos(t)-sin(t)end{pmatrix}$$



How can I simplify my solution?



UPDATE: I used the identity $e^{it} = cos t+isin t$
$$c_1e^{-t}(cos t+isin t)begin{pmatrix}2+i\1end{pmatrix}+c_2e^{-t}(cos t-isin t)begin{pmatrix}2-i\1end{pmatrix}$$






linear-algebra matrices differential-equations systems-of-equations







share|cite|improve this question




















  • 2




    $e^{(-1+i)t}=e^{-t}(cos t+isin t)$
    – Nosrati
    Dec 10 '18 at 4:56






  • 1




    $sin$ and $cos$ are bounded and $e^{-t}to0$ as $ttoinfty$.
    – Nosrati
    Dec 10 '18 at 5:01






  • 1




    At first apply initial condition to renove constants $C_1$ and $C_2$.
    – Nosrati
    Dec 10 '18 at 5:21






  • 1




    Your solution shouldn't has (have) $i$ term!
    – Nosrati
    Dec 10 '18 at 5:29








  • 1




    My method if you interest: $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$ shows $y=e^{-t}cos t$ and $y=e^{-t}sin t$
    – Nosrati
    Dec 10 '18 at 5:44
















1












1








1


1






begin{array} { c } { text { Solve the initial value problem } } \ { mathbf { x } ^ { prime } = left( begin{array} { c c } { 1 } & { - 5 } \ { 1 } & { - 3 } end{array} right) mathbf { x } , quad mathbf { x } ( 0 ) = left( begin{array} { c } { 1 } \ { 1 } end{array} right) } \ { text { and describe the behavior of the solution as } t rightarrow infty } end{array}




My answer so far is:
$$vec{x}=c_1e^{(-1+i)t}begin{pmatrix}2+i\1end{pmatrix}+c_2e^{(-1-i)t}begin{pmatrix}2-i\1end{pmatrix}$$



The textbook has the answer:
$$mathbf{x}(t)=e^{-t}begin{pmatrix}cos(t)-3sin(t)\ cos(t)-sin(t)end{pmatrix}$$



How can I simplify my solution?



UPDATE: I used the identity $e^{it} = cos t+isin t$
$$c_1e^{-t}(cos t+isin t)begin{pmatrix}2+i\1end{pmatrix}+c_2e^{-t}(cos t-isin t)begin{pmatrix}2-i\1end{pmatrix}$$






linear-algebra matrices differential-equations systems-of-equations







share|cite|improve this question
















begin{array} { c } { text { Solve the initial value problem } } \ { mathbf { x } ^ { prime } = left( begin{array} { c c } { 1 } & { - 5 } \ { 1 } & { - 3 } end{array} right) mathbf { x } , quad mathbf { x } ( 0 ) = left( begin{array} { c } { 1 } \ { 1 } end{array} right) } \ { text { and describe the behavior of the solution as } t rightarrow infty } end{array}




My answer so far is:
$$vec{x}=c_1e^{(-1+i)t}begin{pmatrix}2+i\1end{pmatrix}+c_2e^{(-1-i)t}begin{pmatrix}2-i\1end{pmatrix}$$



The textbook has the answer:
$$mathbf{x}(t)=e^{-t}begin{pmatrix}cos(t)-3sin(t)\ cos(t)-sin(t)end{pmatrix}$$



How can I simplify my solution?



UPDATE: I used the identity $e^{it} = cos t+isin t$
$$c_1e^{-t}(cos t+isin t)begin{pmatrix}2+i\1end{pmatrix}+c_2e^{-t}(cos t-isin t)begin{pmatrix}2-i\1end{pmatrix}$$






linear-algebra matrices differential-equations systems-of-equations




linear-algebra matrices differential-equations systems-of-equations






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edited Dec 10 '18 at 7:17









Robert Howard

1,9161822




1,9161822










asked Dec 10 '18 at 4:44









Doldrums

829




829








  • 2




    $e^{(-1+i)t}=e^{-t}(cos t+isin t)$
    – Nosrati
    Dec 10 '18 at 4:56






  • 1




    $sin$ and $cos$ are bounded and $e^{-t}to0$ as $ttoinfty$.
    – Nosrati
    Dec 10 '18 at 5:01






  • 1




    At first apply initial condition to renove constants $C_1$ and $C_2$.
    – Nosrati
    Dec 10 '18 at 5:21






  • 1




    Your solution shouldn't has (have) $i$ term!
    – Nosrati
    Dec 10 '18 at 5:29








  • 1




    My method if you interest: $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$ shows $y=e^{-t}cos t$ and $y=e^{-t}sin t$
    – Nosrati
    Dec 10 '18 at 5:44
















  • 2




    $e^{(-1+i)t}=e^{-t}(cos t+isin t)$
    – Nosrati
    Dec 10 '18 at 4:56






  • 1




    $sin$ and $cos$ are bounded and $e^{-t}to0$ as $ttoinfty$.
    – Nosrati
    Dec 10 '18 at 5:01






  • 1




    At first apply initial condition to renove constants $C_1$ and $C_2$.
    – Nosrati
    Dec 10 '18 at 5:21






  • 1




    Your solution shouldn't has (have) $i$ term!
    – Nosrati
    Dec 10 '18 at 5:29








  • 1




    My method if you interest: $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$ shows $y=e^{-t}cos t$ and $y=e^{-t}sin t$
    – Nosrati
    Dec 10 '18 at 5:44










2




2




$e^{(-1+i)t}=e^{-t}(cos t+isin t)$
– Nosrati
Dec 10 '18 at 4:56




$e^{(-1+i)t}=e^{-t}(cos t+isin t)$
– Nosrati
Dec 10 '18 at 4:56




1




1




$sin$ and $cos$ are bounded and $e^{-t}to0$ as $ttoinfty$.
– Nosrati
Dec 10 '18 at 5:01




$sin$ and $cos$ are bounded and $e^{-t}to0$ as $ttoinfty$.
– Nosrati
Dec 10 '18 at 5:01




1




1




At first apply initial condition to renove constants $C_1$ and $C_2$.
– Nosrati
Dec 10 '18 at 5:21




At first apply initial condition to renove constants $C_1$ and $C_2$.
– Nosrati
Dec 10 '18 at 5:21




1




1




Your solution shouldn't has (have) $i$ term!
– Nosrati
Dec 10 '18 at 5:29






Your solution shouldn't has (have) $i$ term!
– Nosrati
Dec 10 '18 at 5:29






1




1




My method if you interest: $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$ shows $y=e^{-t}cos t$ and $y=e^{-t}sin t$
– Nosrati
Dec 10 '18 at 5:44






My method if you interest: $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$ shows $y=e^{-t}cos t$ and $y=e^{-t}sin t$
– Nosrati
Dec 10 '18 at 5:44












2 Answers
2






active

oldest

votes


















1














Let
$$X=e^{(-1+i)t}begin{pmatrix}2+i\1end{pmatrix}$$
$$X_1=Re(X)=e^{-t}begin{pmatrix}2 cos{(t)}-sin{(t)}\
cos{(t)}end{pmatrix}$$
$$X_2=Im(X)= e^{-t}begin{pmatrix}2 sin{(t)}+cos{(t)}\
sin{(t)}end{pmatrix}$$
Then general real solution is
$$X=c_1X_1+c_2X_2$$
From initial condition we get
$$c_1begin{pmatrix}2\
1end{pmatrix}+c_2begin{pmatrix}1\
0end{pmatrix}=begin{pmatrix}1\
1end{pmatrix}$$
$c_1=1$, $c_2=-1$.



Answer:
$$X=X_1-X_2=e^{-t}begin{pmatrix}cos{(t)}-3 sin{(t)}\
cos{(t)}-sin{(t)}end{pmatrix}$$






share|cite|improve this answer





























    2














    $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$. Hence by characteristic equation $lambda^2+2lambda+2=0$, $lambda=-1pm i$. This shows solutions $y=e^{-t}cos t$ and $y=e^{-t}sin t$.



    Now replacing $x=y'+3y$ gives $x=2e^{-t}cos t-e^{-t}sin t$ and $x=e^{-t}cos t+2e^{-t}sin t$.



    Therefore general solution is
    begin{cases}
    x=C_1(2e^{-t}cos t-e^{-t}sin t)+C_2(e^{-t}cos t+2e^{-t}sin t),\
    y=C_1e^{-t}cos t+C_2e^{-t}sin t.
    end{cases}
    with initial condition $C_1=1$ and $C_2=-1$, we have particular solution
    begin{cases}
    x=e^{-t}cos t-3e^{-t}sin t,\
    y=e^{-t}cos t-e^{-t}sin t.
    end{cases}
    the behaviour of solution is $|mathrm{x}|to0$, because
    begin{cases}
    |x|=|e^{-t}cos t-3e^{-t}sin t|leqsqrt{10}e^{-t}to0,\
    |y|=|e^{-t}cos t-e^{-t}sin t|leqsqrt{2}e^{-t}to0.
    end{cases}
    as $ttoinfty$.






    share|cite|improve this answer





















    • ,Your answer to behaviour of solutions as $trightarrow 0$ contains inequalities. Those inequalities contains$sqrt{10}e^{-t}$ and $sqrt{2}e^{-t}$.My question is why did you select those terms?
      – Dhamnekar Winod
      Dec 10 '18 at 15:21













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Let
    $$X=e^{(-1+i)t}begin{pmatrix}2+i\1end{pmatrix}$$
    $$X_1=Re(X)=e^{-t}begin{pmatrix}2 cos{(t)}-sin{(t)}\
    cos{(t)}end{pmatrix}$$
    $$X_2=Im(X)= e^{-t}begin{pmatrix}2 sin{(t)}+cos{(t)}\
    sin{(t)}end{pmatrix}$$
    Then general real solution is
    $$X=c_1X_1+c_2X_2$$
    From initial condition we get
    $$c_1begin{pmatrix}2\
    1end{pmatrix}+c_2begin{pmatrix}1\
    0end{pmatrix}=begin{pmatrix}1\
    1end{pmatrix}$$
    $c_1=1$, $c_2=-1$.



    Answer:
    $$X=X_1-X_2=e^{-t}begin{pmatrix}cos{(t)}-3 sin{(t)}\
    cos{(t)}-sin{(t)}end{pmatrix}$$






    share|cite|improve this answer


























      1














      Let
      $$X=e^{(-1+i)t}begin{pmatrix}2+i\1end{pmatrix}$$
      $$X_1=Re(X)=e^{-t}begin{pmatrix}2 cos{(t)}-sin{(t)}\
      cos{(t)}end{pmatrix}$$
      $$X_2=Im(X)= e^{-t}begin{pmatrix}2 sin{(t)}+cos{(t)}\
      sin{(t)}end{pmatrix}$$
      Then general real solution is
      $$X=c_1X_1+c_2X_2$$
      From initial condition we get
      $$c_1begin{pmatrix}2\
      1end{pmatrix}+c_2begin{pmatrix}1\
      0end{pmatrix}=begin{pmatrix}1\
      1end{pmatrix}$$
      $c_1=1$, $c_2=-1$.



      Answer:
      $$X=X_1-X_2=e^{-t}begin{pmatrix}cos{(t)}-3 sin{(t)}\
      cos{(t)}-sin{(t)}end{pmatrix}$$






      share|cite|improve this answer
























        1












        1








        1






        Let
        $$X=e^{(-1+i)t}begin{pmatrix}2+i\1end{pmatrix}$$
        $$X_1=Re(X)=e^{-t}begin{pmatrix}2 cos{(t)}-sin{(t)}\
        cos{(t)}end{pmatrix}$$
        $$X_2=Im(X)= e^{-t}begin{pmatrix}2 sin{(t)}+cos{(t)}\
        sin{(t)}end{pmatrix}$$
        Then general real solution is
        $$X=c_1X_1+c_2X_2$$
        From initial condition we get
        $$c_1begin{pmatrix}2\
        1end{pmatrix}+c_2begin{pmatrix}1\
        0end{pmatrix}=begin{pmatrix}1\
        1end{pmatrix}$$
        $c_1=1$, $c_2=-1$.



        Answer:
        $$X=X_1-X_2=e^{-t}begin{pmatrix}cos{(t)}-3 sin{(t)}\
        cos{(t)}-sin{(t)}end{pmatrix}$$






        share|cite|improve this answer












        Let
        $$X=e^{(-1+i)t}begin{pmatrix}2+i\1end{pmatrix}$$
        $$X_1=Re(X)=e^{-t}begin{pmatrix}2 cos{(t)}-sin{(t)}\
        cos{(t)}end{pmatrix}$$
        $$X_2=Im(X)= e^{-t}begin{pmatrix}2 sin{(t)}+cos{(t)}\
        sin{(t)}end{pmatrix}$$
        Then general real solution is
        $$X=c_1X_1+c_2X_2$$
        From initial condition we get
        $$c_1begin{pmatrix}2\
        1end{pmatrix}+c_2begin{pmatrix}1\
        0end{pmatrix}=begin{pmatrix}1\
        1end{pmatrix}$$
        $c_1=1$, $c_2=-1$.



        Answer:
        $$X=X_1-X_2=e^{-t}begin{pmatrix}cos{(t)}-3 sin{(t)}\
        cos{(t)}-sin{(t)}end{pmatrix}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 13:41









        Aleksas Domarkas

        8446




        8446























            2














            $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$. Hence by characteristic equation $lambda^2+2lambda+2=0$, $lambda=-1pm i$. This shows solutions $y=e^{-t}cos t$ and $y=e^{-t}sin t$.



            Now replacing $x=y'+3y$ gives $x=2e^{-t}cos t-e^{-t}sin t$ and $x=e^{-t}cos t+2e^{-t}sin t$.



            Therefore general solution is
            begin{cases}
            x=C_1(2e^{-t}cos t-e^{-t}sin t)+C_2(e^{-t}cos t+2e^{-t}sin t),\
            y=C_1e^{-t}cos t+C_2e^{-t}sin t.
            end{cases}
            with initial condition $C_1=1$ and $C_2=-1$, we have particular solution
            begin{cases}
            x=e^{-t}cos t-3e^{-t}sin t,\
            y=e^{-t}cos t-e^{-t}sin t.
            end{cases}
            the behaviour of solution is $|mathrm{x}|to0$, because
            begin{cases}
            |x|=|e^{-t}cos t-3e^{-t}sin t|leqsqrt{10}e^{-t}to0,\
            |y|=|e^{-t}cos t-e^{-t}sin t|leqsqrt{2}e^{-t}to0.
            end{cases}
            as $ttoinfty$.






            share|cite|improve this answer





















            • ,Your answer to behaviour of solutions as $trightarrow 0$ contains inequalities. Those inequalities contains$sqrt{10}e^{-t}$ and $sqrt{2}e^{-t}$.My question is why did you select those terms?
              – Dhamnekar Winod
              Dec 10 '18 at 15:21


















            2














            $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$. Hence by characteristic equation $lambda^2+2lambda+2=0$, $lambda=-1pm i$. This shows solutions $y=e^{-t}cos t$ and $y=e^{-t}sin t$.



            Now replacing $x=y'+3y$ gives $x=2e^{-t}cos t-e^{-t}sin t$ and $x=e^{-t}cos t+2e^{-t}sin t$.



            Therefore general solution is
            begin{cases}
            x=C_1(2e^{-t}cos t-e^{-t}sin t)+C_2(e^{-t}cos t+2e^{-t}sin t),\
            y=C_1e^{-t}cos t+C_2e^{-t}sin t.
            end{cases}
            with initial condition $C_1=1$ and $C_2=-1$, we have particular solution
            begin{cases}
            x=e^{-t}cos t-3e^{-t}sin t,\
            y=e^{-t}cos t-e^{-t}sin t.
            end{cases}
            the behaviour of solution is $|mathrm{x}|to0$, because
            begin{cases}
            |x|=|e^{-t}cos t-3e^{-t}sin t|leqsqrt{10}e^{-t}to0,\
            |y|=|e^{-t}cos t-e^{-t}sin t|leqsqrt{2}e^{-t}to0.
            end{cases}
            as $ttoinfty$.






            share|cite|improve this answer





















            • ,Your answer to behaviour of solutions as $trightarrow 0$ contains inequalities. Those inequalities contains$sqrt{10}e^{-t}$ and $sqrt{2}e^{-t}$.My question is why did you select those terms?
              – Dhamnekar Winod
              Dec 10 '18 at 15:21
















            2












            2








            2






            $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$. Hence by characteristic equation $lambda^2+2lambda+2=0$, $lambda=-1pm i$. This shows solutions $y=e^{-t}cos t$ and $y=e^{-t}sin t$.



            Now replacing $x=y'+3y$ gives $x=2e^{-t}cos t-e^{-t}sin t$ and $x=e^{-t}cos t+2e^{-t}sin t$.



            Therefore general solution is
            begin{cases}
            x=C_1(2e^{-t}cos t-e^{-t}sin t)+C_2(e^{-t}cos t+2e^{-t}sin t),\
            y=C_1e^{-t}cos t+C_2e^{-t}sin t.
            end{cases}
            with initial condition $C_1=1$ and $C_2=-1$, we have particular solution
            begin{cases}
            x=e^{-t}cos t-3e^{-t}sin t,\
            y=e^{-t}cos t-e^{-t}sin t.
            end{cases}
            the behaviour of solution is $|mathrm{x}|to0$, because
            begin{cases}
            |x|=|e^{-t}cos t-3e^{-t}sin t|leqsqrt{10}e^{-t}to0,\
            |y|=|e^{-t}cos t-e^{-t}sin t|leqsqrt{2}e^{-t}to0.
            end{cases}
            as $ttoinfty$.






            share|cite|improve this answer












            $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$. Hence by characteristic equation $lambda^2+2lambda+2=0$, $lambda=-1pm i$. This shows solutions $y=e^{-t}cos t$ and $y=e^{-t}sin t$.



            Now replacing $x=y'+3y$ gives $x=2e^{-t}cos t-e^{-t}sin t$ and $x=e^{-t}cos t+2e^{-t}sin t$.



            Therefore general solution is
            begin{cases}
            x=C_1(2e^{-t}cos t-e^{-t}sin t)+C_2(e^{-t}cos t+2e^{-t}sin t),\
            y=C_1e^{-t}cos t+C_2e^{-t}sin t.
            end{cases}
            with initial condition $C_1=1$ and $C_2=-1$, we have particular solution
            begin{cases}
            x=e^{-t}cos t-3e^{-t}sin t,\
            y=e^{-t}cos t-e^{-t}sin t.
            end{cases}
            the behaviour of solution is $|mathrm{x}|to0$, because
            begin{cases}
            |x|=|e^{-t}cos t-3e^{-t}sin t|leqsqrt{10}e^{-t}to0,\
            |y|=|e^{-t}cos t-e^{-t}sin t|leqsqrt{2}e^{-t}to0.
            end{cases}
            as $ttoinfty$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 6:13









            Nosrati

            26.5k62353




            26.5k62353












            • ,Your answer to behaviour of solutions as $trightarrow 0$ contains inequalities. Those inequalities contains$sqrt{10}e^{-t}$ and $sqrt{2}e^{-t}$.My question is why did you select those terms?
              – Dhamnekar Winod
              Dec 10 '18 at 15:21




















            • ,Your answer to behaviour of solutions as $trightarrow 0$ contains inequalities. Those inequalities contains$sqrt{10}e^{-t}$ and $sqrt{2}e^{-t}$.My question is why did you select those terms?
              – Dhamnekar Winod
              Dec 10 '18 at 15:21


















            ,Your answer to behaviour of solutions as $trightarrow 0$ contains inequalities. Those inequalities contains$sqrt{10}e^{-t}$ and $sqrt{2}e^{-t}$.My question is why did you select those terms?
            – Dhamnekar Winod
            Dec 10 '18 at 15:21






            ,Your answer to behaviour of solutions as $trightarrow 0$ contains inequalities. Those inequalities contains$sqrt{10}e^{-t}$ and $sqrt{2}e^{-t}$.My question is why did you select those terms?
            – Dhamnekar Winod
            Dec 10 '18 at 15:21




















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