Finding specific solutions of a system of differential equations without computations











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I have a reflection matrix $A=begin{pmatrix}-frac{1}{2} & frac{sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix}$ and a linear, autonomous, and homogeneous system $frac{doverrightarrow{w}}{dt} = Aoverrightarrow{w}$, $overrightarrow{w}(t) = begin{pmatrix} x(t) \ y(t) end{pmatrix}$.



I've noticed that $left{begin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2}end{pmatrix},begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2}end{pmatrix}right}$ is an eigenbasis with eigenvalues $lambda = 1$ and $lambda = -1$. Thus, I thought that a general solution would be somewhere along the lines of $begin{pmatrix} x'(t) \ y'(t) end{pmatrix} = c_1e^tbegin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2} end{pmatrix} + c_2e^{-t}begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2} end{pmatrix}$.



I am unsure about how to go from here to finding specific solutions. I'm asked to find two solutions to this system "without performing calculations", and to indicate where the solutions curves start in the plane. Furthermore, I'm asked to find a solution that starts at $begin{pmatrix} 0 \ 5 end{pmatrix}$.



What am I missing here?










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    I have a reflection matrix $A=begin{pmatrix}-frac{1}{2} & frac{sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix}$ and a linear, autonomous, and homogeneous system $frac{doverrightarrow{w}}{dt} = Aoverrightarrow{w}$, $overrightarrow{w}(t) = begin{pmatrix} x(t) \ y(t) end{pmatrix}$.



    I've noticed that $left{begin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2}end{pmatrix},begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2}end{pmatrix}right}$ is an eigenbasis with eigenvalues $lambda = 1$ and $lambda = -1$. Thus, I thought that a general solution would be somewhere along the lines of $begin{pmatrix} x'(t) \ y'(t) end{pmatrix} = c_1e^tbegin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2} end{pmatrix} + c_2e^{-t}begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2} end{pmatrix}$.



    I am unsure about how to go from here to finding specific solutions. I'm asked to find two solutions to this system "without performing calculations", and to indicate where the solutions curves start in the plane. Furthermore, I'm asked to find a solution that starts at $begin{pmatrix} 0 \ 5 end{pmatrix}$.



    What am I missing here?










    share|cite|improve this question


























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      I have a reflection matrix $A=begin{pmatrix}-frac{1}{2} & frac{sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix}$ and a linear, autonomous, and homogeneous system $frac{doverrightarrow{w}}{dt} = Aoverrightarrow{w}$, $overrightarrow{w}(t) = begin{pmatrix} x(t) \ y(t) end{pmatrix}$.



      I've noticed that $left{begin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2}end{pmatrix},begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2}end{pmatrix}right}$ is an eigenbasis with eigenvalues $lambda = 1$ and $lambda = -1$. Thus, I thought that a general solution would be somewhere along the lines of $begin{pmatrix} x'(t) \ y'(t) end{pmatrix} = c_1e^tbegin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2} end{pmatrix} + c_2e^{-t}begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2} end{pmatrix}$.



      I am unsure about how to go from here to finding specific solutions. I'm asked to find two solutions to this system "without performing calculations", and to indicate where the solutions curves start in the plane. Furthermore, I'm asked to find a solution that starts at $begin{pmatrix} 0 \ 5 end{pmatrix}$.



      What am I missing here?










      share|cite|improve this question















      I have a reflection matrix $A=begin{pmatrix}-frac{1}{2} & frac{sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix}$ and a linear, autonomous, and homogeneous system $frac{doverrightarrow{w}}{dt} = Aoverrightarrow{w}$, $overrightarrow{w}(t) = begin{pmatrix} x(t) \ y(t) end{pmatrix}$.



      I've noticed that $left{begin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2}end{pmatrix},begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2}end{pmatrix}right}$ is an eigenbasis with eigenvalues $lambda = 1$ and $lambda = -1$. Thus, I thought that a general solution would be somewhere along the lines of $begin{pmatrix} x'(t) \ y'(t) end{pmatrix} = c_1e^tbegin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2} end{pmatrix} + c_2e^{-t}begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2} end{pmatrix}$.



      I am unsure about how to go from here to finding specific solutions. I'm asked to find two solutions to this system "without performing calculations", and to indicate where the solutions curves start in the plane. Furthermore, I'm asked to find a solution that starts at $begin{pmatrix} 0 \ 5 end{pmatrix}$.



      What am I missing here?







      differential-equations






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      edited 20 hours ago

























      asked 21 hours ago









      Dominic Hicks

      826




      826






















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          $$A^{-1}=A,$$
          $$e^{At}=A.
          begin{pmatrix}{e^t} & 0\
          0 & {e^{-t}}end{pmatrix}
          .A^{-1}$$

          Final solution is
          $$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
          =begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
          frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$






          share|cite|improve this answer

















          • 1




            Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
            – LutzL
            18 hours ago











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          $$A^{-1}=A,$$
          $$e^{At}=A.
          begin{pmatrix}{e^t} & 0\
          0 & {e^{-t}}end{pmatrix}
          .A^{-1}$$

          Final solution is
          $$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
          =begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
          frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$






          share|cite|improve this answer

















          • 1




            Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
            – LutzL
            18 hours ago















          up vote
          1
          down vote













          $$A^{-1}=A,$$
          $$e^{At}=A.
          begin{pmatrix}{e^t} & 0\
          0 & {e^{-t}}end{pmatrix}
          .A^{-1}$$

          Final solution is
          $$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
          =begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
          frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$






          share|cite|improve this answer

















          • 1




            Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
            – LutzL
            18 hours ago













          up vote
          1
          down vote










          up vote
          1
          down vote









          $$A^{-1}=A,$$
          $$e^{At}=A.
          begin{pmatrix}{e^t} & 0\
          0 & {e^{-t}}end{pmatrix}
          .A^{-1}$$

          Final solution is
          $$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
          =begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
          frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$






          share|cite|improve this answer












          $$A^{-1}=A,$$
          $$e^{At}=A.
          begin{pmatrix}{e^t} & 0\
          0 & {e^{-t}}end{pmatrix}
          .A^{-1}$$

          Final solution is
          $$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
          =begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
          frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 19 hours ago









          Aleksas Domarkas

          6655




          6655








          • 1




            Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
            – LutzL
            18 hours ago














          • 1




            Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
            – LutzL
            18 hours ago








          1




          1




          Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
          – LutzL
          18 hours ago




          Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
          – LutzL
          18 hours ago


















           

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